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Balancing an entire sheet of chemistry equations, memorizing the Periodic Table, and answering tough vocabulary questions is much easier when you have a strong foundation in General Chemistry. Whether you are taking your first college chemistry course or still in high school, knowledge of chemical processes provides insights into a variety of biological and physical phenomena. These phenomena and more are explored in Professor Franklin Ow’s General Chemistry course. In each of his lessons, Professor Ow explains concepts in easy to understand language and follows with plenty of examples you will likely see on homework and exams. Lesson topics include Stoichiometry, Gases, Acid Base Reactions, and Electrochemistry. Professor Ow obtained both his BS and PhD in chemistry from the University of California, Los Angeles. He currently teaches chemistry classes at three universities and has won multiple awards in his academic field.

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Hi, welcome to Educator.com.0000

Today's lesson from general chemistry is on the basic concepts of chemistry.0003

Let's go ahead and take a look at the lesson overview.0009

We are going to first start off with a brief introduction and basically define what we mean by chemistry.0012

Then get on to the three traditional states of matter.0018

After we define the states of matter, we will then get on to classifying matter into a couple of categories.0021

What we call pure substances and mixtures.0027

After discussing what we mean by pure substance and mixture,0031

we will then go on to discuss what we mean by physical and chemical properties and changes.0034

We will finish up the lesson with some sample problems that you can expect0040

when taking this class for real during hot chemistry or high school.0046

These are the types of questions that you can definitely expect you will encounter on some type of quiz or exam.0051

Introduction, what is chemistry?--we typically define chemistry as the following.0058

It is the study of matter, its properties, and the changes that matter can undergo.0063

Within this definition of course we have to define each of these terms that I just underlined.0071

Let's go ahead and first define what we mean by matter.0076

Matter is considered as any substance that has mass; that has some type of weight to it.0079

The conventional forms of matter of course are therefore what we call solids, liquids, and gases.0089

What we are going to do in the next slide is to look at each of these three physical states individually.0096

Solids generally have the following characteristics; they are relatively dense.0105

What that means is that a small volume can weigh a lot, is relatively heavy.0110

They are going to have a definite shape and volume.0125

You can think of any type of solid and yes it has a concrete shape to it and a definite size to it.0127

Number three, another characteristic that is typical for solids is that they have very strong attractive forces.0135

We are going to talk about attractive forces later.0143

But when you think of a solid, you think of an object that is tough.0144

So yes it makes sense that the forces that are holding it together are relatively strong.0149

We also like to look at what is called a particulate-level drawing.0154

The particulate-level drawing allows us to describe the sample on a microscale level.0157

We will say that particulate-level drawing is a microscale description.0167

The basic thing for a solid is that solids are going to be closely packed; closely packed atoms.0174

Basically we are going to represent each atom as a sphere.0187

We are going to have blue, red, blue, red, and that can repeat this alternating pattern below like so.0194

You see that these spheres which represent atoms are physically touching each other and are very very closely spaced together.0207

This is our particulate-level drawing for a solid.0216

Liquids on the other hand are going to have definite volume just like a solid but they have no shape.0221

They tend to have weaker attractive forces than solids.0228

They are less tightly held together than their solids.0231

For a particulate-level drawing, we like to think of liquids as still consisting of the same type of spheres.0235

But because of the spacing, they are now going to be farther apart.0244

This farther apart is going to be indicative of the weaker attractive forces.0249

Weaker attractive forces are farther apart.0256

Finally onto the third physical state which is what we call gases.0263

A gas is going to be the following; they are going to have no shape.0266

Really their volume is determined by the container that the gas is placed into.0273

That is because gases expand entirely.0279

If you release a gas in one corner of the room, it will eventually expand and fill the entire room.0284

Gases tend to have the weakest attractive forces and the lowest density of all three physical states.0292

Something that is unique that we did not talk about with solids and liquids0297

is that gas particles are going to be having very high movement velocities.0300

For a particulate-level drawing, we like to show a gas being enclosed in some type of container.0306

The gas particles are represented by red spheres like so; very far apart.0312

We like to draw little comet tails on the end of them.0316

Showing that they are going to have relatively high movement and that gas particles move in random directions.0320

Now onto the classification of matter; matter can be defined as the following.0329

It can either be what is called a pure substance or a mixture.0333

Pure substances can be either elements or compounds.0338

Mixtures can be either homogeneous or heterogeneous.0343

Now that we have introduced the vocabulary terms, let's discuss each one individually0347

in the next couple of slides followed by a couple of examples.0351

Pure substances are composed of only one type of material.0355

An element refers to matter that is not combined with any other different material.0360

Elements cannot be broken down any further.0366

Something like Fe which is iron; something like Ca which is calcium; something like K which is potassium.0370

When we describe just the basic unit of all which is an element, this is one example of a pure substance.0385

What happens when we have nothing but a pure substance?0394

Nothing but one type of compound which is going to be a combination of elements?0399

Something like CO2 or something like H2O or Cl2 or F2.0404

If you have different samples of each of them, say in separate containers, then each container represents a different compound.0413

Again a compound is simply the combination of more than one element.0423

Mixtures therefore are going to be combinations of more than one compound.0428

In homogeneous mixtures, the compounds are uniformly distributed.0432

The typical example is salt water.0439

In salt water we are going to have the following that is uniformly dispersed.0443

We are going to have water and inside the water we are going to have salt; just like so.0449

This is going to be uniformly distributed.0456

It is going to be looking like that visually too; visually uniformly distributed.0459

However in a heterogeneous mixture we have more than one compound and the distribution is nonuniform.0472

Usually you see physical layers.0481

The typical example of course is going to be a mixture of water and oil.0485

We are going to get water on the bottom and we are going to get a layer of oil on the top.0492

Once again this is known as a heterogeneous mixture.0496

Now onto the final part of the definition of chemistry.0501

We define chemistry as the study of matter and the changes that it undergoes.0505

Let's now discuss what we mean by changes.0509

A physical change is anything that retains chemical composition.0512

Specifically phase changes are going to be your typical examples.0516

When a solid goes to a liquid for example; for melting.0526

Melting, for example as an ice cube melts, is just H2O solid becoming H2O liquid.0535

But it starts off H2O; it winds up H2O; all it is is in a different state.0545

Another type of physical change is when you dissolve something.0550

When you dissolve salt in water, you still have H2O and you still have salt.0555

It is just the salt is now broken up and is not a visible solid anymore.0563

Again these are physical changes that retain chemical composition.0570

Chemical changes however alter chemical composition; so any reaction.0573

Basically how do you know if a chemical change has occurred?0582

Usually you are going to see a change in color.0585

You could see a change in temperature; sometimes you see bubbles appearing.0588

Some typical examples of a chemical change will be burning.0599

Anytime you burn something, you notice that the color always changes to black.0604

You release some type of gas that you smell.0609

Basically you see a release of carbon dioxide and water.0612

So burning is an example of a chemical change.0616

Typically when you cook something, cooking involves a lot of changes at the microscopic level in your food.0619

So cooking is another example of a chemical change.0628

Rust formation, when you see rust forming, it is really iron plus oxygen going on to form rust.0633

We start off with iron and oxygen at first and we wind up with rust in the end.0643

These are examples of where the composition of the reactants literally change.0650

We have talked about physical and chemical changes.0658

There is also what we call physical and chemical properties.0660

Physical properties tend to be measurable traits of an object or relates to its description by your senses.0663

Senses, as an example, we see color; we feel or see texture.0671

Density is going to be a physical property because it is something you can measure.0679

Volume is a physical property; it is something you can measure.0684

Mass of course is going to be a physical property because it is something also measurable.0689

Once again physical properties tend to be measurable traits or relate to a description given to us by our senses.0693

Chemical properties describe an object's reactivity and the types of chemical changes that it tends to undergo.0703

For example is something flammable?--for example is something an oxidizer?--for example is something a radical?0709

We always hear that a lot; how they are very very very reactive; especially inside the body.0720

Once again chemical properties describe something as a chemical change.0725

These may be measurable actually; may be measurable also.0732

Now that we have gone over our brief introduction into the study of chemistry, let's now hit up some sample problems.0744

For example, does step one represent a chemical or physical change and explain.0754

Remember that each of these spheres represents a different element.0759

Before step one, the white spheres are separated from the black spheres.0771

However after step one, you notice that we have a white and a black sphere pair now.0776

They are physically touching, showing the formation of a new compound.0782

This is going to be representative of a chemical change.0786

Number two, does step two represent a chemical or a physical change?0789

These pairs of white and black spheres you see that they are still here next to each other.0794

The only difference is that now they are brought closer together.0802

If they are brought closer together, remember that goes into what we said about solids, liquids, and gases.0807

With solids being very close together and gases being very far apart.0813

So step two most likely represents a phase change of sort.0817

For example, it could be a liquid going to a solid.0822

So we would call step two a physical change.0827

Sample problem number two, we are going to classify each of the following situations as an element, a compound, or a mixture.0833

In part A, we have black, white, and black.0841

That is uniformly distributed throughout the entire area.0845

We know that it is not an element because there is more than one type of colored sphere here.0851

We only see this order of black, white, black.0859

This is entirely one type of compound.0863

Part B; part B, we see nothing but one color which represents just one element.0870

Also you see a pair here; this is going to also represent a compound.0877

Finally, part C, we see a black sphere and a pair of white spheres here.0886

You see that it is not uniformly distributed at all.0893

This is going to be representative of a mixture.0896

Finally sample problem number three, classify each of the following properties as chemical or physical.0905

The hardness of diamond is something that is measurable actually.0912

This refers to also using our senses; this is going to be a physical property.0916

Part B, combustibility of gasoline; combustibility refers to how reactive gasoline can be.0928

Especially in the presence of a flame; this is going to be a chemical property.0937

Part C, elasticity of a rubber band; we see how and we feel how elastic a rubber band can be.0945

Elasticity is something that can be measurable; this is going to be physical.0952

The taste of chocolate; obviously taste is going to be using our senses; that is going to be physical.0958

Finally the darkness of coffee is described as color which we use our senses to detect.0966

Darkness of coffee is going to be an example also of a physical change.0971

That is our opening introduction into the basic concepts of chemistry.0977

I want to thank you for your attention.0982

I hope to see you next time on Educator.com.0983

Hi, welcome back to Educator.com.0000

Today's presentation from general chemistry is going to be on periodic trends.0003

We are going to go ahead and as always start off with a brief introduction.0009

Then we are going to get into something we call electron configuration of atoms.0014

After electron configuration of atoms, we are then going to see how we can use the periodic table0018

to help us determine electron configurations--that is what we call the periodic trend.0023

After that, we are going to go ahead and look at additional periodic trends.0028

Followed by summary and finally a few sample problems.0032

We in the previous presentation discussed quantum mechanics and how it gave rise to what was called an atomic orbital.0038

One of the grand outcomes of quantum mechanics was the establishment of periodicity.0048

What periodicity refers to is the following.0055

That patterns of elements and how they are organized throughout the periodic table can be quite predictable0058

just off of organization in their location on the periodic table.0069

The patterns that we are dealing with have to do with both the physical and the chemical properties of atoms.0075

For today this presentation, we are just going to look at a series of these periodic trends.0082

The first one we are going to look at is what is called electron configuration.0089

Electron configuration of an atom is going to answer the following questions about an atom's electrons.0095

Number one, what type of atomic orbitals are involved--s, p, d, or f?0102

Number two, how many electrons are in each of these orbitals?0111

Electron configuration typically has the following format.0117

It is related to quantum numbers actually.0122

It is always the principal quantum number n, followed by the letter of the atomic orbital.0126

That is going to be s, p, d, or f.0139

There is always going to be a superscript.0143

This superscript is going to be also positive whole numbers--1, 2, 3, etc.0147

Basically this is equal to the number of electrons in that orbital; number of electrons in specific orbital.0156

What we are going to do now, we are going to use the periodic table to help us to determine an atom's electron configuration.0175

Basically when you look at the periodic table, at first it looks like one big mess, but it is incredibly organized.0182

There is so much information we can gather from this very important tool.0189

This first two columns, group 1 and group 2, this is what we call the s block; s block.0196

After the s block comes this whole area in green.0209

It is basically what we call the p block.0212

All of the transition metals are in yellow.0216

That is what we call the d block.0220

The lanthanides and actinides are at the bottom.0224

That is what we call the f block.0229

To read this, the n level, remember n comes first.0233

Basically n is basically just the row that the element is in.0237

n is equal to row number.0241

For example, when you look at hydrogen, you see 1s.0245

When you look at lithium, 2s.0249

When you look at sodium, it is 3s for example.0251

Another part, that s is actually helium.0255

You want to count helium as being right adjacent to hydrogen, just for electron configuration purposes.0259

The d block, this is where n changes.0266

Even though the d block doesn't start until the fourth row of the periodic table,0270

the first n value for the d block is going to be always one less than the row.0274

n is equal to one less than actual row.0279

In the fourth row, the first n value is actually 3.0285

Same thing for the f block; same thing for the f block.0292

It is going to be a little different.0295

You notice here that the f block, we don't start until n equal to 4.0296

But the f block actually starts in the sixth row.0302

The n is going to be two less than the actual row.0308

Basically now what you want to do is you want to count blocks you are in.0320

For example, we can go ahead and let's just start off with hydrogen.0326

Hydrogen has just the one electron; hydrogen is going to be 1s1.0332

When we move over to helium right here, that is going to be a total of two electrons.0343

Helium is going to be 1s2.0350

Lithium, lithium has a total of three electrons.0356

But remember we always start from the beginning; we always start from hydrogen.0360

Starting at the beginning, you get hydrogen 1s1; helium was 1s2.0366

Now lithium coming down through the next row is going to be 2s and just the one block.0372

Remember one block is equal to one electron.0378

When you see something like this, 1s1 1s2, you only count the maximum.0382

You only count the last one when they both have the same n value and the same letter.0387

Therefore I only count the last one; it is just 1s2 2s1 for lithium.0393

I know this may seem a little weird at first.0401

But let's go ahead and look at it more in detail now.0405

How do we fill orbitals?--the filling orbitals is...0407

We are going to base it off of what we call the Aufbau principle; the Aufbau principle.0411

Lithium is what we just looked at; lithium was again 1s2 2s1.0417

Let's go ahead and look at oxygen; oxygen is going to be 1s2 2s2.0427

Now oxygen is in the p block; it is four blocks over.0436

That is going to be 2p4; there is something you always want to check.0441

The superscripts remember tell me the number of electrons.0447

The sum of the superscripts should equal to the number of electrons in that atom.0453

That is how you should always check.0464

When we look at lithium, lithium is element number 3.0466

Indeed, 2 plus 1 gives us 3; that is good to go.0469

Remember how we only count the last one, using Aufbau principle.0473

1s1 1s2 was what we initially had; and then 2s1.0479

If I add those up, I get 4; we know that is not right.0487

That is why you only count the last one; that is oxygen.0491

Let's go ahead and do another example as chlorine.0498

You should follow along on the periodic table.0500

Chlorine, this is going to be... again you always start from the beginning.0502

Again only count the last one; 1s2 completes the first row.0506

2s2 is going to be lithium and beryllium.0511

2p6 completes the p block in the second row.0516

We go down to the next line.0520

It is just like reading sentences in a paragraph.0521

You go from top to bottom, left to right; top to bottom, left to right.0525

After 2p6 comes 3s2; finally now we are into the block for chlorine.0530

Chlorine is in the 3p block five blocks over, 3p5.0536

Again double check, do all of the superscripts add up to the atomic number of the element?0542

Indeed it does.0548

We are going to do electron configuration, two formats; the first way we just did.0551

The second way is to do what we call a noble gas notation.0557

This is going to save us time.0561

Basically what you want to look at is the preceding noble gas; look at preceding noble gas of the element.0563

If I look at lithium, the preceding noble gas for lithium is going to be helium.0577

Helium's configuration is 1s2.0584

If I look at oxygen, the preceding noble gas is also helium.0588

Helium is also represented by 1s2.0595

Finally for chlorine, chlorine's noble gas is going to be argon.0598

Argon is going to be represented... excuse me, not argon... neon.0605

Chlorine's preceding noble gas is neon; that is represented by this entire thing, 1s2 2s2 2p6.0613

The shortcut notation is just to put the noble gas in brackets like this.0623

Followed by all the other electrons to the immediate right; helium in brackets, 2s1.0629

Oxygen is going to be helium in brackets, 2s2 2p4.0636

Finally chlorine, chlorine is going to be neon in brackets and then 3s2 3p5.0644

We want to make a note then; what do these electrons mean?0655

Everything in brackets that is represented by the noble gas, think about this.0662

These are the lowest n values.0666

That means these are the innermost electrons, closest to the nucleus.0672

This is what we call the core electrons.0678

Everything else that is to the right of the bracket has the highest n value, which we just underlined right now.0681

We are going to give these electrons a very important name which is going to come into play later on.0690

These are what we call valence electrons.0695

Valence electrons are the electrons farthest from the nucleus.0699

They are the outermost electrons; outermost electrons.0702

A very important finding is that you can always tell the number of0710

valence electrons for main group elements just by looking at the periodic table.0715

For main group elements, the number of valence electrons equals simply to the column number.0721

Let's go ahead and think about this.0736

For lithium, the valence electron is 2s1.0738

There is only one of them; lithium is in column 1.0742

For oxygen, you have valence electrons of 2s22p4.0746

That is a grand total of six valence electrons; oxygen is in column 6.0751

Finally chlorine has valence electrons of 3s2 3p5; that is a grand total of seven.0756

Look where chlorine is--in column 7 of the periodic table.0763

There is another way of representing the electrons in an atom.0770

This is what we call orbital box diagrams.0779

Basically orbital box diagrams are just schematic depiction to show filling of atomic orbitals.0783

Two rules are essentially followed; very important rules.0792

Pauli exclusion principle, the Pauli exclusion principle states that an atomic orbital can hold a maximum of two electrons which must be opposite spin.0796

Once again the Pauli exclusion principles states that an atomic orbital can hold a maximum of two electrons which must be opposite spin.0807

Hund's rule states that a degenerate set of orbitals are to be filled singly with parallel spin.0818

What degenerate means is that they are going to have the same n value, the same principal energy level.0825

After each orbital is singly filled, you can then go ahead and insert a second electron but with opposite spin.0831

That is a lot to take in; let's go ahead and look at chlorine.0840

Chlorine again is 1s2 2s2 2p6 3s2 3p5.0843

What we are going to do is we are going to use a line or a box to represent each of these orbitals.0855

Remember an s orbital from the quantum mechanics presentation, there is only one per energy level.0860

A p orbital, there are three per energy level.0871

A d orbital means five per energy level.0877

Finally f was seven per energy; again this is only when applicable.0882

For example, a d orbital is not going to appear until n equal to 3.0890

F orbitals won't appear until n equal to 4.0898

That is what I mean by when applicable.0900

Again the periodic table is your best friend in this chapter, in this presentation.0902

You have to use the periodic table and you can see this easily.0906

We are going to go ahead and draw the orbital box diagram for chlorine.0910

That is it; one line represents the 1s orbital; 2s, also represented by one line.0915

2p, remember three per n; one, two, three; this is 1s2s; this is 2p.0923

3s is one line; finally 3p, three lines.0932

Again depending on what textbook you use, sometimes the lines will be instead drawn as boxes.0938

They are the same.0946

What we are going to do right now, we are going to fill electrons using Pauli exclusion principle and Hund's rule.0947

1s2, if I pair electrons, two electrons per orbital, they have to be of opposite spin.0954

That is what Hund's rule tells me.0965

If I put one as spin up, the second electron is going to be spin down.0968

We are representing electrons basically as arrows.0971

2s2, same thing; one spin up, one spin down.0975

The 2p orbital, you notice that there are three of them.0981

That is what we call a degenerate set; they have the same n value.0986

Hund's rule tells me I am going to fill these p orbitals singly first before doubling up.0991

I am going to do this with parallel spin.0996

For example, this is one electron of the six.0999

When I go to the next one and to the next one, that is 2p3 so far.1006

Again I have filled the atomic orbitals singly; now I can double up.1011

They are going to be all of opposite spin--2p4 2p5 and finally 2p6 in that order.1016

Yes it does matter how you write it; very important.1025

Go ahead now onto 3s2; spin up, spin down.1028

Now onto 3p5, once again one electron at a time per orbital before doubling up with opposite spin.1033

One, two, three, now four and five.1041

Once again we have applied Pauli exclusion principle and Hund's rule to what are called orbital box diagrams.1047

The Aufbau principle generally works.1059

There are of course a couple of exceptions that we need to account for.1062

Again you should ask your instructor which one you need to know.1066

But the two most common exemptions are in the first d row.1069

That is going to be copper and chromium.1072

Chromium we would expect to be [argon] 4s2 3d6; expected.1075

But actual configuration is going to be [argon] 4s1 3d5.1086

Copper, copper we would expect [argon] 4s2 3d9; expected.1096

But actual is going to be [argon] 4s1 3d10.1106

We are going to save the reasoning for perhaps a higher level course.1117

But for now, what you want to remember is these exceptions arise from experimental data.1126

The simple explanation is that the configuration is more stable due to what is called a half-filled or a completely filled orbital.1133

You see that this s orbital can hold two electrons.1144

It is holding only one right now; that is half-filled.1149

Here you have five of the d orbitals.1152

Each have two electrons, giving you a maximum of ten.1154

That is going to be a completely filled orbital.1157

This is half-filled; this is completely filled; half-filled and completely filled.1159

Something else to point out, this is going to be true not just for copper and chromium, but for transition metals in general.1172

It is to count the valence electrons.1179

Remember for main group elements, we go by the column number; that is it.1182

But for transition metals, they are not main group elements.1186

What rule do we go by for counting valence electrons?1189

Basically it is a simple rule.1192

You count all of the electrons that come after the noble gas for transition metals.1195

Chromium is going to have a total of six valence electrons.1201

Copper is going to have a total of eleven valence electrons.1207

Once again for transition metals, all electrons coming after the noble gas in brackets will be counted as valence.1212

Let's now take a look at the electron configuration for monatomic ions.1224

We are going to examine the configurations in orbital box diagrams.1230

Aluminum, aluminum is 1s2 2s2 2p6 3s2 and 3p1.1235

What does it mean to be aluminum 3+?1248

3+, that means that we remove three electrons.1252

Is it easier to remove electrons closest or farthest from the nucleus?1257

Electrons that are close to the nucleus, they are being drawn toward the nucleus because of that positive charge.1263

Electrons farther out from the nucleus are not held as tightly so they are easier to remove.1271

When you form cations, you remove valence electrons first.1279

In other words, remove the outermost electrons; 1s2, for aluminum 3+, 2s2 2p6.1288

The first electron I am going to remove is the 3p1 because that is farthest out, followed by 3s2.1304

Aluminum 3+'s configuration is 1s2 2s2 2p6.1312

Let's go ahead and repeat this now for chlorine.1317

Chlorine is 1s2 2s2 2p6 3s2 3p5.1321

Chlorine 1-, that 1- means we have added an additional electron.1330

Once again we are going to add it to the outermost orbital because that is most easily accessible.1335

You add electrons also to valence orbitals.1340

Chlorine 1- is 1s2 2s2 2p6 3s2; then 3p5, you add one more, becomes 3p6.1350

The reason why I wanted to bring this up is because there is something in common that these electron configurations of the ions share.1363

They all end in 2p6 which means they have a completely filled p block.1375

What elements are in that six p block column?--it is the noble gases.1381

This is all noble gas configuration; this is what we call noble gas configuration; noble gas configuration.1388

You can go ahead and check with the periodic table that all of them end in p6; all of the noble gases.1401

The valence electrons for a noble gas, two and six, two and six, which means eight valence electrons now.1409

This number 8 is going to come into play in the next presentation.1423

It is going to become very important.1427

But what I want you to take away from this section is the following.1429

That elements react to form ions and covalent bonds in order to achieve eight electrons.1437

Elements react to achieve eight valence electrons; again just take my word for now.1452

This number eight is going to become hugely important later on.1466

One application of knowing the electron configuration for an atom is we are going to be able to predict an atom's magnetic behavior.1475

Paramagnetism arises when the atom has at least one unpaired electron; one unpaired electron.1483

Why that is important is because paramagnetic species are going to be attracted to an external magnetic field.1491

In other words, they will stick to a magnet if you will.1497

Diamagnetism, the prefix di, you could probably tell what this means already.1501

Diamagnetism arises when an atom or ion contains no unpaired electrons.1505

In other words, all electrons are paired; all electrons are paired.1510

Diamagnetic species are going to be repelled by an external magnetic field.1517

An example of a paramagnetic species, we have a lot of them, maybe like 1s2 2s1.1524

Here we have the one unpaired electron; this will be for lithium.1535

Then let's go ahead and look at an example of a diamagnetic species.1540

Diamagnetic species will be just 1s and then 2s.1544

1s2 2s2, that is going to be for beryllium.1549

Zero unpaired electrons, that is what we mean.1556

Electron configuration was the first big periodic trend.1560

We saw how useful the periodic table came into play to help us predict configuration.1563

Let's now take a look at other periodic trends.1568

Atomic radius, atomic radius literally is the size of the atom.1573

It is the distance an atom's valence electrons are from its nucleus.1577

It is also known as atomic radius or atomic size; it turns out the following.1582

That in a column of the periodic table, atomic size increases top to bottom.1588

The reason is because of the following.1593

As you go down a column, you are increasing the n value.1595

As n increases, so does the size of the atomic orbital, remember the higher the energy.1600

In a row of the periodic table, atomic size decreases left to right.1606

Let's go over the reasoning for that.1613

Within the same row of the periodic table, you are pretty much keeping the same n value.1615

n does not change at all; what does change is the following.1620

That is going to be the number of protons inside the nucleus.1625

As the element gains more protons, its nuclear charge is going to increase1629

which essentially pulls the electrons in to a greater degree.1636

This results in contraction; once again atomic size is going to decrease left to right.1642

Ionic radii, when we look at ionic radii, remember that electrons are added to or taken away from the valence shell.1651

The outermost shell that is; that is very important.1659

For anions, because we are adding electrons to the outermost shell, anions typically are larger than the parent anion.1663

For example, O2- will be greater than just oxygen by itself.1672

Cations, we are removing electrons from the outermost shell.1677

Cations are typically going to be smaller than the parent ion.1682

For example, aluminum 3+ is going to be smaller than aluminum 0.1686

How about for what we call isoelectronic species?1694

Isoelectronic species means we have the same number of electrons.1697

For example, N3- versus O2-, you can look at their electron configurations.1700

They are going to have the same electron configurations, the same number of electrons.1709

The rule for isoelectronic species is just to go by nuclear charge.1712

The element basically with the greater nuclear charge, the greater number of protons,1718

is going to have a greater pull on the electrons, decreasing the size.1724

Therefore N3- which has fewer protons than oxygen is going to be greater in size than O2-.1730

Another periodic trend is what we call ionization energy; let's start off with the definition.1743

Ionization energy is the energy required to remove a valence electron in the gas phase.1750

Again you have to put in energy to overcome the attraction the electron has with the nucleus.1756

Removal of an electron is an endothermic process.1765

Typically then ionization energies are going to be positive values.1770

Think about ΔH in units of kilojoules per mole.1775

Ionization energy is completely opposite to atomic size.1782

Therefore smaller atoms have very large ionization energies; just think about that.1785

When we have smaller atoms, we have the electrons much closer, the valence electrons much much closer to the nucleus.1791

That attraction with the nucleus is incredibly strong.1799

You have to put in more energy to overcome that attraction.1802

For our purposes, fluorine is going to be the smallest of our atoms and therefore the largest ionization energy.1807

There is some additional terminology to get across.1825

An atom's first ionization energy refers to removing a valence electron from the ground state.1828

For example, oxygen goes on to form... excuse me.1834

For example, calcium can go on to form Ca1+ plus an electron when energy is applied.1849

We are just removing the first valence electron.1862

This is what is called the first ionization energy.1865

The second, third, etc, ionization energies then refer to subsequent removal of the electrons from a cation.1868

Let's go ahead and now take Ca1+.1876

Ca1+ now has its turn of being ionized to form Ca2+ plus an electron.1878

Once again this is what is called the first ionization energy.1887

This will be calcium's second ionization energy; very good.1891

The next periodic trend is what we call electron affinity.1899

Electron affinity refers to the energy that is released when an electron is added to an atom in the gas phase.1902

Which means addition of electrons is an exothermic process.1910

We will represent that as for example oxygen plus an electron goes on to form O1-.1915

We can also build on that; O1- plus an electron goes on to form O2-, etc.1922

Again this is what we call electron affinity.1929

Again it is going to be typically an exothermic process which means these are1932

going to be reported in less than 0, negative kilojoules per mole.1937

Electron affinity therefore parallels ionization energy.1945

We are going to find that nonmetals are going to be more apt to1951

gain electrons to form anions which will have the largest electron affinities.1955

What we want to do, we want summarize all of the periodic trends looking at the periodic table.1961

As you go left to right upward, this is a direction of increasing ionization energy and electron affinity.1969

However both of these are opposite to atomic size which increases this way.1982

This is increasing atomic size.1990

Really we use our start point and end point, fluorine being at one corner and then perhaps francium being at another corner.1998

Again fluorine being one of the smallest atoms is going to have the largest ionization energies, a very large electron affinity, etc.2010

These are our periodic trends; of course they come from consequences of quantum mechanics.2024

The periodic table we saw is very organized and informative concerning an element's physical and chemical properties.2033

An atom's electron configuration we learned, we do the filling of the orbitals based off of the Aufbau principle.2040

Finally orbital box diagrams are schematic representations of electrons following two important rules--the Pauli exclusion principle and Hund's rule.2049

Let's go ahead and tackle some sample problems.2063

For each of the following, give the ground state configuration and orbital box diagram.2066

Ground state simply refers to the neutral atom.2071

It is called ground state because it refers to the configuration that represents the lowest energy for the atom; of lowest energy.2078

Iron, the preceding noble gas for iron is going to be argon.2090

Then that is going to be 4s2 3d6.2098

Let's go ahead and draw it out.2102

I am just going to do the valence, the valence electrons; that is 4s.2105

Remember for these, how many d orbitals are there per energy level?2109

There is five of them; you have to draw them always as a degenerate set; degenerate set.2112

Let's go ahead and fill them; 4s2 first; now 3d6.2118

Remember one at a time with parallel spin before pairing up.2122

When you pair up, opposite spin.2127

Finally a phosphorus, phosphorus, you can go ahead and look at the preceding noble gas.2132

Preceding noble gas for phosphorus is going to be neon.2140

After neon is going to be 3s2... it is going to be 3s2 3p3.2143

Let's go ahead and draw that out, 3s.2154

Remember there is three degenerate orbitals for p's; let's go ahead and fill the electrons.2157

3s2 first; now 3p3; one, two, and three; let's just go ahead and complete them.2163

Iron is going to have a grand total of eight valence electrons.2174

Phosphorus is going to have a grand total of five valence electrons then.2179

We expect both of these, because they do have unpaired electrons, we expect both iron and phosphorus to be paramagnetic.2186

Let's go on to now sample problem number two.2200

For each of the following, circle the element and ion that has the larger indicated trait.2203

Basically we are going to use the periodic table all the way.2207

Ionization energy, when we look at sodium and lithium, they are in the same column.2210

Lithium is above sodium; sodium is going to have the smaller ionization energy.2215

Lithium should have the larger one; again the reason is because lithium is smaller.2222

It is going to require energy to remove its valence electron than in sodium.2233

Atomic size for oxygen or nitrogen, oxygen and nitrogen are in the same row.2245

Atomic size decreases left to right; oxygen is going to have the smaller size.2251

Nitrogen is going to have the larger size; remember why?2256

Because nitrogen is going to have fewer protons in its nucleus; so a smaller nuclear charge.2260

Because it is a smaller nuclear charge, electrons are going to be farther out from the nucleus.2270

Last one is atomic size, this time for O2- and N3-.2281

Again anytime you see ions, try to think isoelectronic species.2286

Again for isoelectronic species, you want to just go by the atomic number again; the number of protons.2292

Once again oxygen is going to have more protons.2300

Nitrogen is going to have fewer protons.2306

For the same reasoning as the previous question, N3- should be larger because of the fewer protons.2310

The attraction is less with the electrons; and they are farther out.2319

It was good to see everyone on Educator.com again; thank you for your attention.2325

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is going to be on bonding and molecular structure.0002

As always, we will start off with a brief introduction.0010

After that, we are going to then go into the different types of chemical bonds.0014

After types of different chemical bonds, we are then going to discuss yet another periodic trend0019

which is called electronegativity and how it relates to something we call bond polarity.0024

The next two bullets deal with what we call Lewis electron dot structures.0030

These are going to become quite prominent in our future discussions in general chemistry.0035

After discussing Lewis structures, we will go into something we call resonance and formal charges.0042

After that, we then jump into a type of bonding theory called VSEPR theory or valence shell electron pair repulsion.0049

Followed by molecular polarity and bond properties.0058

As usual, we will finish it off with a brief summary followed by some sample problems.0064

In the previous lecture on molecules and atomic structure and periodic trends, we introduced the concept of valence electrons.0072

Remember valence electrons are an atom's or ion's outermost electrons.0082

That is they are the farthest away from the nucleus and the least tightly held.0088

It turns out that we have highlighted them for a reason.0094

It is because they are the primary electrons involved in chemical reactions and in chemical bonding.0097

In this lecture, we are going on to focus our attention on valence electrons and their qualitative representations.0107

Before going into that, we need to first discuss what types of bonds occur.0116

Really for our purposes for general chemistry, there are two types of bonds.0124

The first type is an ionic bond.0127

This exists between a metal and a nonmetal or between a cation and an anion.0131

We actually touched upon this when we first learned about polyatomic ions and naming ionic compounds.0138

Remember that in an ionic bond, it is really an electrostatic attraction occurring,0145

resulting in complete transfer of electrons always from the metal to the nonmetal.0150

That is not the type of bond we are going to be focusing on in this lecture.0157

We are going to be focusing on what is called a molecular bond.0161

A molecular bond exists between nonmetals.0166

Remember we learned how to name molecular compounds earlier on.0168

In a molecular bond, electrons are not being transferred, instead they are shared.0174

This sharing of electrons occurs mutually or not.0180

Because of this sharing, a molecular bond is also referred to as a covalent bond0184

which is how I am going to start to refer to it from now on, a covalent bond.0189

Next we are going to examine this sharing of electrons and0194

see the extent of how mutual or not the covalent bond is.0197

In order to address the question whether or not the sharing of electrons in0208

a covalent bond is mutual or not, we need to now introduce another periodic trend.0213

This periodic trend is what we call electronegativity.0218

I am going to abbreviate it as EN for short.0222

Electronegativity, it is literally a quantity, it is a value that describes the ability of0227

an atom to attract electrons to itself from a chemical bond in a gas phase.0232

Again it is the ability of an atom to attract electrons to itself in a chemical bond.0240

This is usually determined in the gas phase.0246

Electronegativity is more or less like your GPA which is on a scale of 0 to 4 with 4 being the highest.0250

High electronegativity values tend to be those elements that have very strong nuclear charges.0260

They attract electrons quite readily to themselves.0266

Those atoms that have very strong nuclear charges also tend to be very small.0272

What we are going to do right now, we are going to reexamine our periodic table.0277

Let's go ahead and repeat the periodic trends that we have already been through.0283

Going this way remember is atomic size increasing.0288

Going this way it was increasing ionization energy.0295

If elements that have high electronegativity values have strong nuclear charges and tend to be small,0305

that means the highest electronegativity values are going to be right here.0311

In other words, electronegativity is going to parallel ionization energy.0315

Again electronegativity is going to be parallel to ionization energy and opposite to atomic size.0323

There is a little acronym we can remember to help us with this.0330

F-o-n-c-l-b-r-i-s-c-h, fonclbrisch is the pronunciation of this little silly acronym.0334

But it is a nice crude way of remembering the order of electronegativity0345

in case you don't readily have a periodic table in front of you.0350

Fluorine is going to be the highest electronegative atom.0354

Carbon and hydrogen are going to be the lowest; fonclbrisch.0358

We are going to classify a bond with specific words depending on the0366

difference in the electronegativities of the two atoms that share the bond.0373

If the difference in electronegativity is great, then the sharing of electrons we say is not mutual.0379

That means the atom with the higher electronegativity gets a greater share of the electron density within the covalent bond.0387

Such a bond is what we call a polar covalent bond.0395

We schematically indicate polarity using partial charges and a dipole arrow.0400

Let's go ahead and take a look at an example.0405

We look at the hydrochloric acid and the bond between hydrogen and chlorine.0409

If I look at fonclbrisch, chlorine is going to be more electronegative than hydrogen.0414

That means chlorine is pulling all of the electron density to it.0425

In other words, all of the electron density accumulates around chlorine.0432

It becomes partially negative due to all that electron density.0437

To represent partial, we use the Greek symbol, δ.0442

If chlorine is δ-, that means hydrogen is going to be δ+0447

because all of its electron density is essentially being removed away from it and toward the chlorine.0451

This produces what we call a dipole moment in the molecule.0460

A dipole moment is going to be represented using a dipole arrow.0469

The dipole arrow is always going to point toward the more electronegative atom.0473

Again a dipole arrow points toward electronegative atom; toward the more electronegative atom.0482

Basically what a dipole arrow is, a dipole arrow or dipole moment, it basically shows where the electron density has accumulated.0491

It points toward the center of higher electron density.0503

Shows area / points toward area of accumulated electron density.0510

Let's go ahead and look at the opposite situation.0531

If the difference in electronegativity is negligible, then we have a pretty much equal sharing of electron density.0535

It is more or less mutual.0543

Such a bond is called a nonpolar covalent bond; nonpolar covalent bond.0545

For example, if you just look at typical hydrogen H2, we have a difference in electronegativity between the two atoms to be zero.0551

They are identical; this is a nonpolar covalent bond.0559

You may be wondering what exactly is considered negligible and what is considered great.0563

I am going to leave that up to your instructor and up to your textbook that you are using because0570

everybody can have a slightly different gauge of the difference in electronegativity and0575

what they consider to be negligible and what they consider to be rather significant.0581

Now that we know the different types of bonds and what we are exactly dealing with,0591

we want to now introduce a schematic way to represent valence electrons.0596

We first are going to talk about atoms.0601

Basically we represent valence electrons schematically as dots.0603

If you recall, you can just go by the periodic table to determine how many valence electrons exist for main group elements.0611

For main group elements, the number of valence electrons equals to the column number that the element is in.0621

Let's go ahead and look at sodium.0642

Sodium is in column one of the periodic table; it has one valence electron.0647

All I am going to do, I am going to represent that as one dot.0651

That is it; that is the electron dot structure for sodium.0654

Let's go ahead and look at magnesium.0659

Magnesium is in column two of the periodic table; it has two valence electrons.0661

I am going to put one of those dots there.0666

I am just going to put another dot maybe on another side of the element if you will.0668

Boron is an element in column number three.0675

It is going to have three valence electrons so I am going to show that to have three dots.0681

Carbon, column number four, it is going to have one dot here, here, here, and then there.0685

On to nitrogen, nitrogen is column five; it has five valence electrons.0694

One dot, two dot, three dot, fourth dot; where do I go put the fifth dot?0701

Really there is not too much of a physical significance here, but I am going to start pairing up.0705

Remember when we talked about atomic orbitals and Hund's rule.0711

Remember we singly fill orbitals before pairing up; I am just going to follow that.0715

Let's go ahead and apply that to oxygen.0721

Oxygen has six valence electrons because it is in column six.0722

One, two, three, four, five, and six.0726

On to fluorine now is in column seven.0732

One, two, three, four, five, six, and seven.0738

Finally we can do maybe krypton.0743

Krypton is going to be a noble gas in column eight.0748

One, two, three, four, five, six, seven, and eight.0751

These are the representative elements from their respective column numbers and how we represent them in terms of Lewis dot structures.0757

That was pretty simple, representing valence electrons of atoms, but now let's go bigger.0774

Let's go ahead and look at electron dot structures for polyatomic species.0780

Basically single bonds are going to be shown as a single line each containing two valence electrons.0785

Something like this; this is what we call a single bond.0792

There is going to be exactly two valence electrons in there.0799

I am going to represent valence electrons as VE for short now.0804

Single bonds are not the only type of covalent bonding that we can have.0809

We can also have double bonds.0812

For example, in oxygen, you can have a double bond like that.0815

In a double bond, there is going to be exactly four of the valence electrons.0822

You can also have triple bonds; for example, nitrogen N2 contains a triple bond.0828

In the triple bond, there is just going to be six valence electrons.0835

Not all electrons are participating in the bonding.0841

There are what we call nonbonding electrons.0846

Basically for general chemistry purposes, we are going to represent nonbonding electrons as lone pairs.0848

For example, when I go back to the Lewis structure for oxygen each oxygen contains two lone pairs like that.0855

That is what we call nonbonding electrons.0864

Also for nitrogen just to finish it up each nitrogen also contains one lone pair each.0868

How exactly am I coming up with these Lewis structures?0877

There happens to be a very systematic procedure for us to follow.0881

Let's go ahead and take a look at them right now.0885

Step number one, we are going to determine the central atom.0889

There is a couple of guidelines we can use to help us do this.0892

The first point is the following.0896

It is often the element that appears once in the formula.0897

It is never hydrogen; it is often carbon.0901

If there is more than one element that appears by itself, then you choose the least electronegative element.0907

That is number one is to determine the central atom.0915

Number two, we are going to arrange all other atoms around the center and0917

connect it to the central atom using only single bonds first; using only single bonds first.0921

Let's go ahead and look at water; water is H2O.0928

Here oxygen is the only element appearing by itself.0933

I am going to put it right to the center; step one is done.0938

Now I move on to step two; I surround the center with every other atom.0942

I connect it to the center using a single bond only.0946

Step two is done; let's go ahead and move on.0950

Step three, we are going to complete what is called the octet.0953

Remember we looked at this earlier on that why do atoms gain or lose electrons?0957

It is because they want to achieve eight electrons total, a noble gas configuration.0965

This magic number of eight is what we call an octet.0971

Every atom needs to have an octet in a Lewis structure in order for it to be quote and quote acceptable.0975

There are a couple of exception though--elements that don't need an octet.0982

Hydrogen, hydrogen is completely happy with just two electrons.0987

That is what we call the duet rule for that purpose.0992

Beryllium, beryllium is completely happy with just the four electrons.1004

Boron and aluminum, these elements are fine with just six electrons.1005

Phosphorus and sulfur, phosphorus and sulfur and really elements in period three or more, these actually can have more than an octet.1006

This is what we call an expanded octet.1016

Don't worry, we are going to look at a lot of examples where this occurs.1021

We are going to complete the octet when we read this using lone pairs of electrons.1026

When I go ahead and look at H2O or water, there is two electrons here.1031

There is two electrons here which means there is four electrons.1039

Now I want to give oxygen eight electrons.1042

That means I need four more electrons to get to eight which is two lone pair.1046

I am just going to do it like that.1050

What we need to do now in step four is to perform a valence electron count.1053

Mother nature has given us these elements.1058

We have to limit ourselves to the number of valence electrons provided by each element.1061

We can only use the total number of valence electrons, not any less, not any more.1068

We have to also account for the charge if it is a polyatomic ion.1074

For each positive charge, we are going to delete one valence electron.1079

For each negative charge, we are going to add one valence electron.1083

Don't worry, we are going to do a lot more examples.1087

But let's go ahead and check out water.1089

For hydrogen in water, there is two hydrogens.1092

Each hydrogen is in column one of the periodic table which means I am going to get two electrons just from hydrogen.1096

There is one oxygen.1103

Oxygen is in column six giving me six electrons just from oxygen.1105

The grand total of valence electrons in water is therefore eight.1111

Let's see if we have used eight in this molecule here.1117

I have two from the single bond here, two here, giving me a grand total of four, six, and eight.1123

Yes, this is a valid Lewis structure for water because everybody's octet rule is good to go.1131

Hydrogen's duet rule is satisfied.1139

I have used exactly the allotted indicated number of valence electrons.1142

If too many electrons are used, we need to then resort to multiple bonding.1153

We are going to try double bonds first, followed by triple bonds.1158

But here is the warning.1162

For each multiple bond added, we need to delete a lone pair from each atom sharing the bond.1165

Let's go ahead and look at an example; that is carbonate.1171

Carbonate is CO32-; this is a nice polyatomic ion we can look at.1176

Step one is to determine the central atom.1182

That is going to be the element I am pairing by itself.1184

In this case, it is only carbon.1187

Step two, we are going to surround the central atom with all other atoms.1190

We are going to connect it to the center using a single bond just like that.1195

Step three, we are going to complete everybody's octet using a lone pair of electrons.1201

Oxygen here needs six more electrons because it already has two nearby.1209

Same thing for this oxygen; same thing for this oxygen.1215

The central carbon has six nearby electrons already.1221

It only need two more electrons or one lone pair.1224

Let's go ahead and do a valence electron count now.1231

Don't forget this is a polyatomic ion.1234

I am going to put that in brackets and show the charge.1236

I have three oxygens each contributing six electrons giving me a grand total of eighteen valence electrons just from oxygen alone.1241

I have one carbon contributing four electron.1253

That is going to be four valence electrons just from the carbon.1256

Remember we must account for the charge; a 2- charge means we add two electrons.1259

Remember for negative charges, we add one electron per negative charge.1267

For cations, we take away one electron per positive charge.1273

My grand total of valence electrons is going to be twenty-four valence electrons that I must use in this compound.1278

When we go ahead and look at carbonate now over here, do we have exactly twenty-four?1287

The answer is no; it turns out that we have twenty-six electrons total.1292

We are over our limit.1298

What we are going to do right now is we are going to resort to multiple bonding.1300

We are going to try double bonds first followed by triple bonds.1303

What I am going to do, let's do this in red.1307

I am going to try one double bond first just like that; what else?1310

We have to delete a lone pair from each atom that shares the bond.1316

Let's go ahead and do that right now.1322

There is one lone pair gone from oxygen.1325

The one lone pair gone from the carbon.1327

Let's go ahead and recount.1332

Yes, this Lewis structure, the way we have it here, has exactly twenty-four valence electrons.1334

Every oxygen has an octet; so does the carbon.1340

This is a valid Lewis structure for carbonate.1344

If you follow these steps every time, you will be fine.1349

But let's go ahead and look at one more example.1355

Let's go ahead and look at nitrogen, N2.1359

Nitrogen, let's go ahead and connect it with a single bond just like that.1363

Now I am going to complete everybody's octet using lone pair of electrons.1368

Each nitrogen here needs six more electrons to get eight.1373

Remember they share those two electrons in the bond.1377

When we go ahead and do the count, we have two nitrogen each contributing1380

five valence electrons which means ten valence electrons are allotted to us.1384

When I look at this molecule and when I count the number of valence electrons, we are too many.1390

We have used twelve; we are going to resort to multiple bonding right now.1395

Sorry, we used fourteen; we used fourteen electrons; we only can use ten.1402

I am going to try a double bond.1408

When I do that, I have to delete a lone pair from every atom involved.1410

Let's go ahead and recount; when we recount, this is twelve electrons.1415

I am still way over my limit; double bonds have failed us.1420

Now we resort to triple bond.1424

I have to remove another lone pair of electrons from each atom sharing the bond.1428

Here we go; each nitrogen has an octet.1433

I have used exactly ten valence electrons.1437

Again follow this basic five-step procedure; you will get it right every time.1441

You notice that for carbonate we put the carbon-oxygen, the double bond on the twelve o'clock position if you will.1451

I could have put the double bond along any other one with oxygen.1458

That is what we call resonance structures.1466

Resonance structures are basically different Lewis structure of a same molecule.1468

For carbonate, this is what we drew originally.1472

Again I could have easily put it down here.1480

I am just putting my lone pairs in right now.1489

Or I could have easily put it right here.1492

These are what are called resonance structures.1504

These are Lewis structures of the same molecule.1506

In chemistry, we like to indicate the presence of resonance structures with a double ended arrow like that.1509

We like to keep track of electrons around an atom relative to its nuclear charge.1521

This is what we call the formal charge.1526

The formal charge is equal to the following.1530

It is equal to the number of valence electrons minus the number of lines and dots attached.1532

Let's go ahead and look at this structure for the carbonate ion.1538

Let's go ahead and look at this oxygen here.1547

This oxygen right there, let's go ahead and use this equation.1552

Formal charge here is equal to number of valence electrons.1556

Six minus the lines and dots attached, the number of lines and dots attached.1560

I have six dots and one line attached giving me a grand total of seven.1565

That formal charge is -1.1571

We usually like to indicate the formal charge just as a tiny superscript around the atom.1573

I personally like to circle them like that.1579

This oxygen here is identical; it is three lone pairs and one bond.1583

This is also -1.1588

Let's look at the carbon.1593

The carbon is going to be four valence electrons minus...1594

I have how many lines attached total?--four.1601

This carbon has a 0 formal charge.1606

This oxygen here, that is going to be six minus the number of lines and dots attached.1609

I have two lines attached and four dots giving me a grand total of six.1615

That one is also a 0 formal charge.1620

Notice what do the sum of my formal charges equal to?1626

The sum of my formal charges is equal to -2 which is the overall charge of the ion.1630

The nice thing about formal charges is that when I add up all formal charges,1637

it must equal to the overall charge of the molecule or ion.1641

This is yet another way to determine if the Lewis structure that you have drawn is valid or not.1650

Remember it must equal to the overall charge of the molecule or ion, the sum of the formal charges that is.1658

We have already assigned the formal charge to each element in carbonate.1670

Again the sum of the formal charges equals to the overall charge of the molecule or ion.1677

It turns out that not all resonance structures for a molecule or ion are made equal.1684

In terms of carbonate, all three were identical.1691

We always had one carbon-oxygen double bond and two carbon-oxygen single bonds.1694

All three were identical.1699

But now let's go ahead and look at sulfuric acid.1701

Sulfuric acid is an example of a molecule that can have resonance.1705

But we are going to find that the resonance structures are not identical.1709

For example, when you look at sulfuric acid, one way to draw its Lewis structure is right here with all single bonds.1713

I am just going to go ahead and put in the lone pairs to complete everybody's octet just like that.1721

Yet another way to draw sulfuric acid is with two double bonds like this.1729

Remember that sulfur is one of those atoms that can have more than eight electrons.1739

In this case, it has ten electrons.1744

Everybody else has an octet and hydrogen's duet rule is good.1747

Is it the resonance structure on the left or the resonance structure on the right?1752

Which one best portrays sulfuric acid?1756

We have to go through what makes up a good resonance structure.1761

It is all about formal charges.1765

Good resonance structures are going to maximize the number of covalent bonds.1767

They are going to minimize separation of charge.1773

That means formal charges of zero are ideal.1776

They are going to obey the rule, the octet rule as much as possible obviously.1780

Finally if there is a negative formal charge, we are going to try as1786

best as we can to place it on the most electronegative atoms.1789

If we go ahead and look at it, here we have -1 formal charge, -1 formal charge.1798

Sulfur is actually having a +2 formal charge right here.1804

That is equal to six minus four.1809

But when we look at the Lewis structure on the right, this is 0.1813

This is 0, 0, 0 and now sulfur actually has a formal charge of 0.1818

You cannot beat that; you cannot get better than all 0 formal charges.1826

The Lewis structure on the right, the resonance structure on the right, is what we call the major contributor.1831

The resonance structure on the left is what we call the minor contributor.1842

We have seen how we can use formal charge to determine what is a quote and quote good Lewis structure.1849

Now we are going to use Lewis structures to help us determine how a molecule physically looks like.1856

That is what is its shape?--what is its geometry?1863

The theory that helps us to determine this, that is going to give us some guidelines is what we call VSEPR theory.1870

VSEPR theory stands for valence shell electron pair repulsion.1878

Basically the premise behind this theory is the following.1885

That a molecule is going to adopt a configuration, basically a shape or geometry, which minimizes electron-electron repulsion.1888

Remember electrons are the same charges; they don't want to be close together.1898

In other words, the conformation is going to maximize the distance between thereby electrons.1903

There is different types of electron-electron interactions.1910

Lone pair-lone pair interactions are going to be the worst because it turns out that lone pairs occupy the greatest volume.1913

We want to avoid lone pairs being too close together.1927

After lone pair-lone pair interactions is going to be lone pair-bonding pair interactions and finally bonding pair-bonding pair interactions.1933

Right now in the next slide, we are going to introduce molecular geometries, their names, and their respective bond angles.1946

What I mean by bond angle is if you have a central atom A surrounded by two atoms like this,1955

basically the bond angle is going to be the angle of that.1962

Bond angle just like in a typical geometry class.1967

Let's go ahead; now this looks like a lot of material.1973

This looks very sometimes intimidating; but again it is not too bad.1978

Let's go ahead and just take each molecule one by one.1984

Number one, the steric number.1987

The steric number is just a fancy name for the number of electron groups around the central atom.1990

What I mean by an electron group is basically any type of bond.2002

A single bond counts as one group; a double bond counts as one group.2011

A triple bond counts as one group; or a lone pair.2015

Basically one lone pair counts as one group.2024

The basic geometry, the basic geometry is also known as the electron geometry.2031

Basically the basic geometry or electron geometry is everything in this first column.2043

What they have in common is that the central atom E contains zero lone pairs.2051

When we have a situation where there is going to be only two electron groups around a central atom with no lone pairs,2059

that results in a linear geometry or a bond angle of 180 degrees.2068

When we have three bonding groups and no lone pairs around a central atom,2073

we get what is called a trigonal planar geometry and a bond angle of 120.2077

When we come up to four electron groups around a center and no lone pairs, this is what we call tetrahedral.2083

You are going to see that we have what is called a broken wedge and a solid wedge notation.2092

That implies three-dimensional nature of the molecule.2098

Broken wedge means it is pointing away from you into the paper.2101

Solid wedge means it is pointing toward out of the paper.2107

If you look at this molecular here and imagine its three-dimensional nature, we do get a pyramid.2112

A tetrahedron basically is what it is called; let's go on.2118

Five bonding groups around the center atom and no lone pairs, that is what we call trigonal bipyramid.2124

We have two bond angles going on, 120 and 90 degrees.2130

Finally six bonding groups around the central atom, no lone pairs,2135

we get an octahedron or an octahedral geometry with a 90 degree bond angle.2138

What happens as soon as we place lone pairs on a central atom?2145

As soon as we place lone pairs on the central atom, we get everything here.2150

We are going to use a new name to describe their geometries.2157

That is what we call the molecular geometry or the molecular shape.2161

Please make note, there is a difference between electron geometry and molecular geometry.2166

Sometimes they are identical.2171

When they have zero lone pairs like we just covered, electron geometry is the same as molecular geometry.2174

For steric number three, we saw that the electron geometry was trigonal planar.2182

But as soon as we put a lone pair from one of the Xs,2189

that lone pair is a great area of electron density.2195

It is going to push the Xs away from it.2199

We don't have trigonal planar anymore.2203

Instead the geometry name is called bent or angular.2204

We have a bond angle of less than 120.2208

Steric number four, with one lone pair, we get what is called a trigonal pyramid with a bond angle less than 109.2212

If we have two lone pairs, we have a bond angle of much less than 109.2222

Steric number five, one lone pair is what we call a seesaw geometry.2228

If we have two lone pairs, it is called a t-shape geometry.2236

If we have three lone pairs, it is called a linear geometry with a bond angle of 180.2241

Finally steric number six, one lone pair is called square pyramid.2246

Two lone pairs is going to be called square planar.2252

Three lone pairs is t-shape; four lone pairs is going to be linear.2255

You should definitely consult with your instructor and determine which ones he or she exactly would like you to know by heart.2261

Now that we have gone over the names of the geometries and how molecules can look like,2274

we are going to use a molecule shape to determine if it itself is polar.2280

We have already talked about bond polarity, but what about an entire molecule?2286

A step by step procedure is the following.2290

Step one, we are going to draw the Lewis structure.2292

Step two, from the Lewis structure, we are going to determine the molecular shape.2296

Then we are going to assign all bond dipoles.2300

Remember that is the arrow that points toward the more electronegative atom.2302

Then we are going to do the following analysis.2307

If the dipoles cancelled each other, that means there is no net direction of electron density and the molecule is nonpolar.2310

If the dipoles seem to point in a general direction, that is they don't cancel,2319

then we say the molecule is expected to be polar.2325

Let's go ahead and look at a few general examples.2327

For example, if we have EXX and we have one dipole going this way and one dipole going this way,2331

this is a 180 degree bond angle, those dipoles completely cancel each other.2340

This would be an example of a nonpolar scenario.2346

We can do a couple of more examples.2351

EXXX and we can have one dipole, for example, let's go here, one going here, one going here.2353

In this case, all three dipoles, they point in completely opposite directions at a 120 degree bond angle.2366

This would be also an example of a nonpolar molecule.2373

Let's go ahead and look at another one.2378

Now EXX and X; this would be trigonal pyramid geometry--one dipole here, here, and here.2381

In this case, because this is a three-dimensional pyramid, these dipoles don't cancel.2392

They all point away from the central atom.2397

This would be an example of a polar molecule.2405

Finally one last example, EXX; let's have two lone pairs here.2409

In this case, we can have one dipole going that way; one dipole that way.2418

Remember this is a bent geometry; my net dipole is going downward.2422

This would be also an example of a polar molecule.2427

These types of problems take a little getting used to.2431

But with more and more practice, I am sure... 2433

I know you will get better and better and more confident in it.2435

The final item that we want to look at for covalent bonding is bond properties.2441

There is really three properties that we want to look at now.2447

What is called bond order, bond length, and bond energy; these are the three parameters.2450

The first one is called bond order; basically bond order is a fancy name.2457

A bond order of one is going to be indicative of a single bond.2462

Bond order of two is going to be indicative of a double bond.2466

Bond order of three is going to be indicative of a triple bond.2469

We are going to come across bond order again in a future chapter.2472

I am just going to leave it at that for now.2479

Bond length is literally the length of the bond.2483

It is the distance between two nuclei sharing a bond.2487

The shorter bond, the greater the interaction between bond nuclei.2492

We have greater ability for electrons to be shared.2497

What that means is that we get a stronger bond.2501

The shorter the bond, the stronger it is; shorter means stronger.2504

That is easy to remember because both of the words here start with an s.2511

Bond energy, bond energy is the third parameter.2516

Bond energy is also known as bond enthalpy; that is right.2519

It is going to be yet another type of ΔH.2524

Remember ΔH from thermodynamics; the formal definition is the following.2526

Bond enthalpy refers to the amount of energy required to form a bond between two nuclei in the gas phase.2532

That is it is the amount of energy required; what does that mean?2541

If it is required that means it is going to be an endothermic process which means2545

bond enthalpies are typically going to be reported as positive values and in units of kilojoules per mole.2553

If for example a carbon-carbon single bond has a bond enthalpy of 347 kilojoules per mole.2562

As we go to a carbon-carbon double bond which is stronger, we expect more energy.2567

A stronger bond is going to require more energy to break.2572

A carbon-carbon double bond is 630.2576

Carbon-carbon triple bond is going to be the strongest and have the highest bond enthalpy of around 800 kilojoules per mole.2578

Notice that they are all positive values.2586

Let's summarize what we went over today.2591

First point is that Lewis structures are basic visual depictions of covalent bonding in molecules and ions.2593

A formal charge is a nice tool to help us determine a valid Lewis structure.2600

Using VSEPR theory, we were able to predict the electron and molecular geometries for any given molecule.2606

That is how is it going to look three-dimensionally?2616

Let's now go over a sample problem.2623

This sample problem is basically the following; here we are given three molecules.2627

For each molecule or ion, we want to determine a) the best Lewis structure,2631

b) we want to determine the electron and molecular geometries and determine if it is polar or nonpolar overall.2637

Let's go ahead and do this one.2644

Xenon trioxide, my central atom is going to be xenon.2646

I am going to surround the xenon with the oxygens like that.2652

I am going to connect everything to the center like that.2655

I am then going to proceed and complete everybody's octet.2659

Each oxygen is going to need three lone pairs.2664

The xenon is going to need one lone pair itself.2669

Let's go ahead and see if we have used the exact number of valence electrons allotted to us.2672

Three oxygens contribute six valence electrons; that is eighteen from oxygen.2681

The one xenon contributes eight valence electrons.2691

I have a grand total of twenty-six valence electrons that I can use.2696

Let's go ahead and see if we have used twenty-six.2700

It is two, four, six... two, four, six, eight, ten, twelve, fourteen, sixteen, eighteen, twenty, twenty-two, twenty-four, and twenty-six.2704

This looks alright; everybody's octet is complete.2717

I have used exactly twenty-six valence electrons.2720

Let's look at the formal charges though.2724

The formal charge of the xenon is going to be eight minus five which is +3.2726

Each oxygen is -1.2735

Remember formal charges of 0 are best; +3 is pretty far from 0.2739

There is another Lewis structure that we can draw; it is basically a resonance structure.2745

Remember that xenon is an element in period three or higher.2750

It can actually come up with more than an octet.2756

It turns out that we can come up with a Lewis structure for xenon trioxide like that.2764

When I do this, look what happens to all my formal charges.2771

Oxygens are now 0; xenon has a formal charge of now 0.2775

That is going to be better than our left Lewis structure.2780

Our best answer here is going to be that one.2784

Polarity, oxygen is more electronegative than xenon.2789

I have one dipole that way, one that way, and I have one that way.2793

This is a trigonal pyramid geometry with a bond angle less than 109.5.2798

When I think about this three-dimensionally, these three arrows that we have just drawn do not cancel.2810

This molecule is expected to be polar; the dipole arrows do not cancel.2815

I31-, this is going to be triiodide.2825

In this case, it is going to be iodine, iodine, iodine.2829

Let's go ahead and connect everything to the center of a single bond.2833

Let's go ahead and assign lone pairs to complete everybody's octet now.2836

This iodine on the left needs three; this iodine on the right needs three.2841

The iodine in the middle is going to need two lone pairs.2845

Again this is all going to be a -1 anion.2849

Let's go ahead and do the valence electron count.2853

I have three iodines each contributing seven valence electrons giving me a grand total of twenty-one.2855

I have one additional valence electron from the -1 charge which means I can only use a grand total of twenty-two valence electrons.2861

That is eleven pair--one, two, three, four, five, six, seven, eight, nine, ten.2871

We are too few now; we are two electrons shy.2880

We are not going to try multiple bonding.2886

But remember iodine is in period three or higher.2889

I can actually have another lone pair on the central atom to have more than eight electrons.2893

Let's see if the formal charges though are happy.2901

This iodine here on the left has a formal charge of 0.2903

This one on the right has formal charge of 0.2907

This one in the center has a formal charge of -1.2910

Does everything add up to the overall charge?2916

The answer is yes; this is a valid Lewis structure.2919

Remember iodine is somewhat electronegative in fonclbrisch.2923

It is going to be perfectly defined with a negative formal charge.2927

That is the Lewis structure for triiodide.2932

What is the geometry then?--the geometry is what we call linear.2937

That is going to be a 180 degree bond angle.2941

This molecule is going to also be nonpolar because this molecule is homonuclear which means we have no difference in electronegativity among the atoms.2945

Last example is going to be SF5 for sulfur pentafluoride.2958

Central atom in the middle is sulfur.2964

I am going to surround the center with all fluorines like that.2966

I am going to connect everything to the center using a single bond like that.2972

The next step is to complete everybody's octet using lone pair of electrons.2978

Let's go ahead and do that.2984

You see sulfur has already an octet; in fact it has more than an octet.2993

But that is okay because it is in period three or higher.2997

Everybody seems to be good to go.3002

Let's go ahead and do a valence electron count.3003

I have five sulfurs each contributing six valence electrons giving me thirty... I am sorry, my apologies.3007

I have one sulfur contributing six valence electrons giving me six.3016

I have five fluorines each contributing seven valence electrons giving me a grand total of thirty-five valence electrons.3024

When I add this up, I get a grand total of forty-one valence electrons; forty-one valence electrons.3042

Let's go ahead and see if this is right or wrong.3054

There is a little typo here; I am very sorry about that.3062

This shouldn't be SF5; this should be PF5.3065

I am going to go ahead and change that; my apologies.3072

I am going to go ahead and make that PF5; now it is phosphorus pentafluoride.3075

Phosphorus is going to be in the middle now.3080

I am going to change sulfur to phosphorus.3084

Phosphorus is going to be in column five.3087

That is going to contribute five valence electrons.3092

My total is going to be forty valence electrons.3095

Let's see if we used forty valence electrons or not which is twenty pair.3098

Two, four, six, eight, ten, twelve, fourteen, sixteen, eighteen... excuse me.3104

Two, four, six, eight, ten, twelve, fourteen, sixteen, eighteen, twenty, twenty-two, twenty-four, twenty-six, twenty-eight, thirty.3112

Thirty-two, thirty-four, thirty-six, thirty-eight, and a grand total of forty electrons or twenty pair.3122

Yes, this is the valid Lewis structure for phosphorus pentafluoride.3128

This was molecular shape and geometry using VSEPR theory.3134

This molecule is going to be nonpolar because everything is cancelled with all the dipoles.3141

Everything is going to be cancelled just like that.3148

Thanks everyone for your attention; I will see next time on Educator.com.3154

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be advanced bonding theories.0003

We are going to start off with a brief introduction as to why0010

we need these advanced bonding theories from what we already know.0015

That was really Lewis structures.0019

Pretty much the advanced bonding theories that we are going to be talking about are the following.0022

One is what we call valence bond theory.0026

The second one is what we call molecular orbital theory.0030

We will wrap up the session with a brief summary followed by a pair of sample problems.0033

What you and I have already discussed is the representation of valence electrons schematically.0041

Remember that is what we called Lewis structures.0050

We went through a step by step procedure on how to formulate a valid Lewis structure.0052

Lewis structures are sufficient because using VSEPR theory we were able to0058

come up with some predicted geometries strictly off of the Lewis structure.0067

But there exists really two main problems with Lewis theory that we never talked about before.0076

The first main problem has to do with magnetism.0082

If you noticed that all of the Lewis structures that we looked at, all electrons are paired.0087

They are either in a single bond, double bond, or triple bond for electrons.0093

And they are either in a lone pair.0098

If all electrons are paired, then we would expect everything to be diamagnetic.0101

But as we all know, that is not the case.0108

There is definitely more than enough compounds that are paramagnetic,0110

meaning that they do have at least one unpaired electron.0116

Lewis theory does not account for magnetic behavior.0119

A second problem with Lewis theory is the following.0124

Consider all of the compounds that we looked at.0130

For all of the compounds really, the valence shell has been the p orbital.0133

We looked at p orbitals before.0139

They were pretty much three p orbitals per energy level--px, py, and pz.0141

We actually graphed them.0148

We saw that because of Cartesian coordinates, the px, py,0150

and pz orbitals are going to be perpendicular to each other.0154

If you have nothing but 90 degree angles, how do we get 109.5?0159

How do we get 120, etc?0166

Lewis theory doesn't really account for all of the bond angles that are expected from VSEPR theory.0175

We do need advanced additional bonding theories to make up for the flaws of Lewis theory.0189

Valence bond theory is what we call a localized bonding model.0198

What localized means is the following.0203

It is that an atom's atomic orbitals are going to be0205

centered around that specific atom when forming covalent bonds.0209

In other words, you can think of a specific atom owning0213

those electrons rather than being distributed throughout the entire molecule.0218

Again that is what we call localized bonding model.0224

In order to achieve bond angles that are other than 90 degrees,0227

like 120, like 109.5, what valence bond theory suggests is the following.0231

It is that an atom's valence orbitals begin to overlap each other.0239

When they begin to overlap, they mix.0245

They begin to form these mixtures, these combinations of atomic orbitals which we call hybrids or hybrid orbitals.0248

What we are going to do first is we are going to take a look at one type of hybrid orbital.0260

The first type is what we call an sp3 hybridized carbon atom.0264

We go ahead and look at carbon.0270

What this sp3 means is that we have one s orbital mixing with three of carbon's p orbitals.0272

We can go ahead and show that; let me do that in a different color.0286

If I have a carbon s orbital mixing with a p orbital of carbon, we are going to get a hybrid orbital.0290

A hybrid orbital is going to have both s character and p character.0302

But it is going to be mostly p character because it is one s orbital versus three p's.0307

We are going to get a small lobe representative of the s orbital.0313

We are going to get a much larger lobe that is representative of the p orbital.0319

This is what we call a hybrid orbital.0326

Now that we know what the hybrid orbitals look like, they are going to be centered around the carbon atom.0329

Remember it is what we call a localized bonding model.0335

We are going to have the s orbital right in the middle, closest to the nucleus remember.0338

There is going to be one hybrid here; one hybrid here.0343

One hybrid here; one hybrid just in the back.0347

As you can see the way we have drawn it,0352

if you connect the vertices, we do indeed get a tetrahedron.0355

The concept of hybrid formation allows us to get these bond angles that are predicted by VSEPR theory.0366

Let's go ahead and put that into practice now.0375

Now that we physically have a sense of what a hybrid orbital can quote unquote look like.0378

Sp3 first, we are going to first start paying attention to what the superscripts mean.0385

The superscript tells us actually a great deal.0391

In sp3, there is a superscript of 1 for the s orbital.0395

There is a superscript of 3 for the p orbital.0400

What that translates to is the following.0403

That one s orbital mixes with three p orbitals to form the0405

sp3 hybrids just like we have seen in the previous slide.0412

But the question is how many hybrids do we form?0416

Again we turn to the sum of the superscripts to help us answer that.0419

Basically the sum of the superscripts equals to the number of hybrid orbitals that are going to form.0426

When you look at sp3, the sum of the superscripts is four.0435

That is why we form four hybrid orbitals.0439

For example, let's go ahead and take a look at water0445

and see how hybridization can help to explain for bonding.0452

Water has the following structure--the central oxygen, H, and two lone pair.0457

Once again the number of electron groups, it is going to be the same as the sum of the superscripts.0467

Here we have four electron groups surrounding oxygen.0491

There is going to be a total of four for the sum of the superscripts.0494

That is going to be sp3.0499

We are going to over other hybridization what we call schemes.0500

Sp3 is the first one; this is sp3 hybridization.0505

We can show this in an energy level diagram; energy.0511

Here the sp3 hybridized atom is going to be the oxygen.0517

Oxygen starts off as 2s and then 2p.0522

Let's go ahead and fill in the electrons now.0527

Oxygen is going to be 2s2 and 2p4; this is just atomic oxygen by itself.0532

But as soon as oxygen hooks up with those two hydrogens to form water, its atomic orbitals are going to hybridize.0542

Again we are going to form a total of four sp3s.0550

Where are those four going to form?0554

Are they going to form higher in energy than the 2p's?0556

Are they going to be less than the energy of the 2s's?0559

Or are they going to be intermediate?0564

It turns out that we are going to compromise.0567

We are going to have an energy that is going to be intermediate for the entire system.0570

We form four sp3 hybrid orbitals now.0577

When we go ahead now and fill electrons, let's go ahead and fill it.0581

I have a total of six electrons--one, two, three, four, five, and six.0584

When we do this successfully, the overall picture should agree with the Lewis structure.0593

Guess what?--just like the Lewis structure which has two lone pairs around the oxygen,0599

we have one lone pair here and one lone pair here.0604

What do you think these two electrons are?0608

We have two hydrogens bonded to the oxygen.0612

These are going to be bonding electrons with H right here.0616

These electrons go on to form a single bond with the hydrogen; form single bond with H.0629

We now introduce a new terminology.0642

A single bond in terms of valence bond theory is a specific type of bond we call a σ bond; σ bond.0645

When we go ahead and redraw water, H, H, and O, we are going to see that0657

the lone pairs, they are going to exist in sp3 hybrid orbitals of oxygen.0667

We are going to have another sp3 here and another sp3 here, each containing one electron.0676

Those sp3 orbitals are going to overlap and mix with the hydrogen 1s orbitals which also have one electron.0689

Same thing here; that is what we call a σ bond.0698

Again we get the very important term, what we call a σ bond, from valence bond theory.0706

Let's go ahead and look at one more example of an sp3 hybridized atom.0717

The example I would like to look at is ammonia, NH3.0727

Step one is always do the Lewis structure.0735

Here the Lewis structure for ammonia is going to be here just like that.0742

It is going to be three electron groups plus a lone pair0751

giving you a total of four electron groups around the nitrogen atom.0755

Remember the number of electron groups equals to the sum of the superscripts so this is also sp3.0760

Step two, we are going to now come up with our bonding scheme0771

and show the formation of these hybrids; show hybrid formation.0776

Hopefully again when we are done with this, we can correlate it with the Lewis structure.0785

Draw your energy diagram again; there is energy.0801

Nitrogen, just bare atomic nitrogen, is going to be 2s2 and 2p3.0807

But as soon as the nitrogen starts to form bonds with hydrogen to form ammonia, it is going to mix.0818

It is going to form again four sp3 hybrid orbitals.0827

Let's go ahead and put the electrons in.0833

I have a total of five electrons--one, two, three, four, and five.0835

Let's go ahead and see if we can correlate it with the Lewis structure.0842

The Lewis structure has one lone pair.0846

The lone pair is right there; that is so far so good.0849

You notice that there is three single bonds with hydrogen.0853

Remember we call the single bond a σ bond.0856

σ here, σ here, and σ here; guess what?0859

We have three electrons that are each going to form a σ bond with the hydrogen.0862

Once again they each form a σ bond with a hydrogen 1s atomic orbital.0870

Good, we have a pretty good understanding now of what we mean by sp3 hybridization.0885

The next bonding type, the next hybridization type is what we call sp2.0893

Again let's look at the sum of the superscripts; that tells us a great deal.0900

Sp2, the sum of the superscripts is a grand total of three which means0906

an atom is going to be sp2 hybridized when it has three electron groups around it.0915

In addition remember that the sum of the superscripts also tells us how many hybrids we are going to form.0923

Here we are going to form exactly three sp2 hybrids.0931

One more thing, the superscripts also tells us the number of each atomic orbital that is going to be used.0940

Here we are going to be having one s orbital mixing with exactly two p orbitals.0948

Let's think about that.0959

We know that every energy we said has a px, py, and pz, three p orbitals.0961

But if only two p orbitals are being used out of the three, that tells us the difference.0967

One of the p orbitals remains unused, unhybridized.0974

I am going to put that in; one p orbital is going to be unhybridized.0981

We need to now account for that in our energy diagram.0990

Let's go ahead and do that.0995

A nice example of something that is going to be sp2 hybridized, let's go ahead and look at BF3.0997

BF3 here; here is energy.1013

Regular boron itself is going to be 2s2 and then just 2p1.1019

When that goes ahead and mix, we are going to form exactly three sp2s intermediate in energy.1030

Remember that?--always intermediate in energy approximately.1037

This one p orbital that remains unhybridized, if it remains unhybridized, that means1042

it is going to be the same energy as the initial p orbitals.1048

I am going to put that approximately the same just like that.1056

Again we have exactly three sp2s; and we have a 2p orbital that remains unhybridized.1059

We are going to go ahead and fill the electrons now.1066

In BF3, boron just has three electrons--one, two, and three.1069

Guess what?--there is no lone pairs.1075

There is going to be exactly three σ bonds.1077

When we draw out the Lewis structure for boron trifluoride, that is exactly what we get.1080

Boron has zero lone pairs in this structure and exactly three σ bonds.1087

That looks pretty good right now.1098

But let's go ahead and look at another example.1100

Here we are going to learn now a new type of bonding.1103

An example I want to look at has the following Lewis structure.1106

This is going to be a CH, a double bond, O just like that.1110

This is what we call formaldehyde.1120

Here formaldehyde, the central carbon atom has three electron groups.1125

That is also going to be sp2 hybridized.1131

Let's go ahead and set up the energy level diagram now for the carbon here.1136

Carbon is 2s and 2p; it is going to be 2s2 2p2.1140

That is going to hybridize.1148

Again we are going to form exactly three sp2s--one, two, three.1150

We are going to have one of the 2p orbitals remaining unmixed just like that.1154

Let's go ahead and fill electrons now; carbon has four valence electrons.1159

Let's go ahead and populate the orbitals with four--one, two, and three.1163

I am going to put a big asterisk here because this is where we start to deviate.1169

Where do we put the fourth electron?1173

Going by typical Aufbau principle, we would always populate the lower energy1175

orbitals first before proceeding on to the higher atomic orbitals.1182

But in valence bond theory, we are going to go left to right regardless of energy.1187

We are going to go every orbital one time before doubling up.1193

I am going to put my fourth electron here, not in the sp2.1196

That is this special case for valence bond theory is really1202

a big deviation from what we know from Aufbau principle.1207

Please again keep that in mind.1210

When we do this see, it is going to work out because look at the Lewis structure.1214

The Lewis structure has zero lone pairs for the carbon atom.1218

As you can see in our bonding scheme, there is zero lone pairs.1222

We have the following; we have a σ bond here and a σ bond here; σ, σ.1227

And we have a double bond.1234

We have to now discuss what we mean by double bond.1237

A double bond we learned from Lewis theory has exactly four valence electrons.1241

But what type of bonds?1245

It turns out that a double bond is composed of the following.1248

It is composed of one σ bond and a new type of bond, what we call a π bond.1250

Our σ bond is just going to be the regular σ bond from the hybrid orbitals.1258

Guess what?--this electron that we have right here in the unhybridized atomic orbital,1262

that is going to overlap with oxygen to form a π bond.1268

One σ plus one π is equal to a double bond.1272

We can make a pretty big conclusion right now from these two exercises.1279

Lone pairs and σ bonds are going to come from hybrid orbitals.1288

π bonds are going to come from atomic orbitals that are unhybridized.1300

You may be wondering then what type of bond is stronger, is it σ or π?1311

The lower the energy, the more stable it is going to be.1318

The higher the energy, the more unstable it is going to be.1322

We are going to learn that very soon in our next discussion of thermodynamics.1326

σ bonds are typically going to be stronger than π bonds.1333

The reason why σ bonds are stronger is because once again1341

lower in energy but also we have more overlap of the orbitals.1345

If we have more overlap, the attraction between the nuclei is going to be much stronger.1355

That again is sp2 hybridization.1363

The next one I want to go over with you is what we call sp hybridization.1367

Let's go ahead and take a look at the sum of the superscripts.1372

Here is really s1 p1.1375

Remember if you don't see a number, 1 is always implied.1378

That means we are going to have two electron groups around the central atom.1381

Also this tells us that we are going to form a hybrid here by one s mixing with exactly one p.1392

Also we are going to form exactly two hybrid orbitals; two sp hybrids.1404

Let's go ahead and take a look; let's look at hydrogen cyanide; hydrogen cyanide.1415

Step one, it has the following Lewis structure--H, C, triple bond, N.1424

As you can see, carbon is two electron groups; that is going to be sp.1430

Also nitrogen also has two electron groups around it.1436

That too is going to be sp hybridized.1439

Let's first tackle carbon.1441

Then we will tackle nitrogen to see if it agrees with the Lewis structure.1443

Energy; carbon is again 2s 2p.1447

Put in its electrons; carbon is 2s2 2p2.1454

That is going to go ahead and mix.1461

Again we are going to form exactly two sp orbitals; two sp hybrids.1463

It tells us that to form the two sp's, it is the mixing of one s and one p.1470

But remember there is three p orbitals per energy level; px, py, and pz.1476

If I am only using one of those three, that means I have two remaining unhybridized.1481

We are also going to put that in; where do we put it in?1487

We put it in that it has exactly the same energy level as the initial two p orbitals.1491

Let's fill electrons--one, two, and don't forget this is where we deviate.1497

We go fill left to right all the orbitals first before double up regardless of energy.1502

One electron, two electrons, and now electron number three, and electron number four.1508

So far so good; we don't see any lone pairs at all.1514

The Lewis structure has no lone pairs around the carbon.1518

Let's see what HCN has around carbon; no lone pairs.1523

This bond right here is a σ bond; I am going to label that σ.1531

Now the nature of the triple bond.1535

We saw that a single bond is what we call a σ.1539

We saw that a double bond is one σ and one π.1542

A triple bond is going to be one σ but two π this time.1546

Again a triple bond is going to be one σ and two π.1551

We know that σ comes from hybrid orbitals; there is our σ.1557

We know that π bonds come from unhybridized atomic orbitals right here; π here and π there.1561

As you can see, we have two σ and two π's around carbon which is agreeing with the Lewis structure.1569

Let's see if we can get the same agreement with nitrogen this time.1577

Energy; nitrogen, 2s 2p; nitrogen is 2s2 and now 2p3.1582

Again as we already discussed, this nitrogen atom is also sp hybridized.1595

That is going to be one, two.1600

We are going to form two p's right here; exactly two of them unhybridized.1602

Let's go ahead and fill in the electrons.1606

This time it is going to be five--one, two, three, four...1607

OK, now we can double up again with the opposite spin.1611

According to the Lewis structure for nitrogen in HCN, there is going to be one lone pair.1617

There is going to be a triple bond around nitrogen which is one σ and two π.1626

Guess what?--it agrees perfectly.1632

Here is our lone pair; here is our one σ; here is the two π's.1636

Again we see that valence bond theory helps us to count for the1645

nature of a single, a double, and now a triple covalent bond.1650

There are two other atomic hybridizations that you should be familiar with.1658

So far we saw that sp3 is going to be for four electron groups.1663

That is going to be a tetrahedral electron geometry.1671

We saw that sp2 is going to be three electron groups.1679

That is going to be a trigonal planar geometry; again electron geometry.1684

We just did sp; we saw that sp was two electron groups.1692

That was a linear electron geometry.1696

Now sp 3d; sp 3d, now we start talking about the d block orbitals.1700

That is a grand total of five electron groups.1708

If you recall, five electron groups was a trigonal bipyramid electron geometry.1712

Finally sp3 d2, that is six electron groups.1722

That is going to be an octahedral electron geometry.1728

Again it depends on your instructor and your textbook.1734

But I will just present to you the five basic atomic hybridization schemes.1739

That is the formation of hybrid orbitals and valence bond theory.1746

Valence bond theory as we just saw is successful in explaining for the observed bond angles predicted by VSEPR theory.1753

However we still have that one issue about magnetism.1764

Take oxygen for example; oxygen's Lewis structure is right here.1768

If you look at this, every electron is paired; we would expect diamagnetism; expected.1776

In other words, if we take molecular oxygen, it should be repelled by a magnetic field.1787

But experimentally when you pour liquid oxygen through a magnet,1794

it actually sticks to the magnet showing that O2 is actually paramagnetic.1801

Obviously valence bond theory does not always successfully account or predict the actual magnetic behavior of a compound.1814

Then we need yet another bonding theory.1826

This bonding theory is what we call molecular orbital theory, also known as MO theory.1829

Here is the main difference.1840

In contrast to Lewis and valence bond theory, MO theory is a delocalized bonding model; not localized but delocalized.1842

In other words, the electron density is distributed throughout the1853

entire molecule rather than being centered around any individual atom.1857

You remember our discussion about wave functions.1866

It turns out that MO theory is a highly mathematically based theory.1870

If you recall the nature of a wave function, a wave function is representative of atomic orbitals.1878

The actual function if looked at them contain cosine and/or sine trigonometric functions.1883

If we try to think back to that, your basic cosine or sine wave function is a wave.1891

From your early math classes, you learned that these waves have both1901

positive and negative what are called phases; positive and negative phases.1909

Because atomic orbitals are the visual manifestation of a wave function, we can also infer that1923

atomic orbitals also can be represented by different phases, a + phase and a ? phase.1932

For general chemistry, we are only going to limit our discussion to s and p orbitals.1945

Let's first take a look at how s orbitals can interact.1952

I can have an s orbital of this phase mixing with another s orbital of the same phase.1958

When that happens, I form what is called a σ bonding orbital or a σ bonding MO.1970

How does that look like?1980

If here the dots represent the nuclei and if you have exactly two of these s orbitals1982

mixing with each other, we get a smeared out electron density throughout both nuclei.1990

Here is the nuclei.1997

But now we are going to get something that looks like this.1999

Again this is what we call a σ bonding MO.2005

Instead of two separate s orbitals, they now mix and form one entity.2009

As you can see from this diagram, it is completely distributed among both nuclei.2014

This is what we call delocalized.2023

The other possibility can occur.2026

What happens when we have s orbitals mixing that are of opposite phases?2028

To represent opposite phases, I am going to have now shading in one of these s orbitals just like that.2034

Now when I have s orbitals of opposite phases interacting,2044

this is going to be equivalent to destructive interference.2049

Remember we can interpret orbitals as wave-like properties.2056

When we have destructive interference, we actually get the following this time.2062

We are going to get something that looks like this now.2069

This is actually a node.2075

Remember we talked about nodes where we have zero electron probability.2077

This type of MO is not a σ bonding anymore.2085

But it is called a σ anti bonding MO; a σ antibonding MO.2089

For our purposes right now, something you have to remember is that2099

σ bonding will always be lower in energy than σ antibonding.2102

Once again σ bonding is going to lower in energy to σ antibonding.2109

Remember mother nature is always going to favor lower energy.2114

If an orbital is going to result in bonding of the molecule, it is going to be lower in energy.2117

Once again we went over σ bonding and σ antibonding.2124

σ antibonding will typically be represented as a σ*.2129

We talked about s orbitals; let's now talk about p orbitals.2135

P orbitals can interact in one of two ways.2139

They can interact head on or in a parallel fashion.2143

Let's talk about head on first.2152

When I mean head on, if we have a molecule AB like that, the bonding axis is going to be z-axis.2154

I am going to represent this as x; then represent that as y.2166

Once again the z-axis is going to be colinear with the bonding axis.2169

When we have a head on overlap of p orbitals, it is going to be the pz orbitals.2174

Here is one pz; here is another pz interacting head on.2182

We can have the same phases of the pz orbitals overlapping with each other like that.2189

When that happens, we form a σ bonding orbital.2203

Again we form a σ bonding orbital.2210

The MO is going to be distributed throughout the entire molecule around both nuclei.2214

We are going to get something that looks like this now2221

where the two nuclei are here, here, here, and here.2230

Again this is what we call a σ bonding MO.2238

You may ask then what is the other possibility?2242

The other possibility is for the pz orbitals now to interact with their opposite phases like that.2244

When the opposite phases interact, it is just like we saw last time with the s orbitals.2266

That is going to be antibonding; this is going to be σ*.2270

That is going to be represented the following way where the two nuclei again are here and here.2276

Let's go ahead and draw the nodes in, shall we?2295

Here you see that there is one node here, one node here.2298

You see here for the antibonding one node, one node, and one node.2302

I want you to compare and contrast.2307

You notice that the antibonding MO for the p orbitals has three nodes2310

while the bonding MO for the p orbitals has two nodes.2316

The same pattern occurs for the s orbitals.2320

You notice that the antibonding MO for the σ has one node2324

while the bonding MO for the σ has zero nodes.2331

The antibonding MOs typically have more nodes than the bonding MOs.2338

Let's go ahead and look at the parallel interaction.2346

Parallel interaction is going to occur for the remaining two p orbitals, basically px and py.2349

If I have parallel interaction this time, I can have p orbitals occurring like that.2355

When we have parallel interaction, the overlap is not as great.2365

This is not a σ MO anymore.2370

Instead it is going to be a π MO.2372

Because it is the identical phases interacting with each other,2375

this is going to be what is called a π bonding MO.2378

We are going to get an MO that looks like this now distributed among both nuclei just like that.2381

Here we have exactly one node.2395

The other possibility is for parallel interaction but now of opposite phases like that.2401

When we have opposite phases again that is what we call an antibonding MO.2409

This is going to be π*.2413

π* is going to look like this distributed throughout the entire compound.2415

Let me go ahead and erase that.2428

As you can see once again the antibonding MO has more nodes than the corresponding bonding MO.2434

One other possibility is having an s orbital directly combined with a p orbital in a head to head fashion.2448

That is going to be an s orbital combining with the pz orbital just like that.2455

Again that is going to be σ bonding.2464

Again we can also have s and p opposite phases.2467

That is going to be σ antibonding.2472

We pretty much covered all our bases and the different possibilities of the different types2477

of MOs that can form from s and p and how they look like.2482

Now that we have the combinations down, let's put it all together.2490

Let's go ahead and construct what we call a typical MO diagram.2494

The simplest type of compounds we are going to be looking at again are going to be homonuclear diatomics.2499

Remember two things; σ bonding is going to be less than σ antibonding.2507

Similarly π bonding is going to be less than energy of π antibonding.2518

σ versus π, σ bonding versus π bonding, that is going to become a major issue pretty soon.2527

But let's just go over the small ones first.2535

We will come across this issue here.2540

Hydrogen H2, molecular H2, two hydrogen atoms are going to combine.2542

Hydrogen is very simple.2550

We just have the hydrogen 1s orbital here and a hydrogen 1s orbital here.2551

Each of them contains one electron.2555

When hydrogens come together, we get the formation of MOs.2560

When an s orbital combines with an s orbital, remember what the two possibilities were?2566

It is going to be σ bonding.2571

The other possibility at higher energy is going to be σ antibonding.2574

Again bonding is going to be lower in energy than antibonding; that applies there.2580

There is some things I want to point out.2591

The number of MOs that you form must always equal to the number of atomic orbitals that you initially start with.2593

That is another way you check your work.2601

Here I start with a total of two atomic orbitals.2603

I wind up with two MOs--one bonding, one antibonding.2606

Let's go ahead and fill electrons going by the standard Aufbau principle.2609

That is going to be one electron here and then one electron here; spin up, spin down.2615

Now so what?--what is the big deal?--the big deal is the following.2621

We can now calculate something we call bond order.2626

Remember bond order?--it was one of those characteristics of a bond.2629

The equation for bond order is going to be the following.2634

It is going to be equal to the 1/2 number of bonding electrons minus the number of antibonding electrons.2638

When we do it for H2, we get 1.2654

Remember what bond order of 1 means?2657

A bond order of 1 means we have a single bond formed, expected to form.2659

We know H2 exists, we know hydrogen exists as a diatomic gas.2668

It is one of those several elements that occur diatomically.2673

Yes, hydrogen is expected to form.2678

Again all of these experiments are done in the gas phase.2680

Yes, we know that hydrogen exists as diatomic hydrogen in the gas phase.2686

A bond order of 1 is what we expect.2690

Remember if we have a bond order of 2, we expect double bond.2693

We have a bond order of 3, we expect a triple bond.2698

What happens if we have bond order less than 1?2703

If we have a bond order of less than 1, that is less than a single bond which means2706

we don't expect the molecule to exist at all in the gas phase; should not form.2709

You see the power of MO theory already.2716

It allows us to predict if a molecule is going to be stable or not in the gas phase.2718

Let's talk about helium; helium is a noble gas.2724

We know that noble gases exist as monoatomic gases, not diatomically.2729

Expectation for He2 is it should not exist.2734

In this case, we expect a bond order of less than 1.2741

BO less than 1 expected.2746

Let's go ahead and look at the energy diagram here.2750

Once again helium here, helium here, and again 1s, 1s.2754

Each helium is 1s2; now let's go ahead and make the MOs.2760

Again anytime we have 1s combining, one possibility is going to be σ bonding.2766

The other possibility is going to be σ antibonding.2772

Now let's go ahead and fill the electrons; one, two, three, and four.2777

When we go ahead and look at the bond order equation for He2,2782

this is going to equal to 1/2, 2 minus 2, which is going to be 0.2787

Again a bond order of less than 1 is what we expected.2793

That is what we actually get.2797

We are going to keep on moving down.2800

Now we are going to look at diatomic lithium.2802

Let's see if we expect diatomic lithium to exist in the gas phase.2805

Here lithium and lithium; lithium is going to be 2s, 2s.2810

Remember we only care about valence electrons; here 2s1 2s1.2823

Once again when we have s orbitals, we can expect formation of2830

a σ bonding orbital and then of a σ antibonding orbital.2835

Let's go ahead and put the electrons in--one electron here, one electron here.2841

Lithium, Li2, this bond order is going to be 1/2 times 2 minus 0 which is going to be 1.2845

Yes, Li2 is observed experimentally in the gas phase.2853

A bond order of 1 does make sense.2858

Let's move on to the following.2864

It turns out that not only can we do MO diagrams for2865

neutral species but we can also do it for cations.2869

Now this is Li21+; Li, Li, Li21+ in the middle.2874

Again lithium is a 2s atomic orbital; they are going to mix.2882

We are going to form σ bonding and then σ antibonding.2888

Lithium is 2s1.2897

Again because this is going to be 1+, that means we lose one of the electrons.2900

I am going to leave this one blank.2904

I am going to leave this one blank to show that we remove one electron.2906

Its MO is just going to be that, just one electron in the bonding MO.2913

BO or the bonding order is going to be equal to 1/2 times 1 minus 0.2917

We get an order of 1/2.2922

This suggests that the Li2 cation should not occur in the2925

gas phase because here the bond order is less than 1.2931

Good; we now move on to B2.2938

The reason why this is now going to be slightly different is2943

because as soon as we enter boron, we enter the p block.2946

We have not have not talked about how the p orbitals are going to be expected to be looking.2953

What do you get more overlap with?2960

Is it going to be s and s?--or is it going to be p and p?2963

The s orbitals are going to be closer to the nucleus.2967

We expect them to be lower in energy; we know that from Aufbau.2974

For example, 2s is usually less in energy than 2p.2978

It turns out that the σ bonding here is going to be less than the π bonding.2982

What we expect because typically we have more overlap with the σ's than we do with π's.2993

Remember σ is more direct overlap; π is parallel interaction.3000

We expect σ bonding to be less than π bonding; let's see what happens.3005

Boron, boron; boron is going to be a 2s orbital here, 2s orbital here, and 2p's, 2p's.3016

Let's fill the electrons in for boron; each boron is 2s2 2p1, 2s2 2p1.3028

Let's go ahead and tackle what we know so far already.3035

We know that the 2s orbitals can mix.3038

They are going to form our typical σ bonding and σ antibonding.3042

But what about the p orbitals?--remember the p orbitals.3048

Remember that when pz interacts with pz, we can get σ bonding and σ antibonding.3053

Remember that px and py are the same.3062

Those guys are going to give us π bonding and π antibonding.3071

Notice that there is two sets.3080

It turns outs that when we talk about diatomic boron, it turns out that the π orbitals are3086

going to be lower in energy than the σ orbitals just like that and3098

that the π*s are going to be lower in energy than the σ*.3116

That is not what is expected.3125

Remember we expect σ bonding to be less than π bonding.3126

What happens is the following; the reason why σ is above π.3132

It is due to the phenomenon that we call sp mixing.3139

What happens is the following.3145

That a boron 2s orbital is actually going to mix or interact with the boron 2p orbital.3147

What that does is it results in a lowering of the σ bonding MO from 2s3162

but a raising in energy of the σ bonding orbitals from the 2p's.3176

Essentially one goes down lower in energy; one goes up.3186

It makes sense that the 2s goes down because it is more stabilized due to the mixing.3191

Remember 2s's are going to be closer to the nucleus.3195

They get affected more than 2p's due to this interaction.3198

For B2, again the big take-home message is that the π bonding MOs are going to be3203

lower in energy than the σ bonding MOs of the 2p's due to sp mixing.3209

That is our MO diagram; let's go ahead and fill it up now.3216

I have a total of six electrons--one, two, three, four, five, and six.3220

When we go ahead and calculate now the bond order, the bond order here is going to be equal to 1/2...3226

number of bonding electrons is 4, two antibonding electrons.3232

That is going to be a bond order of 1.3236

Yes, we do expect diatomic boron to exist in the gas phase.3238

Let's look at another example, N2.3246

N2 is something we have looked at already in terms of Lewis structure.3249

Look at that; that is our Lewis structure of N2.3253

We expect a bond order of 3; let's see if we get that.3255

Nitrogen, nitrogen, this is 2s, 2s; then here is nitrogen's 2p orbitals just like that.3266

N2 is right in the middle.3276

Once again the 2s orbitals are going to mix to form σ and σ*.3279

Again the π orbitals from the 2p's are going to be lower in energy than the σ orbitals from the 2p's.3288

Then followed by π* and σ*.3299

Let's go ahead and populate the orbitals with the electrons.3307

Nitrogen has a configuration of 2s2 2p3, 2s2 2p3.3311

Now let's go ahead and fill; that is a total of ten electrons.3321

One, two, three, four, five, six, seven, eight, nine, and ten.3324

When we go ahead and calculate the bond order, 1/2 number of bonding electron which is going to be 8.3335

Number of antibonding electrons is 2.3342

Look at that; we get a bond order of 3 which is what we expect.3345

The accepted Lewis structure for N2 is a triple bond.3351

The next compound is going to be molecular oxygen.3359

Molecular oxygen, that is the one where we get paramagnetism from experiment.3363

Paramagnetism expected which means the MO diagram should have at least one unpaired electron.3373

Let's go ahead and look at the MO diagram; oxygen here--2s, 2s, and 2p.3397

O2 right in the middle; let's go ahead and form our MOs.3413

Again the 2s's can form σ bonding and σ antibonding and 2p's.3420

Let's just for now go by what we have already done3433

for the previous compounds where the π's were less than σ.3436

This is π; this is σ; and then this is π* and then σ*.3445

Let's put the electrons in--2s2 2p4, 2s2 2p4.3455

That is a grand total of twelve electrons that we have to use.3467

Let's fill them--one, two, three, four, five, six, seven, eight, nine, ten, and eleven and twelve.3470

Let's go ahead now and see do we have lone pair expected?3482

Yes, we have lone pairs right there; paramagnetism is good to go.3487

But let's go ahead and look at the bond order.3490

The bond order is the following.3492

1/2 times the number of bonding electrons minus the number of antibonding electrons.3496

That is going to be number of bonding electrons is 83501

minus number of antibonding electrons is going to be 4.3505

That is going to be 2; yes, double bond expected.3510

So far this sample diagram does work.3514

However experimentally it turns out that this is not the correct MO diagram for O2.3517

It turns out that in the actual MO diagram, these orbitals here,3525

the one we had that little controversy with is flipped.3537

It turns out that in the actual MO diagram, the σ is going to be less than π.3542

Let's go ahead and draw that.3547

I will explain why in a second why that makes sense.3549

2s2 and then 2p4; let's go ahead and fill everything.3552

Here we have our σ bonding, σ antibonding.3575

Here is the actual order; σ on the bottom, π on the bottom, then π* here, then σ* here.3585

Let me go ahead and label that; σ, π, π*, and σ*.3595

Let's go ahead and fill electrons right now.3601

One, two, three, four, five, six, seven, eight, nine, and ten, and then eleven and twelve.3604

It turns out that this is going to be the accepted one where σ is less than π.3618

The reason is because of the following--the sp mixing that we discussed earlier stops.3624

The reason why it stops is because it turns out when we reach oxygen3635

in the periodic table, it is a relatively heavier element than say nitrogen.3638

What that does is the following.3645

It is that the s orbital is now too low in energy to mix with the p orbital.3647

As the atom gets heavier and heavier, most of the interaction is going to come with the s orbital.3662

That is going to if you will benefit first.3674

That is going to be much lower in energy.3677

Because of that, the s orbital of oxygen is just too low in energy compared to the p orbital of oxygen.3682

Sp mixing stops.3689

We return to what is expected where the σ bonding MO orbital from the p orbitals is3692

going to be less than the energy of the π bonding MO orbitals of the 2p's.3698

Again this is always going to be the case; please make a note.3710

This is always the case for O2 and higher; F2 counts, Cl2 counts, etc.3714

Really everything up to and including N2 is one situation.3727

Then O2 and after is going to be this situation.3732

So far we have looked at only homonuclear diatomics.3738

We can do MO diagrams also for heteronuclear diatomics too.3741

Let's consider hydrochloric acid; hydrochloric acid has the following Lewis structure.3746

Here is hydrogen; here is chlorine; hydrogen is 1s.3756

Because the chlorine is more electronegative than hydrogen, its 2s orbital is going to be lower in energy than the 1s.3765

Then chlorine's 2p is going to be relatively higher in energy there.3774

Hydrogen is 1s; chlorine is 3s 3p.3780

Chlorine is going to be 3s2 and then 3p5 just like that.3792

In the Lewis structure for hydrochloric acid, we have a σ bond right there.3806

But chlorine has three lone pairs here which means...3816

These three lone pairs means that the six electrons are not involved in bonding, are nonbonding.3823

In terms of MO purposes, nonbonding electrons will occur not in a3832

bonding MO, not in an antibonding MO, but in a nonbonding MO.3841

That is a third type of molecular orbital that we now introduce.3849

For our purposes in GChem, nonbonding MOs are going to be at3855

the same energy as their atomic orbitals; same energy as atomic orbitals.3858

Let's go ahead and see what happens then.3871

It turns out that the 3s orbital of fluorine is just too low in energy to interact with anything.3879

Here is our first nonbonding MO.3894

I am going to go ahead and put those two electrons there just like that.3898

The next item is now the 1s orbital interacting with the 3p orbitals.3907

When this happens, it turns out we are going to get the following experimentally.3916

The 3p and the 1s are going to mix.3922

They are going to form a σ.3925

Of course if they form σ, that means they must also form a σ*.3927

Remember we have to conserve orbitals; we have a total of five atomic orbitals.3936

We have only used three so far in the MOs.3943

That means I have two remaining.3946

Those two remaining are going to be additional nonbonding orbitals just like that.3948

Let's go ahead and fill everything up.3955

We have a grand total of eight electrons; two is already accounted for.3957

One, two, three, four, five, six, seven, and eight.3964

Guess what?--our MO diagram agrees with our Lewis structure.3969

Here is one lone pair nonbonding; here is another lone pair nonbonding.3974

Here is another lone pair nonbonding; there we have our σ bond.3980

When you calculate the bond order, the bond order is equal to 1/2 number of bonding electrons which is 2.3986

You notice that there is no electrons in antibonding MOs.3993

You do not count nonbonding electrons in this equation.3996

There we go--a bond order of 1 which agrees with the Lewis structure of3999

hydrochloric acid where we just get a single covalent bond.4004

That is how we tackle heteronuclear diatomics.4008

To summarize, valence bond theory was the first bonding theory we talked about today.4013

It explains for observed bond angles but doesn't always account for magnetic behavior.4018

MO theory is pretty much the one that is going to supersede everything else.4022

That is the delocalized bonding model which views electrons as being distributed throughout4028

the entire molecule rather than just being centered around any individual atom.4033

Let's go ahead and look at a pair of sample problems.4041

In this one, determine the atomic hybridization around the central atom in the following molecules.4044

I picked these because these were Lewis structures that we have previously seen in the last lecture.4050

Here is the Lewis structure for XeO3, for xenon trioxide.4057

It is going to look like that.4061

Here we have a total of four electron groups around the xenon.4063

That is going to be sp3 hybridization.4068

SF6, SF6 is going to look like this.4071

There are lone pairs around each of the fluorines.4080

Remember sulfur is one of those elements that can have more than an octet.4086

This is completely valid Lewis structure.4092

Here we have six electron groups around the sulfur.4094

That is going to be a sp3 d2 hybridization.4097

Now I31-, I31-, its Lewis structure is going to be this.4102

When we go ahead and look at this, this is a total of five bonding groups around the iodine.4124

That is going to be sp3 d.4133

That is atomic hybridization prediction for a central atom from a Lewis structure.4137

Now moving on to the final sample problem, sample problem two.4145

Draw a qualitative MO diagram for F21- and comment on its expected stability.4149

Let's draw it--energy; fluorine, fluorine, and then F21-.4156

Fluorine has the 2s orbital and then its 2p's; let's go ahead and populate.4165

Fluorine is 2s2 2p5 and again 2s2 2p5.4177

Don't forget we have a 1- charge.4190

I am just going to go ahead and add that to here.4192

It doesn't matter which one; they are the same.4195

2s orbitals are going to mix.4198

They are going to form a σ MO and a σ* MO.4202

Here fluorine is after nitrogen; that is when the sp mixing does not occur anymore.4208

In this case, from the 2p orbitals, we are going to get σ first4218

followed by the π's and then π* and now σ*.4224

σ, π, π*, σ*; let's go ahead and fill the electrons in.4233

I am going to have a grand total of fifteen electrons to use.4239

One, two, three, four, five, six, seven, eight, nine, ten, eleven, twelve, thirteen, fourteen, and fifteen.4244

Let's go ahead and calculate its bond order in order to comment on the stability.4258

The bond order is equal to 1/2 number of bonding electrons--that is going to be 8.4262

Minus number of antibonding electrons--that is going to be 7.4270

In this case, we get a bond order of 1/2 which is less than 1.4278

Therefore we do not expect F21- to occur in the gas phase.4282

That is another qualitative MO problem.4292

Thank you all for your attention; I will see you all next time on Educator.com.4297

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is on gases.0002

Going to start off with a brief introduction followed by the following series of topics.0009

The first is what we call the kinetic molecular theory of gases0014

which is basically a bunch of postulates that describe gas behavior.0018

We are then going to go over the parameters that are used to characterize gases--namely pressure, volume, temperature, and moles.0024

When we combine those, we are going to get a series of gas laws which relates all four of those parameters.0035

All of these simple gas laws are then going to culminate into what we call the ideal gas law.0042

After we study the ideal gas law, we are then going to go over some applications of the ideal gas law0048

because we can come up with several additional parameters straight from that law.0054

A unique topic then is going to be gas mixtures and what we call partial pressures.0061

Finally the last topic is going to be stoichiometry and applying it to reactions that involve a gas.0067

Then of course as always, we will finish off with a summary followed by some sample problems.0077

Basically there are five postulates to the kinetic molecular theory of gases.0085

The first one is the following.0092

It deals with gas motion and basically tells us that gases travel in straight lines obeying Newton's laws.0095

They have straight trajectories; they are in constant motion.0106

Number two, the molecules in the gas occupy no volume.0110

That is we treat them as individual points.0114

In other words, if you look at a gas sample, most of it is actually empty air.0119

Number three, when gas molecules collide, we say that they follow elastic collisions.0124

That is upon collision, there is no loss of energy.0131

There is no transfer of energy.0136

You can imagine a bunch of billiard balls colliding with each other.0138

After they collide, they bounce off each other and go their separate ways.0143

Imagine that but pretty much going on infinitely with no loss of energy.0147

Number four, there are no attractive or repulsive forces between gas molecules0153

which explains why gases are so diffuse if you will.0159

Finally when we talk about the kinetic energy of a gas,0166

the kinetic energy of a molecule is really related to its kelvin temperature.0171

Any gas that follows these five postulates of the kinetic molecular theory, we call that gas an ideal gas.0178

In reality, there is no such thing as an ideal gas.0190

But by applying this model, it allows us to make a lot of simplifications and0195

a lot of assumptions which allows us to further study the gases and0200

use models which help us to describe gas behavior pretty well actually.0206

That is the kinetic molecular theory of gases.0215

We now next turn to the parameters that are used to characterize gases.0219

The parameters are basically pressure, volume, temperature, and moles.0224

From physics, pressure is formally defined as the amount of force per unit area.0231

Pressure is equal to force per unit area.0237

If we had for example a flat surface and we put a box on it,0246

that box is applying a downward force on the surface.0252

In other words, the box is applying a certain pressure on the surface.0256

If we take the same box but we stand it upright this time, same box,0260

my area of contact is now a lot smaller than in the first picture.0266

In this case, because the area is smaller, the pressure is going to be larger.0273

That is our formal definition of pressure as defined from physics.0284

But in terms of gases, we are going to describe pressure from a particulate level diagram.0290

Imagine a container.0297

Basically we have gas particles that are moving in random directions once again following0301

the kinetic molecular theory of gases and moving in straight lines, constant motion.0307

Not only do the gas particles collide with each other but the gas particles0314

also collide with the container wall; collision with container wall.0319

When that collision occurs with the wall of the container, that itself0331

generates a force just like billiard balls colliding with each other.0334

It is this force of impact that we tend to relate to gas pressure.0339

Force of impact is proportional to gas pressure.0348

For chemistry, for gases, this is our interpretation of pressure.0362

Now that we have defined pressure, let's go ahead and talk about0370

the common units of pressure that are used to make measurements.0373

From physics, the typical units of pressure are going to be psi which is pounds per square inch and pascal.0380

All of us see psi in tires; we also see psi for water pressure too.0388

We are not going to use psi and pascal too much.0400

Really in chemistry, we are going to use these three--millimeters of mercury, atm, and torr.0402

Millimeters of mercury is what we see on a barometer for the weather report.0407

We also see that whenever you take your blood pressure reading.0414

Atm stands for atmospheres or one atmosphere of pressure.0419

Basically one atmosphere of pressure is going to be the pressure roughly at sea level.0430

Once again at sea level, we are at roughly one atm of pressure.0440

Finally torr, torr is named after Torricelli who invented the barometer.0445

Given at the bottom, we have the relationships for each of these units.0462

One atm is equal to 14.7 pounds per square inch which is equal to 760 millimeters of mercury0467

which is equal to 760 Torr which is equal to 101.325 kilopascals.0475

Of course, you see all of these equivalent statements.0481

From equivalent statements, we can then use them as conversion factors.0483

Now that we have talked about pressure, let's continue on.0490

From the kinetic molecular theory, we are told that gases travel in straight paths.0494

This implies the following.0500

That gases are going to travel in straight paths until they collide with something,0502

either with each other or the wall of the container.0505

What that says is that gases are going to expand to fill their entire container.0508

Hence the volume of a gas is strictly determined by the container that it is placed into.0514

The reason why we can do this is because gases are compressible.0522

Remember that most of a gas is empty air, that they have negligible volume.0526

I could take the same amount of gas in a ten liter bottle0533

and compress it easily to a smaller bottle, no problem.0536

The common units of volume are going to be liters of course and milliliters.0541

Sometimes milliliters, you also see cc or cubic centimeters.0545

Another postulate from the kinetic molecular theory is the kelvin temperature.0553

We say that the kinetic energy of a gas is proportional to its kelvin temperature.0559

In other words, temperature of a gas is directly related to how fast these particles are moving.0563

Because it is directly related, this says the following.0570

The faster a gas is, the hotter its kelvin temperature.0573

We can also interpret this at the particulate level.0577

We are saying that kinetic energy is going to be proportional to the temperature in kelvin.0583

If you consider a gas sample, if we apply heat to this, of course that is going to result in0593

much faster motion because the gases will get more kinetic energy; faster motion, higher kinetic energy.0606

Finally the last parameter is the mole amount of a gas.0625

For gases, the amount of a gas is going to be related to moles of course.0632

Those are the four parameters--pressure, volume, temperature in kelvin, and moles.0637

Now that we have introduced the four parameters, we now get into what is called the simple gas laws.0645

The simple gas laws basically do the following.0650

They are a series of equations that relate the four parameters of a gas that we just covered.0653

There are restrictions for these gas laws to work.0660

First we assume ideal behavior.0663

That is the gas is going to follow all five postulates of the kinetic molecular theory.0665

All other parameters are held constant.0671

If I compare pressure and volume for example, that means temperature and moles are being held constant.0675

All else is held constant.0680

Let's go ahead and tackle each of these gas laws now.0682

The first gas law is called Charles's law.0685

Charles's law states that volume and temperature are directly related, holding pressure and moles constant.0688

Volume is directly related to temperature; let's take a look at this.0696

If I have this container and these gas particles are moving in random directions,0702

if I heat it, the gas particles are going to have more kinetic energy.0712

We are going to result in an expansion of the container if the container is flexible.0719

This is going to be larger volume.0728

That is why if you try heating a balloon up, you see the balloon expanding.0733

It is because the temperature is going to increase the kinetic energy of the molecules.0739

They are going to push outward on the container, on the wall of the balloon.0746

The equation to do this, to quantify Charles's law, is V1 over T1 is equal to V2 over T2.0751

That is if we know the initial volume and initial temperature, we can get0762

either the final volume or the final temperature, whichever is not given.0765

The restriction for this equation is that this must be in kelvin.0771

Temperature must be in kelvin.0775

Volume, it can be in any units as long as they are identical units.0777

That is a pretty straightforward equation to use.0787

Volume and temperature are directly proportional to each other.0790

The next gas law is what we call Boyle's law.0794

Boyle's law states the following.0796

That pressure and volume of a gas are inversely related when temperature and moles are held constant.0799

That is pressure is inversely related to volume.0805

Let's think about this; let's say I had two gas particles in a small container.0810

All of a sudden, let's say the size of the container has increased.0819

I am holding everything else constant.0824

When this happens, I am going to have a smaller rate of collision with the container wall.0828

Because my rate of collisions with the container wall is going to be smaller, my force drops off.0844

Therefore my pressure is going to drop off.0853

Once again pressure is inversely related to volume.0859

If you ever go to the higher elevation, you notice that a potato chip bag or a snack bag is always larger.0862

It is because at the higher elevation, the outside pressure is much smaller.0870

To compensate, the air inside the bag is going to expand.0875

This is why balloons also tend to pop the higher they go.0880

Because as the elevation increases, the air pressure gets lower.0884

The air molecules inside the balloon expand against the walls of the balloon.0887

The equation for this, for Boyle's law is P1V1 is equal to P2V2.0894

Once again for this equation to work, P1 and P2 must be identical units.0902

V1 and V2 also must be identical units.0913

Once again this is a rather straightforward equation to use.0917

We can calculate any final pressure or volume given the other three parameters.0919

That is Boyle's law.0926

The next gas law is what we call Gay Lussac's law.0929

Gay Lussac's law is pressure and temperature.0933

It tells us that pressure and temperature of a gas are directly related when volume and moles are held constant.0936

That is pressure is directly related to temperature; that just makes sense.0943

When we have gas particles just like this, let's say this is colder.0950

That is going to generate some pressure.0958

But if we take the same volume, the same box, the same amount, and I heat this sample up, that is going to0960

result in obviously a higher rate of collision with the walls of the container; more collisions with container wall.0969

That is going to increase my force.0982

Because my force goes up, my pressure goes up also.0985

On extreme temperature differences, a car tire is always going to be at a lower pressure when it is colder.0992

Later in the day when it gets much hotter, the pressure will slightly increase because of this difference.1002

The equation for Gay Lussac's law is the following.1009

P1 over T1 is equal to P2 over T2.1011

Once again the units of pressure must be identical.1016

But remember that our temperature is always related in kelvin whenever discussing a gas.1021

That is Gay Lussac's law.1029

The last simple gas law is what we call Avogadro's law.1033

Avogadro's law is the following.1037

That the volume and moles of a gas are directly related when temperature and pressure are held constant.1039

V is proportional to n.1046

Just think about maybe a car tire.1049

You put more air into it; you increase the amount of air.1054

What happens?--the volume increases.1059

The equation for Avogadro's law is V1 over n1 is equal to V2 over n2.1064

Once again the volume units must be identical.1073

n1 and n2 will always be in moles.1080

That is Avogadro's law.1084

If you look at the four gas laws, Charles's law, Boyle's law, Gay Lussac's law,1086

and Avogadro's law, really Boyle's law is the only one that stands out.1091

It is the only where we have something times something is equal to the product of something else.1096

Every other gas law is division on the left side of the equation and on the right side of the equation.1102

Please make a note of that, Boyle's law is definitely the one that stands out.1110

In case your instructor ever requires you to memorize these gas laws.1119

When we put all of these simple gas laws together, they culminate into one equation.1124

This grand equation is what we call the ideal gas law where PV is equal to nRT.1132

When we do this, there are a couple of restrictions.1138

That pressure must be in atm; volume must be in liters; n is simply moles.1140

The temperature must be in units of kelvin as we always have said.1147

There is something we haven't introduced yet; that is what R is.1152

R is what we call the universal gas constant.1155

It is equal to 0.08206 liters atmosphere K mol.1158

Once again you may or may not have to memorize this.1164

Definitely refer to your instructor for that.1166

That is a relatively straightforward equation to use.1169

Pretty much for an ideal gas, I can determine the pressure, volume, moles,1171

or kelvin temperature given any of the other three parameters.1177

Again this is the ideal gas law.1183

Probably something you want to be comfortable with is to solve for a single variable.1187

Pressure here is going to be equal to nRT over V.1193

Volume is equal to nRT over P; n is equal to PV over RT.1198

Temperature is going to be equal to PV over nR.1206

That is again the ideal gas law.1211

The ideal gas law is relatively straightforward to use.1214

But another important aspect of it is that we can derive and come to many conclusions using this law.1217

The first relationship that we are going to get from the ideal gas law is what is called standard temperature and pressure.1226

It becomes very difficult to compare gases because there is so many parameters--pressure, volume, temperature, and moles.1235

A set of universal conditions has been defined.1242

This set of universal conditions is called standard temperature and pressure or STP for short.1246

Standard temperature is 273.15 kelvin; standard temperature is 1.0 atm.1252

When these values are plugged into the ideal gas law, we can go ahead and solve for the ratio of volume to moles.1258

When we solve for this ratio of volume to moles, we get exactly 22.4 liters per mole.1265

This is what we call molar volume; its significance is the following.1271

That at STP, one mole of any ideal gas regardless of its identity occupies a volume exactly 22.4 liters.1277

One mole equals 22.4 liters; that is an equivalence statement.1288

From that, we can use that as a typical conversion factor.1294

Once again molar volume at STP only, 22.4 liters per mole.1298

Another application we can derive from the ideal gas law is gas density.1304

Gas density is going to be measured in grams per liter.1312

We are not going to be in its derivation.1315

But the density of a gas in grams per liter is equal to the following.1317

It is equal to the molar mass of the gas in grams per mole times1322

the pressure in atm divided by the universal gas constant times the kelvin temperature.1330

You can convince yourself that when all the units cancel, we are left with grams per liter.1337

This equation once again is relatively straightforward to use.1342

However there is an important thing that cannot be overlooked.1345

We now have a relationship between density and temperature for gases.1348

You see here that density is going to be inversely related to the kelvin temperature.1353

That means the following.1359

That as temperature of a gas goes up, the gas density decreases.1360

As temperature goes up, gases tend to become lighter.1366

Therefore they tend to rise; this explains why hot air balloons rise.1373

As you heat the gas within the walls of the balloon, the gas becomes less dense than air.1379

It results in a lower density and results in rising of the hot air, bringing the balloon upwards.1385

Once again density is inversely related to the kelvin temperature of a gas.1394

A final gas law that focuses on pressure, this is called Dalton's law of partial pressures.1401

Dalton's law of partial pressures refers to gas mixtures.1409

It tells us the following.1412

Pretty much that the whole is equal to the sum of the parts.1413

The sum of the individual pressures of each gas component is equal to the total pressure of the gas mixture.1419

These individual pressures, the technical term is called partial pressures.1426

Basically very simple--the total gas pressure of a mixture is equal to the partial pressure of 1431

the first gas plus the partial pressure of the second gas, etc.1443

Once again the total pressure is equal simply to the sum of all individual pressures.1448

We now come back to stoichiometry.1457

Stoichiometry is something that we spend a great deal of time on.1463

At the basis of stoichiometry was the following.1468

We want to go from moles of A to moles of B.1472

To do this, we use the conversion factor, the mole to mole ratio.1475

From moles of B, you can go to grams using molar mass.1481

You can go to atoms and molecules using Avogadro.1489

You can go to liters if it is a solution using molarity.1501

The same thing applies on the other side to go to moles of A for example.1507

We spent a deal of time doing mole to mole conversion and also mass to mass conversions.1512

All we are going to do now, we are going to apply our knowledge of stoichiometry to gases.1517

If everything is about pretty much getting to moles first, we have an ideal gas law that helps us do that.1524

Typically for gas stoichiometry problems, we are going to use ideal gas law where n is equal to PV over RT.1532

If we are at standard temperature and pressure, we could take a shortcut.1540

We can just use the molar volume to get to moles because we know that one mole equals 22.4 liters.1544

In this case, the ideal gas law not needed.1552

But again this is only at STP.1557

Let's go ahead and do a sample problem then; the question is the following.1561

How many liters of oxygen gas at standard temperature and pressure are needed to form 10.5 grams of water vapor?1566

As soon as I see the letters STP, I know that I am dealing with 1 atm pressure and 273.15 kelvin.1573

I also know that one mole of a gas is going to be equal to exactly 22.4 liters.1582

The first thing you always do in stoichiometry is to make sure the chemical equation is balanced like we have always done.1590

Here we are going to need two hydrogens and two waters.1598

What do we have here?--we have the mass of the water vapor.1605

Somehow we want to go from mass of water vapor all the way to liters of O2 gas.1613

Because I am at STP, the liters of O2 gas is going to1622

come from molar volume which is one mole is equal to 22.4 liters.1627

But in order to get the moles of O2, I first need the moles of H2O.1641

Before moles of H2O, we then have our mass of H2O which is given.1651

There is our basic flow chart; it is pretty much three main steps.1656

Let's go ahead and do this.1660

10.5 grams of water vapor times 1 mole of water divided by its molar mass of 18.016 grams of water.1664

That gives me moles of A.1679

Now from moles of A to moles of B using the mole to mole ratio1681

which is 1 mole of O2 over 2 moles of H2O.1684

Finally now that I am at moles of O2, I can go ahead and1690

use molar volume as a conversion factor to go and get volume.1693

22.4 liters for every one mole of O2.1697

When all is said and done, we get a volume of 6.5 liters that are1703

required at STP for this reaction to make 10.5 grams of water vapor.1707

Let's now go ahead and summarize this lecture.1715

We started off today's lecture with the kinetic molecular theory of gases.1719

It is basically five postulates which describe ideal gas behavior.1723

We then proceeded to tackle the simple gas laws which basically relates the1728

four parameters used to characterize gases--pressure, volume, kelvin temperature, and moles.1732

When we culminated all of these simple gas laws, we arrived at the ideal gas law.1740

The ideal gas law allows us to come up with many applications including density and its relationship with temperature.1745

Finally all of our stoichiometry skills that we established previously can easily apply to gas problems.1754

That is our summary; let's go ahead and do a series of sample problems.1764

Here is sample problem one; you have a 827 milligram sample of a gas.1769

It occupies 0.270 liters when measured at a temperature of 88 degrees Celsius and a pressure of 975 millimeters of mercury.1775

Calculate the molar mass of the gas; let's take it one by one.1784

Here we have mass; here we have volume, temperature, and pressure.1788

The question is asking for molar mass.1797

Molar mass, we all know to be in units of grams per mole.1800

We have the grams already; that is the 827 milligrams or the 0.827 grams.1805

All we have to get then is the moles.1812

Once we have that, we can divide the two numbers to give us the molar mass.1815

We need to get the moles of this gas which is n.1821

We are given pressure, volume, and temperature; that is three out of the four parameters.1826

We can go ahead and use the ideal gas law to help us do this.1834

Moles is equal to PV over RT.1838

Pressure is 975 millimeters of mercury.1845

We have to then go ahead and convert this to atm remember.1851

That is our restriction.1853

We are going to multiply this by the volume in liters.1857

It is already in liters.1860

We are going to divide this by the universal gas constant.1863

We are going to then multiply this by the kelvin temperature.1871

88 plus 273.15; this gets us 0.012 moles.1876

Now we can go ahead and proceed to solve for the molar mass.1889

0.827 grams over 0.012 moles.1892

That is going to be equal to roughly 69 grams per mole for molar mass.1899

This is another nice application of the ideal gas law.1907

It can be used to determine the molar mass of a gas that follows ideal behavior.1911

Let's go ahead and now proceed on to sample problem two.1919

What mass of silver(I) oxide is required to form 388 milliliters of O2 gas1923

when measured at 734 millimeters of mercury and 25 degrees Celsius?1930

Mass is what we want to get; we are given volume.1937

We are given pressure; we are given temperature; guess what?1940

You have chemical equation here which pretty much means you have a stoichiometry problem.1945

Always the first step is to balance.1951

When we go ahead and balance this, we are going to need 2 silver oxides and 4 silvers.1955

We want to go from basically the following.1962

We are given the pressure, the temperature, and the volume of O2.1967

That is three out of four parameters.1971

We can go ahead and get the moles of O2.1973

n of O2 is equal to PV over RT.1976

That is going to be equal to 734 millimeters of mercury times 1 atm divided 760 millimeters of mercury.1981

Going to multiply that by the volume in liters which is 0.388 liters,1994

divided by the universal gas constant, 0.08206 liters atmosphere K mol,2000

times the temperature in kelvin, 25 plus 273.15.2009

Then the moles of O2, we get 0.015 moles of oxygen gas.2015

We want to go from moles of O2 which is what we have.2023

Somehow we want to go all the way to the mass of silver(I) oxide.2027

We know how to do that.2032

This is really now just a matter of doing something we have already learned.2033

We are going to go from the moles of O2 to the moles of silver(I) oxide using the mole to mole ratio.2038

Then on from there is to the mass of silver(I) oxide using the molar mass.2048

Let's go ahead and finish this up.2055

You have 0.015 moles of O2.2058

The mole to mole ratio is going to be 2 moles of silver(I) oxide for every 1 mole of oxygen gas.2062

Then we are going to go ahead and multiply this by the molar mass of silver(I) oxide to get to grams.2072

Its molar mass is 231.74 grams for every 1 mole of silver(I) oxide.2078

We get roughly 7.0 grams of silver(I) oxide that are required.2089

That is another stoichiometry problem that involves gases.2097

Thank you all for your attention; I will see you all next time on Educator.com.2102

Hi, welcome back to Educator.com.0000

Today's session in general chemistry is going to be concerning intermolecular forces and liquids.0002

As always we are going to start off with a brief introduction.0012

Then we are going to talk about the intermolecular forces first of polar compounds and finally of nonpolar compounds.0015

After we talk about the intermolecular forces in these classes of compounds, we will then0024

get into properties of liquids and how intermolecular forces affect these properties.0030

As always we will go ahead and do a brief summary of the presentation followed by some sample problems.0038

For the introduction, what we are looking at is basically the attractive forces that hold molecules together.0048

If you have a molecule and an identical molecule comes along right beside it,0056

what is responsible for those two molecules interacting with each other or attracting each other?0063

We call these forces of attraction intermolecular forces.0067

For my presentation, I am going to be abbreviating intermolecular forces as IMF.0073

What we are going to see is a good application of IMF.0080

We will see that the strength of a liquid's IMF can easily influence how it behaves physically.0085

Let's now get into the intermolecular forces for polar molecules.0092

The first type of IMF for polar molecules is what we call an ion-dipole force.0098

An ion-dipole force is going to occur between a polar molecule and an ionic compound.0104

For the example that I want to look at on the particulate level, this is going to be table salt dissolved in water.0111

Here I have water and table salt.0118

As soon as I dissolve table salt in water, the sodium chloride is going to break up into ions.0126

We have Na+ and Cl1-.0131

We have already discussed polarity previously.0136

We know that water is partial positive here, partial positive here, and partial negative there.0139

Something interesting is going to happen.0146

The partial negative end of oxygen is going to get attracted to the full positive ion that is sodium.0148

In addition, the partial positive end of water is going to get attracted to the full ion that is chloride.0160

This is what we call an ion-dipole force.0171

It is basically the attraction between an actual ion that has a full charge and a compound that has a dipole moment.0175

Coulomb's law describes the force of attraction between ions and/or dipoles.0188

Basically Coulomb's law tells us that this attraction is going to0194

be proportional to the following: q1q2 over r20200

where q1 and q2 are the relative charges of the cation and anion; charges of cation and anion.0210

r is just the distance between the two nuclei; ionic distance.0227

We see a very important proportionality.0237

That the force of attraction is going to be directly proportional to the magnitude of the charges.0242

It is going to be inversely proportional to 1 over r2 or inversely proportional to the atomic size.0254

Once again this is what we call Coulomb's law.0264

It tells us, it really quantifies the degree of attraction between charges and/or dipoles.0267

There is one important physical parameter that describes what we just illustrated.0280

That is an ion interacting with a solvent.0285

This is what we call the enthalpy of solvation which is also known as the enthalpy of hydration.0290

Basically what it is is the following.0299

It is another type of ΔH like we learned before.0302

It is also a ΔH of solvation.0307

It describes the energy change when an ion becomes solvated.0311

In other words, when an ion becomes surrounded by solvent molecules.0323

What we want to know, remember mother nature favors low energy.0336

The more negative this energy, the more likely an ion will be dissolved or solvated by water.0339

An ion is more likely to be solvated by water the larger it is and the more positive the charge.0346

Let's go ahead and move on then.0358

Ion-dipole was the first type of IMF for a polar compound.0364

Let's now move on to the second type.0369

The second type of IMF that can occur in liquids is what we call dipole-dipole forces.0372

Dipole-dipole forces occur between molecules that have a permanent dipole moment.0379

Then the strength of the dipole-dipole force tends to increase with overall polarity.0386

For example, if we take a carbon-hydrogen bond versus an OH bond,0391

the OH bond is the more polar of the two and therefore will have a stronger IMF.0397

How does this all relate to boiling point?0411

Basically the stronger the IMF, the more energy required to overcome the attraction.0415

When we think of energy, we can think of boiling point.0421

Boiling point is the temperature at which the boiling process is going to occur.0425

Remember temperature is the measure of average kinetic energy.0428

Really when we say the more energy required to overcome the attraction, we really mean a higher boiling point.0432

This was again dipole-dipole forces.0442

The next type of intermolecular force is what we call hydrogen bonding.0445

Hydrogen bonding you can think of as a very strong type of dipole-dipole.0450

This is going to occur in a molecule that has hydrogen that is directly bonded to nitrogen, oxygen, or fluorine.0454

You will recognize these three elements being the most electronegative.0462

Therefore their bond with hydrogen will be the most polar.0467

Hydrogen bonding is important of course biologically because we are going to find out that liquids that0474

have hydrogen bonding tend to be number one, nonvolatile, and number two, have relatively higher boiling points.0480

You can think of water as our typical example.0487

Hydrogen here, partial positive, partial positive, partial negative.0493

We can have another water molecule come into play just right there.0497

We can have an interaction between the partial negative oxygen with the partial positive oxygen.0502

You can imagine another water molecule just like that; we can have even more attraction.0508

What we see is basically that we get a network of hydrogen bonding; network of strong hydrogen bonding.0520

Why is this biologically important?0537

We know water because of its strong hydrogen bonding is going to be nonvolatile.0539

Imagine if all the world's oceans all of a sudden evaporated.0546

Life would be unbearable; our planet would be too hot.0551

Because of hydrogen bonding, water does not evaporate.0554

That allows us to have a good cooling effect by the ocean.0557

Again that is also going to lead to a high boiling point.0563

Once again it is in our best interest to have water to be a very stable compound.0568

Again it is due to hydrogen bonding.0575

What we are going to go into now, now that we are done with polar molecules,0579

let's get into the intermolecular forces for nonpolar molecules.0583

We are first going to look at the solubility of a nonpolar gas in a polar solvent.0588

Wait a second, I thought polar and nonpolar are not supposed to mix.0596

But it turns out that for gases, there can be a slight degree of mixing.0599

The typical example is going to be carbon dioxide in water which is basically your average carbonated beverage like soda.0605

We all know that CO2 is nonpolar.0614

Let's go ahead and look at its Lewis structure.0619

Each oxygen is partial negative; the carbon is partial positive.0623

What can happen is the following.0629

A water molecule can come into play right here just like that.0632

Partial positive, partial positive, and partial negative.0641

Even though carbon dioxide is nonpolar, at any one point in time, at any point in time,0646

an electron cloud can get slightly distorted; electron density can distort.0660

As soon as we get a distortion, that instant we form a temporary dipole moment.0672

It is that temporary dipole moment which can then interact with the polar water molecule.0685

Let's go ahead and take a look at the interaction between two nonpolar molecules.0698

The interaction between two nonpolar molecules is what we call a London dispersion force or a van der Waals force.0703

Another name for it is induced dipole attraction.0712

Let's go ahead and take a look at molecular fluorine as an example.0716

Here is our typical Lewis structure for a molecular fluorine.0722

Because it is homonuclear diatomic, it is nonpolar overall.0726

However, at any point in time, we can have distortion of the electron cloud once again; distortion of electron density.0730

You can imagine then that at any one point in time, if what I am drawing represents the F2 molecule,0746

we can get one side to be partial positive and we can get one side to be partial negative.0754

This is not going to last forever.0760

It is a temporary dipole moment just like it was for carbon dioxide.0761

Basically if I have this temporary dipole moment, another F2 molecule can come into play, can come right alongside it.0770

It too can also have a temporary dipole moment.0781

We can get an interaction between the partial positive end of one F20784

molecule with the partial negative end of the second F2 molecule.0789

Because of the temporary nature though of this interaction, we can also0794

infer that London dispersion forces are going to be incredibly weak.0799

In fact, dispersion forces are the weakest type of intermolecular force due to temporary nature of attraction.0806

This brings into play a very important concept.0829

You are going to see this concept in this chapter.0832

You are going to see this concept in organic chemistry quite a bit actually when you get to that level.0835

To what extent does an electron cloud or electron density get distorted?0841

The extents to which this happen is known as polarizability.0848

Basically the more polarizable the molecule, the easier it is to0854

have an induced dipole moment and therefore the stronger this attractive force.0860

Let's go ahead and look at the molecular halogens as our example.0866

F2, CL2, Br2, and I2.0873

Under standard conditions, fluorine is a gas; chlorine is a gas.0880

Bromine is a liquid; I2 is a solid.0885

Why is that though?--why are we going from gas to liquid to solid?0890

In other words, as we go down this column, why does the intermolecular forces increase?0894

Because that is what explains for going to a more condensed state from gas to liquid to solid.0906

Why does the IMF increase?--the answer is because of polarizability.0911

Iodine is the largest out of these molecules.0921

Because it is the largest, the electrons are not held as tightly.0927

Because the electrons are not held as tightly, we get easier distortion.0939

More likely for electron cloud distortion to occur.0945

Because it is more likely for the electron cloud distortion to occur, we can form stronger dispersion forces.0955

In other words, polarizability increases with molar mass.0976

Once again polarizability tends to increase with molar mass; that is it.0991

That pretty much explains why we change physical state as you go down the column of0997

the halogens from fluorine to iodine, from a gas all the way to a solid.1003

We finished our discussion on the attractive forces that can occur for liquids.1012

Let's go ahead and apply them now.1019

We are going to apply these to the physical properties of liquids.1021

There are several of them.1026

The first physical property that we are going to be talking about is what we call standard molar enthalpy of vaporization.1027

It is yet another energy; it is another type of enthalpy.1034

The definition of ΔH of vaporization is the energy required to vaporize one mole of a liquid under standard conditions.1038

But in order to vaporize a liquid, you have to overcome the intermolecular forces,1046

the IMF whether it be dipole-dipole, hydrogen bonding, or dispersion.1051

In other words, the stronger a liquid's IMF, the more energy required and the larger the ΔH of vaporization.1056

Let's go ahead and look at the trends and boiling points of representative liquids.1064

Here we will examine two inorganic liquids, water and hydrogen sulfide.1069

For water, let's look at the IMF; we have hydrogen bonding here.1075

For hydrogen sulfide, there is no hydrogen bonding; only dipole-dipole and van der Waals.1080

Because hydrogen bonding is going to be the strongest type of intermolecular force here,1089

we conclude that water has the stronger IMF and therefore the higher ΔH of vaporization.1094

That is how we apply intermolecular forces to solving these types of problems.1110

Let's go ahead and take a look at some representative organic liquids.1118

The first type of organic liquids I want to talk about are what we call aliphatic hydrocarbons.1122

Aliphatic hydrocarbons are basically straight-chain alkanes.1126

Here we have CH3CH2CH3 and CH3CH2CH2CH3.1131

First of all, if you recall the electronegativities of carbon and hydrogen are very similar.1137

In fact, they are nonpolar bonds; both of these molecules are nonpolar.1142

Because both of these are nonpolar, the only type of intermolecular force are van der Waals in each of them.1149

What is the only difference then?--the only difference is chain length or molar mass.1156

We see here that this has the longer chain which means higher molar mass.1162

Just like we said previously, as the molar mass increases, the polarizability1171

also tends to increase, therefore the strength of the intermolecular force.1176

Therefore this molecule here, CH3CH2CH2CH3, has the stronger IMF1180

because of intermolecular forces... excuse me... of the higher molar mass and stronger IMF.1186

The next two, CH3CH2OH versus CH3OCH3.1197

In this left molecule, we have hydrogen bonding and dipole-dipole.1202

In the right molecule, there is no hydrogen bonding; only dipole-dipole and van der Waals.1210

We are going to obviously pick the one with the hydrogen bonding.1221

That is going to give us the stronger IMF and therefore the higher ΔH of vaporization.1224

These are again just examples of what we call aliphatic hydrocarbons, straight-chain alkanes.1247

We now move on to branched out hydrocarbons.1252

For branched hydrocarbons, it is important to go ahead and draw the Lewis structures out.1257

This one we did already, CH3CH2CH2CH3.1263

Nothing is going on here; this is aliphatic.1268

But when we move to its counterpart on the right side, we are1271

going to get a new type of structure that looks like this.1275

This is what we call a branched alkane; this is no longer aliphatic.1283

What we have to take away from this is the following.1290

That any type of branching is going to lower the surface area of your compound.1292

If the surface area is lower, that means there is less area of the molecule to be exposed to an adjacent molecule.1304

In other words, the IMF is going to be lower also.1314

This molecule here that is branched has the weaker IMF and therefore the lower ΔH of vaporization.1320

Next another physical property is what we call vapor pressure.1331

Vapor pressure is going to be commonly measured in our traditional units1337

for a gas pressure which is atm, torr, and even kilopascal.1341

What vapor pressure is is the following.1347

It is going to be the pressure of a vapor that is directly above the surface of its liquid in dynamic equilibrium.1349

If you take a small beaker and you cover the beaker and you close it, let's say you have a liquid in here.1358

That liquid is going to go into a gas phase naturally.1368

But not all of it because as soon as the gas forms, it is going to go back and condense back down to a liquid.1375

This is in constant motion and not a static condition.1381

This is what we call dynamic equilibrium; it is a dynamic process.1387

Basically whenever we have this type of closed system, if we measure the pressure of1393

the vapor right above the liquid, that is what we call vapor pressure.1399

Now that we have a little drawing to help support the idea of vapor pressure, we can now put it into play.1408

Vapor pressure is actually going to parallel what you and I know as smell.1415

Basically a liquid with a strong odor, we are detecting its gas phase with our nose.1421

A liquid with strong odor means it vaporizes easily.1427

Its vapor pressure is going to be relatively high.1430

If you compare gasoline versus water, gasoline we know smells very strongly.1433

Water has no odor; we can conclude that gasoline has a higher vapor pressure.1445

The technical term for this with a higher vapor pressure is what we call volatile; volatile.1456

We say water is therefore is nonvolatile.1463

We do have an equation that allows us to quantify the relationship between vapor pressure and temperature.1473

This is what we call the Clausius-Clapeyron equation.1484

The Clausius-Clapeyron equation is basically the natural log of P2 over P1 which is equal to...1488

ΔH of vaporization over R times 1 over T1 minus 1 over T2.1507

All this is saying is the following.1515

If T2 is greater than T1, that automatically implies that P2 is greater than P1.1519

In other words, vapor pressure increases with temperature; P increases with temperature.1527

Just think about that.1535

A gas station is going to smell a lot worse on a warmer day.1537

It is because the vapor pressure of any gasoline that is dropped is going to be much higher than on a cooler day.1541

That is again the Clausius-Clapeyron equation.1550

The next physical property of liquids is what we call normal boiling point.1553

Normal boiling, we all know what boiling point is.1557

It is the temperature at which a liquid is going to boil.1560

But normal boiling point is specifically the process that occurs at atmospheric pressure.1562

Since boiling requires energy, a high boiling point implies that lots of energy is needed to boil a liquid.1569

A high boiling point means very strong IMF which also means1577

higher ΔH of vaporization which also means lower vapor pressure.1588

Let me go ahead and summarize that.1601

Strong IMF, high boiling point, high ΔH of vaporization, lower vapor pressure.1603

Simply knowing the intermolecular force, we can take an educated guess at a lot of the physical properties of a liquid.1618

Now onto the last physical properties we will cover.1630

This one is what we call surface tension.1634

Surface tension is something measurable.1636

It basically quantifies the tendency of a liquid to minimize its surface area.1639

To do this, it is going to form spherical drops.1644

A liquid with a relatively high surface tension easily forms spherical droplets1649

such as water which has very strong IMF due to hydrogen bonding.1655

Liquids with low surface tension, they tend not to form spherical droplets; nonspherical droplets.1663

Finally we reached the last physical parameter of liquids.1686

It is what we call viscosity; viscosity is also measurable.1693

It basically quantifies a liquid's resistance to flow.1698

What that really translates to is how thick a liquid is.1702

If you think of something like honey or like syrup, we say those are very viscous liquids.1706

They are very resistant to flow.1712

Basically if an IMF is very large, the viscosity tends to also be large.1714

That makes sense that the intermolecular force of attraction is large.1720

Liquid is not going to want to separate from each other.1724

It is going to want to stay intact and stay together.1726

In other words, it is going to be somewhat thick if you will or not as flow-like as water or something like that.1728

Again this is what we call viscosity.1741

We now reach our summary slide.1746

Basically in this chapter, we introduced the types of intermolecular forces that are relevant to liquids.1750

Number one is what we call ion-induced dipole; number two is what we call dispersion.1757

Number three is what we call dipole-dipole; number four is what we call hydrogen bonding.1762

Remember ion-induced dipole is going to be between a free ion and a polar molecule.1768

Dispersion is going to occur between nonpolar compounds.1780

Dipole-dipole and hydrogen bonding are going to be for polar compounds.1784

After we went over all of the different types of intermolecular forces, we then applied them1791

and used them to make educated guesses on how a molecule is going to behave physically.1796

We saw that the IMF can exert strong influence on a liquid's physical properties such as boiling point and vapor pressure.1803

In other words, by using IMFs, we can make a strong educated guess on the relative volatility of a liquid.1813

For sample problem one, let's go ahead and determine which of the following liquids will have the lower vapor pressure.1823

As soon as I see this molecule here which is OH, I see hydrogen bonding immediately.1830

That tells me this molecule is going to have relatively strong intermolecular forces.1836

Here CH3OCH3, that is going to look a lot like this.1841

I don't have any hydrogen bonding at all.1847

But I do have dipole-dipole and of course van der Waals or dispersion.1849

For here, CH3CH2CH3, there is none of that.1858

There is no dipole-dipole; there is no hydrogen bonding.1864

This molecule is completely nonpolar.1867

I am only left with van der Waals or dispersion.1871

What low vapor pressure means is that we are going to have strong intermolecular forces.1878

Therefore the molecule that is expected to have the lower vapor pressure is going to have the strongest intermolecular forces.1886

That is the CH3CH2OH right there.1893

Finally sample problem two, we are going to take the same compounds.1899

But now we are going to determine which of the following will have the largest ΔH of vaporization.1903

As soon as I see largest ΔH of vaporization, that means a lot of1912

energy is required to vaporize one mole of a liquid under standard conditions.1916

That tells me I am dealing with a liquid with relatively strong IMF.1920

Using the same rationale, it is going to be the CH3CH2OH again expected1927

to have the largest ΔH of vaporization because this has hydrogen bonding.1931

This one in the middle has only really dipole-dipole and dispersion.1938

This one here has only dispersion.1946

What I encourage you to do is when you tackle these types of problems,1951

you should always try to from the formula get the structure, the Lewis structure that is,1955

because that will really help you see any of types of bonds.1963

Step two is then go ahead and identify the types of IMF present.1966

Hopefully that is enough to distinguish the liquids from each other.1979

But if IMF identical, then what you need to go by is really1984

the size of the molecule or its molar mass; go by molar mass.1993

Again what was the word that describes the effect of molar mass and size on distortion of electron cloud?1999

This is all related to polarizability; good.2006

That was our general chemistry lecture on intermolecular forces and liquids.2015

I want to thank you guys for your attention.2021

I will see you next time on Educator.com.2024

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is going to be on the chemistry of solids.0002

We are going to go ahead and start off with a very brief introduction followed by the basic structure of solids.0008

We will get into a few thermodynamic properties, namely the lattice enthalpy,0015

followed by different types of crystalline solids that we can encounter.0022

Then we will something characteristic of solids, liquids, and gases--that is phase changes.0027

Phase changes are going to then lead us into namely phase diagrams.0035

We will finish up the lesson talking about our summary followed by a few sample problems.0039

We have briefly alluded to what we meant by a solid a long0050

time ago when we first did our first one or two lessons.0054

If you recall, solids have the following characteristics; that they are relatively dense.0060

They have a definite shape and a volume; they have very strong attractive forces.0065

If we look at it on a particulate level, we can imagine solids0070

as being atoms or spheres that are closely spaced together like so.0075

The separation that we have in gases and liquids, we basically do not have in solids.0085

That is due to the very strong attractive forces.0090

Now that we have talked about and we reviewed what we0096

know about solids, let's get into how solids look three dimensionally.0101

The basic part of a solid is what we call the unit cell.0107

The unit cell is basically a pattern of a solid that is repeatable in all directions, in all three directions.0113

As you can see from this representative diagram, the unit cell is the basic cube here.0132

You see that this cube is repeatable in all directions.0139

That is what we call the unit cell.0144

A lattice point is basically a fixed location of an atom that composes a solid; a fixed location of an atom.0146

You see that because it is part of a unit cell, this fixed location is reproducible also in all three directions.0162

We are going to learn that there are three basic types of unit cells.0171

The first is what we call primitive cubic.0175

We are going to find that primitive cubic is actually very rare.0178

There is only really a handful of elements, really single-digit number of elements that have this structure.0181

But primitive is your basic cube like so; it is if you will completely empty.0187

Once again primitive cubic is right here.0199

You basically have lattice points at all corners of the unit cell.0205

Each corner actually contributes only one-eighth of the sphere.0220

Because there are eight corners here, we say that we get one net atom per unit cell in primitive cubic structure.0227

Let's now jump into body centered cubic; body centered cubic is what we abbreviate bcc.0238

Basically for body centered cubic, we take that primitive cubic like so.0245

Let me go ahead and redraw it here.0251

We are going to take that primitive cubic.0253

But in addition to having the spheres at the corners, we also have0257

one sphere that is right in the middle of this cube itself.0268

Not only do we have one-eighth by eight, we also have the entirety of the0274

one sphere in the middle giving us two net atoms per unit cell for bbc.0281

Bcc solids have the characteristic of being brittle.0293

The final unit cell is what we call fcc.0304

For fcc, I am going to go ahead and redraw that primitive cubic again.0309

But for fcc which is called face centered cubic, we are going to0323

have a sphere at the center of each face of this unit cell.0326

There is going to be one right here.0330

There is going to be one right here sticking inward.0333

There is going to be one right here sticking inward.0336

One right here sticking inward; one on the bottom sticking inward.0338

Of course one on the back sphere sticking inward.0342

We have one-eighth by eight from the corners.0347

It turns out that each of the spheres in the center of each face contribute one-half.0351

There is going to be six sides, six faces.0357

We get four net atoms per unit cell.0362

It turns out that what is characteristic of fcc solids is that these solids tend to be the opposite of bcc.0370

They tend to be malleable.0379

Now that we have our brief introduction into crystal lattices, let's go on to a thermodynamic approach.0384

This is what we call lattice enthalpy.0391

Lattice enthalpy is the energy that is released.0394

Again this is released when a solid crystal forms from its constituent0398

elements in the gas phase represented by the following chemical equation.0403

If we take a crystalline solid maybe of the form MX,0407

it is formed from M+ gas and X- gas.0417

The correct equation will be M+ gas plus X- gas goes on to form MX solid.0423

We are going to see that this is what we call the lattice enthalpy0433

which is therefore ΔHlat--is how some textbooks abbreviate it.0436

Basically because mother nature favors low energy, the lower or the more negative0442

the lattice enthalpy, the more favorable the formation of this crystalline solid.0447

Let's examine the equation to calculate its lattice enthalpy.0451

You are going to find that the lattice enthalpy is basically going to be analogous to Coulomb's law.0455

This is going to be analogous to q1q2 over r where q1 and0463

q2 are the charges of the cation and anion respectively; cation/anion charges.0471

r is going to be the internuclear distance between cation and anion adjacent; internuclear distance.0482

We see that as the magnitude of the charges increase...0495

Once again as the magnitude of q1 and q2 goes up,0505

ΔH is going to become more negative which is favorable.0511

That makes sense.0516

As the charges goes up, there is a stronger attractive force between opposite charges.0516

It should be more favorable that the solid is going to form.0522

We also see that as r goes up, ΔH is actually going to become less and less negative.0526

It is actually going to go up.0535

Basically this is saying that the farther apart ions are,0537

the weaker the attractive force between the two nuclei which makes sense also.0541

Basically ΔH negative is going to be given by the following0547

combination of parameters--smaller ions and also ions with greater charge.0556

Smaller ions, greater charge tend to be indicative of a solid that is going to form in the gas phase.0567

We also now are going to focus on crystalline solids.0578

Crystalline solids are those that have these different unit cells such as bcc and fcc.0582

There are different types of crystalline solids; number one is molecular solids.0589

Molecular solids consist of molecules held together by relatively weak0594

intermolecular forces that influence the physical properties, especially melting point.0598

Common examples include ice and dry ice.0603

When you think of ice and dry ice, these don't have really high melting points.0606

Ice melts relatively at a pretty low temperature; dry ice readily sublimes.0611

Molecular solids tend to have relatively low melting and boiling points.0619

Like ice and dry ice, they tend to be brittle.0623

Network solids are essentially the opposite.0627

They consist of molecules that are held together by very strong covalent forces and covalent bonds.0630

They form a framework extending throughout the crystal.0637

Common examples include diamond and graphite.0639

Make sure you guys understand the importance of diamond is that it is the hardest known naturally occurring substance.0644

Network solids tend to be hard and rigid with very very high melting points in the thousands of degrees Celsius.0653

Now that we have talked about the different types of crystalline solids, let's go into phase changes that involve solids.0663

Melting, melting is basically the process in which the intermolecular forces that hold a solid together are overcome and0669

broken, resulting in greater separation between the atoms themselves such that we turn the solid into a liquid.0676

The enthalpy is typically abbreviated as ΔH of fusion under standard conditions.0684

This is going to be an endothermic process because this requires energy.0693

This tends to be positive.0698

Because these are all enthalpies, these are typically in units of kilojoules per mole.0703

Freezing, freezing is basically the exact opposite.0710

It is such that we lower the temperature of our liquid such that atoms can come into contact with each other.0713

There is less motion of the atoms.0719

When they touch each other, they don't have as much time to escape even partially from neighboring atoms.0722

We go into the solidification process.0729

Basically ΔH of freezing is going to be the same magnitude but opposite sign of ΔH of fusion.0732

We see that ΔH of freezing is going to be negative.0744

ΔH of freezing is an exothermic process whereas ΔH of fusion is an endothermic.0750

That is melting and freezing.0758

Let's now go on to the next two; this is sublimation.0760

Sublimation is the process in which a solid coverts directly to a vapor0763

from the solid state at a cold enough temperature and low enough pressure.0769

Because this is going solid to a gas directly, this is going to be an endothermic process0774

because we have to require energy to overcome the very strong attractive forces in the solid.0782

ΔH of sublimation is going to be a negative value.0787

Finally deposition is the process when a vapor converts directly to a thin solid layer.0794

The thermodynamic value is going to be ΔH of deposition.0801

This is going to be equal to negative ΔH of sublimation.0806

We see that deposition is going to be an exothermic process.0811

That is sublimation and deposition.0817

Now that we have talked about the different phase changes that can occur and we have really0821

completed our discussion on all three of the physical state solids, liquids, and gases, let's culminate everything.0826

When we try to culminate everything, what we get is a phase diagram.0833

We can graphically predict when a substance will be a solid, liquid, or gas at any given temperature and pressure combination.0837

When we go ahead and graph this for a substance, when we graph pressure0845

as a function of temperature, we get what is called a phase diagram.0848

Let's take a look at the diagram for water; there is several points of interest.0853

Number one, the melting point; this is called the normal melting point.0857

Normal melting point, anytime you see the word normal, that is always at 1 atm, at standard pressure.0862

We see that at a pressure of 1 atm, we can go ahead and translate it over.0871

When the solid becomes a liquid, that is what we call melting.0877

The melting point is the 0 degrees Celsius that we know for water.0883

The normal boiling point, we just translate along.0888

You see that the liquid to gas transition is 100 degrees Celsius.0893

That is where water boils at standard pressure.0899

The sp, sp is what stands for sublimation point.0904

The sublimation point as you can see for ice is right here.0911

As you can see, this is going to be at a very cold temperature and a very very low pressure0919

which is why we don't see ice readily sublime under normal ambient conditions; sublimation.0922

There is another point of interest here that we want to talk about--that is point C.0932

Point C is the absolute highest temperature where a liquid can exist.0936

This is what we call the super critical point.0944

At the super critical point or the critical point, you cease becoming a liquid and you have nothing but gas.0950

We say that gas overtakes a liquid state.0957

Finally there is a really nice point of interest right here which is0965

what we call the intersection of the solid, liquid, and gas phase.0968

That is called the triple point.0973

Only at this one unique set of pressure and temperature is where we get coexistence of all three physical states.0975

Once again that is what we call the triple point.0984

That is what we call a phase diagram; that is for water.0988

Let's look at another phase diagram, this time with dry ice.0992

Dry ice we know readily sublimes.0995

Anytime you go to see maybe a theatrical production, you sometimes see a fog being generated.0997

This is basically one source of the fog generated is going to be dry ice being blown out to the stage.1003

Dry ice we expect to readily sublime.1009

Let's see if we can detect that in the phase diagram.1012

At 1 atm which is what we are typically at, yes we can go ahead and translate over.1016

Boom, we hit point A.1023

At point A, represents the normal sublimation point for dry ice.1025

You see that when you go down, the temperature is only 197.5 kelvin.1032

That is definitely less than room temperature.1039

Dry ice readily sublimes; that is why it produces a high vapor pressure.1043

Just by looking at a phase diagram, you can easily see that dry ice readily sublimes.1050

Also look how large the sublimation area is; compare that to that of water.1056

The sublimation interface between the solid and gas is quite prominent in the phase diagram compared to that of water.1062

Sublimation, a solid to gas interface curve very prominent in dry ice.1073

That is the phase diagram of a different compound versus water.1092

Now let's go ahead and summarize our brief session on the chemistry of solids.1096

The common unit cells for solids are primitive cubic, body centered cubic, and face centered cubic.1102

The unit cell structure we see can influence a solid's density and texture.1107

We saw that lattice energy quantifies the extent or tendency for a molecular1112

or network solid to form from its ions in the gas phase.1121

Finally we see that phase diagrams plot pressure versus temperature.1125

They allow us to predict how a compound is expected to exist,1130

as a solid, a liquid, or as a gas at any set of these conditions.1136

Let's go ahead and look at sample problem one.1142

Of the group 1 fluorides, which should have the highest lattice enthalpy?1144

When we mean highest lattice enthalpy, we really mean most negative or the most exothermic.1149

Of the group 1 fluorides, LiF, NaF, KF, etc; the only difference is atomic size.1160

Lithium is going to be the smallest cation here.1172

The smallest cation results in a smaller r value which results in a more negative ΔH.1178

The lithium fluoride is definitely your answer for that one.1191

Of the lithium halides, which should have the lowest lattice enthalpy here?1195

Lowest lattice enthalpy is going to be...1199

Sorry about that, this should be the most negative.1206

Let me go ahead and rephrase that; most negative lattice enthalpy.1209

This should also be the most negative; very sorry about that.1213

Lithium halides, LiF, LiCl, LiBr, and LiI.1220

Once again the only difference is the atomic size.1227

Fluorine is going to have the smallest r value.1231

The smallest r value is going to give you the more negative ΔH.1236

This is a nice sample problem which illustrates the periodic trends and its effect on the value of ΔH.1244

Sample problem two, how many joules of energy is required to melt 546 milligrams of ice at standard pressure?1256

You are given a ΔH of fusion for ice to be 6.01 kilojoules per mole.1263

Basically we are going to use kJ per mole as a conversion factor.1268

Basically we want to go from milligrams of ice to moles of ice and1279

then from moles of ice to what the question is asking for--to joules.1287

As you can see, we can use kilojoules per mole as a nice convenient conversion factor for the moles to cancel.1292

Let's go ahead and do this.1301

546 milligrams of ice times 10-3 grams divided by 1 milligram times1303

1 mole of the ice which is water divided by its molar mass 18.016 grams.1315

Then times 6.01 kilojoules divided by 1 mole.1323

Finally we have to go from kJ to regular joules.1329

We are going to take all of that.1333

We are going to multiply it by 103 joules over 1 kilojoule.1334

That is going to give us our answer in units of joules.1338

There is nothing really special about this problem.1342

It is basically just using the dimensional analysis with ΔH of the phase change as a conversion factor.1344

Dimensional analysis is something we have done so many times.1353

Again this is just nothing new.1357

We are just introducing a new type of conversion factor.1358

Finally onto sample problem three which involves a phase diagram.1362

Below is the phase diagram for helium; we know helium is a noble gas.1367

At standard temperature and pressure, we expect a gas to exist.1376

We can also spot that; it makes sense of that.1381

What is the normal bp for helium?--let's locate 1 atm or 1 bar.1384

The boiling point is when a liquid goes into the gas.1391

When we translate over, here is our transition between liquid to gas right there.1393

When we translate down, our normal boiling point is right there which is 4.22 in this case kelvin.1401

The next question is can solid helium sublime?1411

What we want to look for is the interface between a solid and a gas.1416

You see that it is right there going from a solid to a gas.1422

Absolutely liquid helium can sublime.1426

However it is going to be not readily occurring because we as you can see from1429

this curve, it only occurs at super low pressures, at only very low pressures.1435

What is the maximum pressure at which solid helium can exist?1447

Let's go ahead and look at the solid curve.1452

The solid curve here is strictly ending right there.1457

The highest pressure at which solid helium can exist is only at 100 bar.1463

Anything after that, we are going to become a liquid.1471

Finally at what set of conditions do all three states coexist?1476

Again that is what we call the triple point.1481

At the triple point which is right there, we are going to be1484

at a pressure maybe approximately 0.05 bar and a temperature of 2.17 kelvin.1488

That is a nice problem that illustrates a phase diagram.1501

I want to thank you all for your attention.1507

I will see you next time on Educator.com.1510

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is concerning solutions and their behavior.0003

Here is the unit lesson overview.0012

We are first going to do the introduction into the different units of solution concentration,0015

basically what we call molarity, molality, weight percent, and parts per million.0023

After we get into the different concentrations of units of solutions, I then want to go0032

into what you may have learned in high school chemistry which the rule, like dissolves like.0038

We are going to see how solutions can interact with each other.0045

The third one is then solubility.0050

What is the factors that influence how well a solid or a gas dissolves into a solution?0054

We are going to learn that there is really two factors.0063

The first one is pressure; the second one is temperature.0066

After we talk about the factors that affect solubility, we will then go into what we call colligative properties.0071

The colligative properties that we are going to be discussing are more or less four of them.0079

The first one is vapor pressure.0085

The second and third are together, boiling point and freezing point.0087

Last but not least is osmotic pressure.0092

As always we will get into summary slide and sample problems.0096

Let's go ahead and now jump into the units of concentration.0105

Whenever you buy a beverage or something like that from the supermarket,0109

they usually give you maybe like the percentage of fruit juice suppose in your beverage.0114

That percentage is a unit of concentration.0122

It tells you how much of that ingredient is in the entire beverage.0124

Of course for chemistry, we also need a way of defining a solution concentration.0131

The most common solution concentration unit that we see in the chemistry laboratory is what we call molarity.0138

Molarity is formally defined as the following.0147

This is going to be equal to moles of solute for every liter of solution.0150

Again moles of solute for every liter of solution.0165

Molarity is typically symbolized with a capital M.0168

There is a couple of ways we can actually verbalize this too.0174

Let's say we had 0.67M NaCl.0177

If we were to translate this and verbalize it, we would say 0.67 molar NaCl solution.0184

Again this is the language that we use.0197

Anytime you see capital M, it is molar.0200

Again this is the most common unit of concentration we see in the chemistry lab for solutions.0204

The second one is related to molarity; it is just a different variation.0210

This is called molality; molality is going to be equal to the following.0215

This is going to be equal to moles of solute for every not liter of solution but kilogram of solution.0224

Moles of solute for every kilogram of solution.0238

Typically molality is symbolized with lowercase m.0241

0.67M NaCl, we would verbalize this as 0.67 molal NaCl solution.0246

Molarity followed by molality.0263

The next two we don't see too much.0267

But you do see this third one called weight percent.0270

You do see it quite often commercially.0274

For example, that beverage you buy in the supermarket.0279

What weight percent is is the following.0281

This is going to be equal to the mass of solute divided by the mass of solution times 100.0285

Weight percent is commonly abbreviated m/m percent; that is what we commonly see.0301

Again this is what we call weight percent, mass of solute divided by the mass of solution.0310

Finally the fourth and final concentration is ppm; ppm stands for parts per million.0315

Really we reserve the use of ppm for solutions that are very dilute or0327

the concentration of solute is incredibly small; used for very dilute solutions.0333

What parts per million is is basically the following.0347

This is going to be equal to milligrams of solute for every liter of solution0350

which is also equivalent to milligrams of solute divided by kilograms of solution.0360

You can see why this is reserved for very small quantities.0370

You see that the mass that we use for the solute is intentionally fixed to milligrams.0373

These are the four concentration units to express a concentration for a solution.0380

I now want to go into how solutions can interact with each other.0390

The way solutions can interact with each other is an old rule of thumb we have heard before.0395

It is called like dissolves like.0402

This refers to something that we have discussed previously; this refers to polarity.0404

Basically polar solutions will mix and be miscible.0416

Nonpolar solutions will also mix and be miscible.0436

However polar and nonpolar do not mix.0446

You can see the importance of when we talked about VSEPR theory and using VSEPR to determine molecular polarity.0459

You can see why this would come in handy.0466

For example, an alcohol and H2O; both of these are polar.0470

These two will mix and be miscible with each other.0491

When you go to the supermarket and you purchase rubbing alcohol, rubbing alcohol is typically 70 percent.0496

It will say that on the bottle.0507

What that means is that it is 70 percent alcohol and the rest is just water, 30 percent.0508

Again that is going to be miscible.0513

The next one that we can see is the following.0521

We can take water and fat; water we know is polar already.0526

But the fats and oils are going to be nonpolar.0539

We know that when we combine water and oil, we get a heterogeneous layer, a heterogenous mixture.0550

The two do not mix; immiscible.0555

Again anytime you are predicting if two solutions are going to mix, just use the rule of thumb0562

like dissolves like to determine if it is going to be homogeneous or heterogeneous.0569

Once two solutions mix, the next question is how well do they mix?0579

In other words, what are the factors affecting solubility--how well two components combine together?0586

The first factor is the effect of pressure; this is what we call Henry's law.0608

Henry's law tells us that solubility is going to be directly proportional to the following--something we call Henry's constant and P.0618

Capital P is going to be the vapor pressure of the solute.0637

Let's imagine the following; let's say we had a can of soda.0653

In this can of soda was H2O and CO2 gas.0660

Let's say that this was open versus the same can of soda closed.0667

Now the question is in which of these cans is the CO2 going to be more in the water?0682

The answer of course is going to be where the can is closed.0690

When the can is closed, the vapor pressure of CO2 is greater than when0693

the can is open because in the can that is open, CO2 can escape.0701

Basically we see the relationship that solubility of the CO2 is directly proportional to its pressure.0709

Henry's constant helps make this proportionality into the equation.0720

Henry's constant is going to be unique for each substance.0725

Once again Henry's law tells us that solubility is directly proportional to vapor pressure.0737

That is the first factor.0747

The second factor is going to be the effect of temperature.0749

The effect of temperature is different on a gas versus a solid.0753

Let's go ahead and look at that can of soda again.0759

Let's have both of these cans open.0765

Once again we have H2O, CO2, H2O, and CO2.0771

Let's make the left can hot; let's make the right can cold.0778

Now the question is the following.0786

In which cans of soda is the CO2 going to escape more?0788

In other words, is a hot can or cold can of soda going to fizz more?0795

It turns out that the warmer the can, the more fizz you get.0801

But what is this is fizz that you and I hear?0815

This fizz that we hear is CO2 gas escaping which means that the CO2 gas is not as soluble in the water.0819

It is less soluble; it is escaping instead.0834

The effect of temperature on gas solubility is an inverse relationship.0838

That is as the temperature increases, the solubility of a gas goes down.0845

That is as the temperature increases, the solubility of a gas goes down.0854

Less of it stays in solution.0859

More of it escapes which explains why a warmer can of soda fizzes more than a colder can.0861

The effect of temperature on solid solubility is now we are going to find is going to be opposite; opposite to gas solubility.0872

This one we know; it is a little more intuitive.0885

Let's say we are brewing a tea bag.0890

Which of the situations are you going to get the tea to brew faster, in cold water or hot water?0894

It is in hot water; you see that happen before your eyes.0899

That is because as temperature goes up, solids dissolve more easily.0902

Just think about it.0914

When are we able to clean better, our dishes that is, in cold water or hot water?0917

It is hot water because all the fats and oils are going to dissolve more easily in the water and in the soap.0921

As temperature goes up, solids dissolve much more easily.0928

The effect of temperature on solid solubility is completely opposite to that of gas solubility.0936

Again the examples that we did was a tea bag brewing faster in warmer water.0946

Another sign from everyday life is the appearance of your tap water.0962

Doesn't tap water appear cloudy when it is warmer?0967

The answer is yes because there is more dissolved ions and minerals coming out.0971

Warm tap water is not as clear.0977

It is cloudier due to more due to more dissolved ions and minerals that are present and abundant in tap water.0984

Again these are the factors that affect solubility; it is pressure and temperature.1002

We now move on to what are known as colligative properties.1011

Colligative properties of a solution are defined as the following.1016

These are properties that depend on the relative quantity of solute particles and not on the chemical identity per say.1020

We are interested in solute amount; again that is going to be relative to solution.1029

There is a law that helps to quantify the relationship; this is called Raoult's law.1045

Raoult's law is P is equal to x times P0 where P is the vapor pressure of the solute in solution.1052

P0 is the vapor pressure of the pure solute just by itself.1074

x is going to be the mole fraction of solute.1088

As you can see, as x increases, that is as solute amount goes up, so does the vapor pressure.1100

That makes sense; you toss more of a solute in.1123

It is going to have just a higher vapor pressure right above the surface of the liquid.1126

What is x again?--x is going to be the mole fraction; net mole fraction of solute.1136

This is going to be the moles of solute divided the moles of solute plus moles of solvent1145

which is also can be rewritten as moles of solute divided by the moles of total solution.1159

Of course the mole fraction of solute is going to be less than or equal to 1.1171

Once again this is Raoult's law.1179

Basically the vapor pressure of the solute is directly proportional to what the quantity is in solution.1183

The next two colligative properties also depend on amount.1195

This is a boiling point elevation and freezing point depression.1200

Basically as solute is added to a solution, the freezing point of the solution decreases.1206

That is what we call a depression.1227

The boiling point of solution is going to go up.1230

That is what we call elevation.1236

If we were to look at a phase diagram of pressure versus temperature, we are going to get something like this.1239

For example, let's go ahead and do this for water.1250

For water, this is 0 degrees Celsius; this is 100 degrees Celsius.1256

Let me go ahead and do a blue line.1262

The blue line is now going to represent the water this time with salt added.1264

This is with NaCl.1271

What happens is as you can see the freezing point now decreases.1274

The boiling point now increases.1280

This is bpnew which is greater than 100 degrees Celsius.1286

This is freezing pointnew; that is going to be less than 0 degrees Celsius.1291

Freezing point depression, boiling point elevation; let's go ahead and explain why.1298

Basically this is telling me the following; that more energy needed for vaporization to occur.1305

As I toss sodium chloride in the water, we are introducing more attractions, more attractive forces.1322

This is really due to more attractive forces between solute and solvent.1330

If all of a sudden the attractive forces shot up, we have to supply more energy in the1343

form of heat to overcome those attractive forces to induce vaporization, hence a higher boiling point.1348

For the freezing point, we actually need a cooler temperature.1362

The colder temperature needed because what happens is the following.1368

When we reach a colder temperature and try to induce freezing, like molecules tend to interact with each other.1377

What I mean by that is the following.1398

That water is going to want to interact with itself.1399

NaCl is going to want to interact with itself.1406

In order for the solute and solvent to separate out like this, we need to1411

remove the thermal energy so that interactions like that will be minimized.1417

A cooler temperature helps to minimize what we call unwanted interactions.1424

Water sticks with water; sodium chloride sticks with sodium chloride.1440

Freezing process can occur where we eventually result in solidification.1444

Cooler temperature minimizes unwanted interactions so that solidification can occur.1451

Again this is boiling point elevation and freezing point depression.1469

We actually have two equations that can help us through this.1475

They are the following.1480

The change in the freezing point is going to be equal to some constant K times lowercase m times i.1483

I am going to call this KF.1496

The change in boiling point is equal to some constant KB times m times i.1499

What these are is the following.1507

i is what we call the van't hoff factor.1510

This is going to be proportional to number of ions from solute.1518

m is just the molality.1529

Finally K's are just going to be constants unique to the solute.1535

But what you see is that the temperatures are directly related to the amounts.1542

Temperature is directly related to the amount.1551

The more solute you have in solution, the greater the change.1555

The more solute you have, the lower the freezing point.1560

The more solute you have, the higher the boiling point, the greater the ΔT.1563

Once again these are called boiling point elevation and freezing point depression.1569

The final colligative property, that is the final property that is dependent on relative solute amount is what we call osmotic pressure.1575

Before we get into osmotic pressure, let's first go ahead and define osmosis.1588

Osmosis is the flow of solvent through a semipermeable membrane into a more concentrated solution.1593

In other words, solvent naturally flows from a dilute area to a more concentrated area.1599

The typical diagram to illustrate this is the following.1632

Let's say we had a container; in the container is this barrier.1637

This barrier is what we call a semipermeable membrane; what this is is the following.1642

It allows only certain sized particles through.1652

Once again it allows only certain sized particles through.1663

In this case, it is going to be solvent molecules only because we are going to assume that a solvent1667

like water is going to be relatively smaller to a much bigger and heavier solute compound such as sodium chloride.1673

Let's say initially that these two regions which are separated by a semipermeable membrane have equal levels of liquid.1685

However let's say that one side was just H2O.1698

On the other side, we had Na+, Cl-, and H2O.1702

That means this side represents my concentrated side.1709

This side represents my dilute side.1717

If we allow this to proceed, after some time, we are actually going to get a change in water levels because of osmosis.1721

Solvent is going to flow from the dilute area to the more concentrated area.1734

In other words, I am going to get solvent going this way.1739

What that results in is my dilute side is going to drop in volume.1742

My concentrated side is going to increase in volume because of the presence of more solvent molecules.1752

This is what is known as osmosis.1762

The colligative property is what we call π; this is osmotic pressure.1767

Pretty much what osmotic pressure is, it is the pressure needed to be1776

applied to prevent the flow of osmosis; to prevent osmosis from occurring.1785

π is equal to i times M times R times T where i is once again that van't hoff factor that we talked about.1801

Again that is going to be proportional to the number of solute ions.1815

Big M is the molarity.1823

R is going to be our universal gas constant, 0.08206 liters atmosphere K mole.1827

Finally temperature is going to be the kelvin temperature.1836

But as you can see, that π is directly proportional to M and to i.1841

Basically if you have an area of great concentration where you have a1851

lot of solute, osmosis is going to happen very easily and very readily.1857

I am going to need much more pressure to stop that process from occurring.1862

Again this is called osmosis.1868

Let's go ahead and summarize our presentation on solutions and their behavior.1873

Solution concentration we found is often expressed in four ways--molarity, molality, weight percent, and as ppm.1882

Again weight percent is really what we see commercially.1891

Ppm is really for dilute solutions, very very dilute solutions.1897

We saw the rule that like dissolves like.1903

In other words, polar solutions are going to be miscible with other polar solutions.1908

Nonpolar solutions are going to be to miscible with other nonpolar solutions.1913

Simply put, polar and nonpolar do not mix, forming heterogeneous mixtures.1918

What we saw was the traditional portrayal of fat oil plus water giving us a heterogeneous mixture.1923

We also discussed factors that influence the solubility.1932

The solubilities of gases and solids in a solution we found are influenced by pressure and by temperature.1936

Remember that for temperature, it is going to be opposite for solids and gases.1943

It is an opposite effect.1952

Finally we also discussed colligative properties.1956

We saw that colligative properties are not really dependent on the identity of the compound itself but really just the relative solute amount.1958

Now that that is our summary for solution behavior, let's get into a pair of sample problems.1970

Calculate the vapor pressure of water at 20 degrees Celsius in a solution1978

prepared by dissolving ten grams of sucrose in 100 grams of water.1981

You are told that the vapor pressure of pure water at this temperature is 17.5 torr.1986

Let's go ahead and write out the equation.1992

This is pressure is equal to mole fraction times P0.1994

This is going to be the vapor pressure of solution.1999

This is the mole fraction of solute.2004

This is the vapor pressure of the pure solvent.2014

The vapor pressure of the pure solvent is given to us to be 17.54 torr.2025

x is the mole fraction... I am sorry... not of the solute.2033

But it is going to be the mole fraction of the solvent.2038

What we see is the following.2042

That as the mole fraction of solvent, as x goes down, the vapor pressure also drops.2043

As we toss more and more sucrose into this water, the water is going to be less and less volatile.2055

Its vapor pressure is going to drop.2062

All we have to do is plug it in, fill in the equation.2065

What we need therefore is the mole fraction of sucrose and the mole fraction of water.2069

x of sucrose is going to be equal to the following.2075

The moles of sucrose divided by the moles of sucrose plus moles of H2O.2082

Let's get the moles of sucrose.2097

That is going to be 10 grams of sucrose times 1 mole divided by the molar mass of sucrose.2098

Sucrose is given to us right here, C12H22O11.2106

That is just going to be molar mass of sucrose.2113

That is going to be divided by the total moles.2118

That is going to be divided by 10.0 grams over molar mass of sucrose plus the moles of water2121

which is going to be 100.0 grams of water divided by the molar mass of water, 18.016 grams per mole.2131

That is going to give us the mole fraction of sucrose.2141

To get the mole fraction of the water therefore, every time I have a fractional counterpart, the sum has to equal to 1.2146

The mole fraction of water is just 1 minus the mole fraction of sucrose.2156

That is it.2162

All I do, I plug that directly into P is equal to xP0 to get the vapor pressure of the water.2163

As you can see, it is going to drop because of the added sucrose.2173

P is less than P0.2179

That is going to be sample problem one, vapor pressure lowering.2182

Let's now move on to the final sample problem.2189

What is the molality of C6H12O6 in the solution prepared by dissolving 90.5 grams in 250 grams of water?2192

Remember molality is lowercase m.2201

That is equal to the moles of solute divided by kilograms of the solution.2203

This is going to be equal to the solute is the C6H12O6.2212

The water is the solvent.2219

The moles of solute, let's go ahead and get that.2223

That is 90.5 grams of C6H12O6 times 1 mole divided by its molar mass which is approximately 180 grams.2226

I am going to take that. I am going to divide it by the kilograms of solvent.2238

The total amount of solvent we have is 90.5 plus the 250.0 grams.2242

I want kg; I want kilograms.2251

I am going to take this; I am just going to multiply by 10-3.2252

When all is said and done, we get our answer in blank molal of C6H12O6.2259

That was sample problem two and simply using the molality equation to calculate the concentration.2270

This concludes our lecture and presentation on solution behavior.2278

I will see you next time on Educator.com.2283

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is going to be on chemical kinetics.0002

Here is our brief overview of the lesson today.0008

As always we will go ahead and start off with our introduction.0010

Then we are going to get into what is really at the heart of chemical kinetics.0013

That is the rate of chemical reactions; that is how fast does a reaction occur?0017

We are then going to see what are the mathematical equations which actually attempt to quantify the rate of an equation.0021

That is what we call a rate law.0029

Our objective is to then try to derive these equations for any given reaction.0031

There is two sets of experiments that you can do to perform this.0037

Number one is what we call the method of initial rates.0040

Number two is what we call the integrated rate laws.0043

We will then jump into how a reaction proceeds, the step by step process which is0045

called a reaction mechanism, followed by the effect that temperature has on reaction rates.0051

We will go ahead and wrap up the lesson with a brief overview of0059

catalysis followed by our summary and a pair of sample problems.0062

Chemical kinetics basically is the area of chemistry that examines the factors which can influence the rate of a reaction.0070

What do we mean by factors that can influence the rate?0081

There are several--temperature, pressure, the reactant concentration, the addition of a catalyst, and mechanical force.0084

We are going to see that basically as temperature goes up, the rate goes up.0095

We are going to see that in general as pressure goes up, the rate goes up.0104

We are going to see that as the concentration of reactant goes up, the rate goes up.0112

When we add a catalyst, we are going to define what that exactly means.0120

But this also increases the rate.0123

Finally mechanical force, this is what you probably do in lab.0126

All we mean by mechanical force is something as simple as stirring or shaking.0129

Of course this will increase the rate of the reaction.0137

This is what we mean by factors which can influence the rate of a reaction.0142

Let's take a look though how a typical reaction progresses through time.0150

Consider the reaction A plus 2B goes to C.0154

This is telling me that for every one of A, two Bs are required.0158

One C is going to be produced.0165

Remember we are only looking here at the forward direction only.0167

When we go in the forward direction, we say that reactants are consumed which means concentration of reactant go down.0176

The products are made which means that the concentration of product goes up.0190

It looks like because twice the amount of B is going to be consumed as A,0201

the rate of consumption of B is going to be double that of A.0206

It is very typical for us in chemical kinetics to graph this.0221

We can go ahead and graph the concentration of the reactant with respect to time and the concentration of product too.0225

At time zero, I have no product; I only have reactants.0237

Down here right at the origin, this is the concentration of C initial.0240

The way that concentrations change with respect to time is not linear.0247

But instead it is going to be characterized by a simple curve just like that.0251

You notice that at some time t, the concentration of C stops changing.0260

We are going to plateau; it is not going to increase forever.0264

How about the concentration of A and B?0269

The concentration of A let's say is here and the concentration of B.0272

Let's say we had equal amounts.0278

Let me go ahead and draw now a curve that represents the concentration of A changing with respect to time.0280

It is going to be a mirror image of the concentration of C changing because they are both 1:1 ratio.0287

But how is the concentration of B changing?0293

The concentration of B, it is going to drop at double the rate.0296

The concentration of B is going to drop much much much... quicker than the concentration of A.0302

You notice that for all of the reactions, there reaches a point where the curves plateau0315

which means that the concentration of reactants and the concentration of products no longer changes.0325

We are going to talk about this region in another lecture.0337

This is what we call the equilibrium region.0340

In chemical kinetics, what we are interested in is really the start of the reaction, the early part0346

of the reaction where we actually can monitor change in the concentration of reactants and products.0351

That is the representation graphically; how about mathematically?0360

It turns out that the rate of change is going to be equal to the following.0364

The change in the concentration of A, Δ[A]Δt, this is going to be equal to0368

1/2 Δ[B]Δt which is equal to the change in concentration of C Δ[C]Δt.0382

We have to put a negative sign in front of the reactants such that the overall number is0392

going to be positive because Δ[A]Δt is going to be slope of the curve.0401

This is going to be slope of the curve you see.0408

This is going to be the slope of that curve too.0411

This gives us a general equation for relating the rates of change of each reactant and product concentration with respect to time.0416

Now that we have gone through a brief introduction, let's go now into0429

what is at the core of chemical kinetics--that is the rate law.0434

A rate law is going to be a mathematical equation which relates the rate of reaction to the concentration of the reactants.0439

Basically the rate law tells us that the rate of a reaction is proportional to the concentration of reactants typically.0447

When we write out the rate law, all rate laws have the following equation0458

Of the form rate which is going to be usually in molarity per second...0461

is equal to some constant k times the concentration of A raised to0467

some power times the concentration of the B raised to some power, etc.0472

Let's go ahead and define what each of these mean.0478

k is what we call the rate constant.0484

The rate constant is going to be unique to a reaction at a certain temperature.0491

In addition, we are going to see that the rate constant, the units of k vary.0508

We are going to see that very soon.0519

x and y are what we call rate orders.0522

Once again x and y are what we call rate orders.0529

If x is equal to 1, we say reaction is first order with respect to the concentration of A.0537

If x is equal to 2, we say the reaction is second order with respect to the concentration of A.0554

In addition, if the sum of the rate orders x plus y is equal to 0, we say zero order overall.0564

If the sum of the orders is equal to 1, we say first order overall.0577

If x plus y is equal to 2, we say second order overall.0585

This is just some terminology that we want to introduce and clarify.0592

Let's go ahead and get back to the units of k.0598

The units of k will vary.0601

For zero order, the units of k is just going to be reciprocal seconds.0604

For first order, the units of k is going to... excuse me.0619

For zero order.. my apologies.0627

For zero order, the units of k is going to be molarity per second.0629

For first order, the units of k is just reciprocal seconds.0636

For second order overall, the units of k is going to be inverse molarity inverse second.0640

You can easily plug that back into the equation for k and see that the units will cancel.0649

This is the general rate law.0657

Rate is equal to some constant k times the concentration of A raised to0659

the x power times the concentration of B raised to the y power.0662

But let's go ahead now and see what the significance of the rate orders are.0666

Consider the following reaction: A plus 2B goes to C.0673

The rate law is given to us to be k times A squared times the concentration of B.0678

Let's go ahead and study what happens if we change the concentration of one of these reactants.0686

If the concentration of A doubles holding B constant, then we see that the rate is going to increase by 4.0692

It is going to quadruple.0710

If the concentration of A triples holding B constant, we see that the rate is going to increase by a factor of 9.0714

What if the concentration of B doubles holding concentration of A constant?0728

If that happens, the rate is just going to increase by 2.0739

If the concentration of B triples holding A constant, we see that the rate is going to increase by a factor of 3.0746

You see what the significance of the rate order is.0756

The rate order is really proportional to the sensitivity of a reaction rate on0760

the concentration of a specific reactant molecule, on the concentration of a specific reactant.0778

Pretty much we see that the larger the value of these rate orders, the more sensitive0788

your reaction is to a change in concentration of a specific reactant.0795

Basically what the kinetics deals with is focusing on this rate law and solving for x, y, and k.0802

In other words, our goal is to solve for the rate constant and rate orders.0811

We can do this in one of two ways.0820

The first experimental way to derive a rate law is what we call the method of initial rates.0822

In the method of initial rates, you basically do what we just did.0828

You change one reactant concentration holding all else constant and seeing how the rate varies.0834

Let's go ahead and take a look at the following.0857

You are usually given some table of data.0859

The table of data is going to list different concentrations, different molarities of each reactant and the rate that was measured.0865

For example, in experiment number one, this concentration of iodide was found.0873

This concentration of thiosulfate was found; this is the initial rate.0881

In experiment number two, we see that the concentration of thiosulfate was fixed.0886

The concentration of iodide was tripled; we see that the rate also tripled.0894

What does that mean?--basically as the concentration of iodide tripled, the rate tripled.0901

This is a 1:1 correspondence.0913

Anytime you have this 1:1 correspondence, it is a rate order of 1; rate order of 1.0917

Let's now see what the rate order is for S2O82-, thiosulfate.0927

For thiosulfate, we are going to look at experiments two and three.0933

You see that as the concentration of thiosulfate tripled... I'm sorry.0941

We have to do experiments one and three, not two and three.0953

Here as the concentration of thiosulfate tripled, iodide was held constant.0956

What happened to the rate?--the rate tripled.0966

As S2O82- tripled, the rate tripled holding everything else constant.0968

This was also a 1:1 correspondence; the rate order is also 1.0975

Therefore we have the following rate law.0986

The rate of this reaction is equal to some constant k times the concentration of I-0989

raised to the first power and the concentration of S2O82- raised to the first power.0994

The next thing now that we have the rate orders, we can go ahead and now solve for k.1002

We can solve for k by simply plugging in any experiment--one, two, or three--directly into the equation.1011

Solve for k using any experiment number.1019

For example, let's use experiment one; the rate was 0.044 molarity per second.1029

That is going to be equal to the rate constant k times the concentration of I- which is1037

0.125 molar times the concentration of thiosulfate, 0.150 molar, all raised to the first power.1041

Then you can just use your algebra to go ahead and solve for the value of k for this second order overall reaction.1050

Here we are going to get units of inverse molarity inverse seconds.1063

That is how we use the method of initial rates--very straightforward experiment to determine the rate law.1070

The second way of determining a rate law is to use what we call integrated rate laws.1078

Integrated rate laws, they quantify the relationship between the reactant concentration and time; and time.1085

Basically these are all derived mathematically.1095

You should always ask your instructor if you need to know how to derive it or not.1100

For zero order rate law, the rate is equal to the rate constant k.1104

The integrated rate law is the following.1108

The concentration of A at any given time is equal to the initial concentration of A minus kt.1111

For the first order overall, the integrated rate law is natural log of A0 minus natural log of A equal to kt.1116

Finally the second order integrated rate law is 1 over A minus 1 over A0 is equal to kt.1125

The nice thing about these equations is that they are all linear.1133

If several concentrations are determined at different times, you can get the rate constant graphically.1139

For a zero order overall reaction, we are basically going to graph the concentration of A as a function of t.1146

That is going to give us a nice straight line with slope equal to ?k.1156

For first order overall, we can go ahead and graph the natural log of the concentration of A versus t.1162

We are also going to get a straight line whose slope is equal to ?k.1174

For second order overall, we are going to plot 1 over the concentration of A which is equal to time.1179

Here we are going to get a nice straight line with a positive slope equal to k.1189

Once again this is a graphical determination of the rate constant anytime1194

you have data from several different time intervals and measured reactant concentration.1198

Another thing we like to talk about too is the half-life.1210

For zero order, the half-life is equal to the initial concentration of A over 2k.1214

For first order, the half-life is equal to the natural log of 2 over k.1221

For second order, the half-life is equal to 1 over k times the concentration of A0.1228

Once again you should ask your instructor to make sure if you have to know the derivation or not.1235

Basically the half-life is very important because it tells us the time required to reach 1/2 of the initial concentration of A.1241

Again that is what we call integrated rate laws.1261

The next thing we are going to look at is what we call a reaction mechanism.1267

A reaction mechanism basically represents the step by step reactions which when combined give you the overall net reaction.1271

For example, let's say we had the following given, step one.1278

Step one was 2NO gas going on to form N2O2 gas.1285

This tells us that the reaction is occurring both in the forward and reverse directions.1293

This is what we call the reversible reaction; this is usually very very fast.1299

Step number two is going to be O2 gas plus N2O2 gas going on to form 2NO2 gas.1305

You are told that this reaction is very very slow.1318

The overall reaction is going to be 2NO gas plus O2 gas going on to form 2NO2 gas.1323

You notice that N2O2 gets cancelled out.1336

Any item in a chemical reaction mechanism that gets cancelled is what we call an intermediate.1342

That is it is both formed and consumed during the course of a chemical reaction; formed and consumed during a reaction.1351

The reason why we care about what the slow step is is because of the following.1370

The slow step, which is in this case step two, is like the weakest link in your chain.1376

It is like the slowest person on your track and field relay team.1382

The slow step determines, it limits how fast a reaction can go.1387

We call this the rate limiting step.1392

The rate limiting step is equal to the rate of the overall reaction.1399

If we were to write out the rate law for this, we would get the following.1408

The rate is equal to the rate constant k2...1412

This is k2, of the second step.1417

Times the concentration of O2 raised to the first power1419

times the concentration of N2O2 raised to the first power.1422

We can always use the coefficients as the orders if the reaction you are1427

looking at is a part of the mechanism, what we call an elementary reaction.1438

Coefficient is equal to the rate orders for elementary steps only.1444

Otherwise you would have to go through integrated rate laws or method of initial rates1451

to go through the whole process again to find what x and y are.1454

When we look at this, we have a problem.1458

We have N2O2 appearing in this rate law.1460

This is the intermediate.1463

You can never have an intermediate appearing in the rate law.1465

We have to do something about this; what we do is the following.1468

We use what is called a steady state approximation.1473

You utilize the fast equilibrium step where the k1 times the concentration of NO squared1481

equals to k-1 times the concentration of N2O2 where k1 represents the forward.1492

k-1 represents the reverse.1502

What we do then is we solve for the intermediate.1505

Concentration of N2O2 is equal to k1 over k-1 times the concentration of NO2 squared.1507

We then plug this back into our experimental rate law.1518

Rate is equal to k2 times the concentration of O21522

times k1 over k-1 times the concentration of NO squared.1526

You can collect all of the constants together and just call that what we call kobserved.1534

We get left with O2 times the concentration of NO squared.1539

Here we have our final rate law that is going to be the rate law for the overall reaction.1546

It is just by coincidence here that the rate orders are the coefficients.1555

That is not usually the case.1559

But that is how we solve for it where kobserved is equal to k2k1 over k-1.1560

Once again this is how you deal with reaction mechanism problems.1571

Let's now move on to the next topic.1577

This is the relationship between temperature and reaction rate.1579

Basically in general, as temperature increases, so does the reaction rate; just think about this.1583

You know the tea bag is going to brew faster in warm water than cold water.1588

You can see that visually happening right before you.1594

In order for a reaction to occur, reactant molecules must do two things.1597

They must collide; they must collide with sufficient energy.1602

They must collide in the proper orientation; sufficient energy and proper orientation.1606

Basically we can look at a sample here.1613

Let the y-axis be fraction of sample; let the x-axis be temperature.1616

At any given temperature, I am going to have a bell curve distribution of molecules just like that.1628

Let's call this T1.1639

What happens to T2?--what happens when we have a hotter temperature?1641

When I have a hotter temperature, my bell curve is going to shift just like that.1649

I call this T2; T2 is greater than T1.1652

Let's say that in order for the sufficient energy, in order for the1658

reaction to occur, let's go ahead and make that as a dotted line.1663

I am going to call this dotted line EA; EA is equal to activation energy.1669

What this is, it is the minimum energy required for the reaction to proceed, for collisions to occur.1677

Basically anything below EA, anything less than EA, you get zero collisions and no reaction.1691

Anything greater than or equal to EA, you get collisions; therefore a reaction will occur.1704

Basically you see that as you go from T1 to T2,1712

the fractional molecules with an energy greater than EA significantly increases.1717

At T2, larger percent of molecules with an energy greater than or equal to EA.1723

This is why as temperature goes up, so does the rate of a reaction in general.1739

We have a nice equation which can actually quantify this.1749

This is called the Arrhenius equation.1754

The natural log of k1 over k2 equals to EA over R times 1 over T2 minus1756

1 over T1 where R is our universal gas constant in terms of energy, 8.314 joules per K times mole.1766

Temperatures T1 and T2 are kelvin temperatures; what this basically says is the following.1779

If I do a series of reactions at different temperatures and I calculate the1789

rate constant, I can then do graphical determination of the activation energy.1794

If I graph natural log of k1 over k2 as a function of 1 over temperature,1800

I am going to get a nice straight line with slope equal to ?EA over R.1809

Again the Arrhenius equation is very useful because it gives us graphical approximation of the activation energy for a reaction.1815

Again this is the relationship between rate and temperature.1829

Finally the last factor we are going to study is a catalyst.1833

Basically a catalyst's job is to do the following.1837

A catalyst assists reactant molecules to be in the proper orientation for proper collision to occur.1840

What that does is that the activation energy is lower.1847

If this reaction represents without a catalyst, the activation energy is going be basically right here.1853

This is the energy that you must overcome for the reaction to go form products.1867

I can then proceed and draw another curve where this is a catalyst now.1873

With the catalyst, you see that the activation energy EA is much lower.1878

Activation energy catalyst is going to be always much less than the activation energy no catalyst.1884

Because of that, with the lower activation energy, that means a faster reaction.1892

A catalyst again speeds up a reaction by lowering the activation energy.1900

We see that of course the nice about catalysts is that they can be reused over and1907

over again because during the course of a reaction, they are recovered; they are recovered.1913

That is catalysis.1921

Let's now get into our summary before we jump into our sample problems.1923

Kinetic studies the factors that can influence the rate of a chemical reaction.1927

We saw that we can have two main experiments to help us determine1932

the rate law--the method of initial rates and the integrated rate law.1936

We found that in a reaction mechanism, that the slowest step dictates the overall reaction.1940

That is what we call the rate limiting step.1946

Finally we introduced the Arrhenius equation which gives a mathematical relationship between the reaction rate and the temperature.1952

The nice thing about this equation again, this gave us graphical estimate of EA.1963

Let's now get into a pair of sample problems.1975

A certain first order reaction has a half-life of twenty minutes.1977

Calculate the rate constant; that is part A.1980

Part B, how much time is required for this reaction to be 75 percent complete?1984

Let's go ahead; you are told that the reaction is first order.1989

For a first order reaction, T1/2 is equal to the natural log of 2 over the rate constant k.1992

You are told that the rate constant is 20.0 minutes.2001

Here the rate constant is simply going to be equal to the natural log of 2 divided by 20.0 minutes.2007

That gives us our answer in units of reciprocal minutes; that is part A.2014

Let's go ahead and do part B now.2024

How much time is required for this reaction to be 75 percent complete?2026

As soon as you see the word time, you should immediately, immediately, think integrated rate law because2029

method of initial rates does not have time in it; only integrated rate laws does.2036

For the first order integrated rate law, it is the natural log of the initial concentration2041

of A minus the natural log of the concentration of A is equal to kt.2047

We know what k already is because we already solved for that in part A.2056

We are good to go on that.2060

The question is asking for how much time is required.2062

This is what we are trying to solve for.2065

All that matters is is what is the identity of A0 and what is the identity of A?2068

You are told that how much time is required for this reaction to be 75 percent complete?2074

Let's say A0 is going to be 100.2079

If the reaction is 75 percent complete, that means only 25 percent of A is remaining.2083

A is going to be 25.2090

Again we now have enough information to solve for t.2092

We are going to get our final answer of t in units of minutes.2097

How do you know if you have done something wrong?2105

You should always check your answer because if you get a ?t, again that just doesn't make physical sense.2107

You know you have done something mathematically wrong.2119

Always check your answer for a negative time.2121

That is sample problem number one.2125

Let's go ahead and move on to sample problem number two.2127

Consider the following reaction.2130

2N2O5 gas goes on to form 4NO2 gas plus O2 gas.2132

Here we are given several rate constants that were found at several temperatures.2137

The only equation that we know that deals with this is the Arrhenius equation.2145

The natural log of k1 over k2 is equal to2150

EA over R times 1 over T2 minus 1 over T 1.2154

We know that we are going to be using this equation quite easily.2161

Here we can calculate EA from there.2165

The nice thing about this is because this is going to give us a nice straight line,2170

we can use any pair of k and T data points to go ahead and solve for EA.2174

Once again this is going to be using the Arrhenius equation to solve for EA.2193

Our units of EA is going to be in kilojoules per mole.2199

Next one is what is the order of the reaction.2206

This is kind of a trick question; this is something I have asked students before.2208

This is something I have seen asked by other instructors before.2213

The order of the reaction, you don't have to do any work for that.2217

The reason is because they already give you the units for k.2219

The units of the rate constant tell us the rate order.2226

It is only first order where the units of k is reciprocal time; first order overall.2234

Again just watch out for that when you do problems.2248

Again the units of k tell us a great deal of information without doing any work.2251

That is our lecture from general chemistry concerning kinetics.2257

I want to thank you for your time.2262

I will see you next time on Educator.com.2263

Hi, welcome back to Educator.com.0000

Today's lecture from general chemistry is on the principles of chemical equilibrium.0003

Let's go ahead and take a look at the lesson overview.0010

We will first start off with a brief introduction and then get in right to0012

the core of everything which is basically what we call the equilibrium constant.0017

We are going to first define this equilibrium constant and then go into different0021

types of equilibria including what we call homogeneous equilibria and heterogeneous equilibria.0026

We will then go and see how we can change the value of Keq.0033

We will also introduce something called the reaction quotient.0039

We then jump into a very fundamental principle in all of general chemistry.0042

That is called Le Chatelier's principle.0046

After that we will get into the quantitative part of the chapter which involves what we call ICE tables.0050

We will wrap everything up with a very brief summary followed by a pair of sample problems.0056

Chemical equilibrium, exactly what this is.0065

This refers to the simultaneous occurrence of a forward and a reverse reaction at the same rate.0068

We see that equilibrium is a dynamic process, not a static process.0075

For example, let's say you had two beakers here; these beakers are closed.0079

Let's say that this one beaker had H2O in it.0086

Let's just say this is time zero.0091

At time zero, I have nothing but pure liquid water.0094

But we know this from everyday experience inside a water bottle.0097

If you let some time progress, we notice that the water level is going to drop a little.0101

That is because some of the liquid water has entered the gas phase.0108

But there is also going to be a point in time where this water level is not going to drop forever.0116

It is going to reach a minimum.0125

It reaches a minimum because as soon as the vaporization process occurs, the condensation process also occurs.0129

We say for this case that the rate of evaporation is equal to the rate of condensation.0142

If we were to write this out in a chemical reaction, this would be H2O liquid.0158

Now we introduce a new type of reaction arrow which is this.0164

That goes to H2O gas; this is our equilibrium arrow.0169

It basically shows that the forward direction is happening simultaneously with the reverse direction.0180

Let's now get into how we can represent equilibrium numerically.0190

Consider the following reaction.0198

Small a big A plus small b big B, equilibrium sign and then small c big C small d big D.0199

In this reaction, let the lowercase letters represent the stoichiometric coefficients.0207

You know the moles after we balance the chemical equation.0222

It turns out that after a chemical reaction has reached equilibrium, it is experimentally determined that the ratio of0226

product to reactant concentration raised to the stoichiometric coefficients is actually constant at any given temperature.0235

Basically the concentration of C raised to the c power times the0243

concentration of D raised to the d power over the concentration of A0249

raised to the a power times the concentration of B raised to the b power.0256

This whole ratio of products to reactants raised to the coefficients is equal to some constant that we call K.0262

Sometimes you are going to see this called Keq.0272

This is formally what we call the equilibrium constant.0275

The only thing that can change the value of Keq is temperature.0285

The equilibrium constant is temperature dependent.0290

We can typically represent K in two different ways.0302

Kc is when molarities are used; Kp is when partial pressures are used.0307

Every textbook is a little different.0326

But for the partial pressures, the typical units are going to be atmosphere or bar.0328

Once again Kc and Kp are just the equilibrium expressions when0338

molarities and partial pressures are used for Kc and Kp respectively.0343

For solutions, the equilibrium constant can be expressed in units of molarity just like we discussed.0353

However it turns out that the equilibrium constant is unitless.0361

Kc, how do we go ahead and do that?0364

Mole over liter raised to some power divided by mole over liter raised to some power.0368

It turns out that the equilibrium expression is always referenced to 1 molarity.0378

This is really moles over liter per 1 molar.0385

This is moles over liter per one molar raised to the y power and raised to the x power.0394

As you can see that after this is done, we see that all units cancel.0402

Keq is actually one of the few unitless values in all of your general chemistry studies.0411

That is Kc.0420

It turns out that if we choose to do our problem with Kp, maybe this is going to be in atmospheres.0422

This is also relative to 1 atm, raised to some power divided by atm raised to some power.0429

We see it again that the units cancel; Keq again is unitless.0438

What is the relationship between Kc and Kp?0451

Kc and Kp are directly proportional to each other.0455

Depending on the reaction, sometimes they are nearly identical.0461

But the exact relationship between the two is the following where0465

Kp is equal to Kc times RT over Δn where Δn is0469

equal to the moles of gaseous product minus the moles of gaseous reactant.0479

Once again Kp and Kc can be very similar.0489

However they are not quite the same thing.0495

You should really check with your instructor to see what he or she prefers.0499

For chemical equilibria, we are usually dealing with very small concentrations.0507

Because of this, we assume that pure solids and pure liquids are going to remain relatively unchanged.0512

If you have the following reaction, A solid plus B aqueous goes on to form C aqueous plus D solid,0520

it is only the aqueous species that are going to affect the equilibrium, affect Keq.0532

Hence pure solids and pure liquids are going to be assigned an arbitrary value of 1 when incorporated into the expression for K.0546

That is they have no effect.0555

Kc for this reaction here would be simply the concentration of C times 1 divided by the0557

concentration of B times 1 which is just equal to the concentration of C over concentration of B.0566

Once again pure liquids and pure solids do not appear in the expression for K, in the Keq expression.0573

Again this is going to be specifically for heterogeneous equilibria.0593

What are some ways where we can manipulate K?0599

The first way of manipulating K is by multiplying an entire chemical equation by a factor.0602

For example, let's take A aqueous plus 2B aqueous goes on to form 3C aqueous.0608

In this case, K as you see is equal to the concentration of C cubed0621

divided by the concentration of A times the concentration of B squared.0627

Let's go ahead and take this chemical reaction and multiply everything through by 2.0633

2A aqueous plus 4B aqueous goes on to form 6C aqueous.0638

It turns out therefore that this Knew is going to be equal to concentration of C to the sixth0647

divided by the concentration of A squared times the concentration of B raised to the fourth power.0654

We see very nicely that Knew is simply equal to the original Kc squared.0661

The rule of thumb is the following.0670

That when multiplying a reaction by a factor, K is going to be raised to that factor.0672

K is raised to this factor.0695

That is the first way of algebraically manipulating K.0699

Once again this is by multiplying through a chemical reaction by a factor.0702

The second way is by taking the reverse reaction.0709

For example, let's take not A plus 2B going to 3C0712

but 3C aqueous goes on to form A aqueous plus 2B aqueous.0717

In this case, K is equal to concentration of A times the concentration of B squared divided by the concentration of C cubed.0726

We see that this is going to be 1 over kforward.0739

kreverse is simply the reciprocal of kforward.0745

The third way of changing K is by adding a series of individual chemical reactions together to form a net balanced chemical equation.0753

Let's go ahead and see.0760

For example, A aqueous plus B aqueous goes on to form C aqueous.0765

C aqueous plus D aqueous goes on to form E aqueous.0777

Let's go ahead and add these two together.0788

When we add these two together, we are going to get A aqueous plus B aqueous plus D aqueous goes on to form E aqueous.0790

You see that C is going to be cancelled out because it is going to be formed and consumed simultaneously.0803

What we want to look at now are the expressions for k for each of these.0815

Here k, I will call this k1 for reaction one, is equal to the0819

concentration of C divided by the concentration of A times the concentration of B.0823

Here k2 is equal to the concentration of E divided by the concentration of C times the concentration of D.0830

Here I will call this third one knet.0841

That is equal to the concentration of E divided by the concentration of A times the concentration of B times the concentration of D.0845

We see that when we add individual reactions together to get a net reaction, the equilibrium constant0855

of the net reaction is simply equal to the product of each individual equilibrium expression constant.0864

That is the third way of algebraically manipulating Keq.0874

Let's now go on to another aspect of the equilibrium constant.0884

It is what we call the reaction quotient.0887

The equilibrium constant is good when we actually have the values at equilibrium.0890

But what happens if we use values not at equilibrium?0894

When concentrations or pressures are inserted into the expression for K that are not at equilibrium,0899

the ratio of products to reactants is now what we call the reaction quotient symbolized Q.0906

Q is going to be equal to products raised to some power divided by reactants raised to0911

some power except that these molarities and partial pressures are not at equilibrium.0916

The significance is the following.0942

Q can be used to predict which way a reaction will shift to reobtain a state of equilibrium.0944

Basically we are going to have the following general rules.0976

If Q is greater than K, that means we have too much product, not enough reactant.0980

We are going to shift left.0987

If Q is less than K, we have too much reactant relative to product.0992

We are going to shift right.1001

Finally if Q is identical to K, we are at a equilibrium state.1005

Shifting left is the same as making more reactant.1015

Shifting right is the same as making more product.1028

At equilibrium, there is no net change; neither direction is favored over the other.1036

Once again Q can be very useful for determining what direction a reaction will shift if any.1044

Now we have come into one of the most fundamental principles from all of general chemistry.1055

This is called Le Chatelier's principle.1060

Le Chatelier's principle tells us that when a system at equilibrium is disturbed,1063

it will react to counteract the disturbance in an attempt to restore equilibrium.1068

One of the best examples we can think of is from us.1078

When we bleed, when we lose blood, what is the natural thing that our body does in order to counteract this?1082

Your body is going to try to make more blood.1091

When we lose blood, our bodies try to make more in order to counteract the lost.1092

This is a nice example of Le Chatelier's principle.1112

But now we are going to apply this to chemical reactions.1114

In Le Chatelier's principle, you notice that there is the word disturb.1119

There are several ways of disturbing a chemical system that is at equilibrium already.1124

Number one is a change in reactant or product concentration.1127

Number two if we are dealing with gases, it is going to be a change in reactant or product partial pressure.1132

The third one is the change in reaction volume.1137

The fourth one is going to be a change in temperature.1140

Let's now study each of these.1144

Basically if the concentration of reactant goes up, then we are going to shift away from it.1150

We are going to shift right.1163

If the concentration of product goes down, we don't have enough of it.1165

We are going to shift right; that is one situation.1170

If the partial pressure of a reactant goes up, partial pressure is just the same as concentration.1178

We know from ideal gas law that pressure is proportional to amount.1189

If partial pressure of the reactant goes up, we are going to shift right.1194

If the partial pressure of the product goes down, we are going to shift right.1204

For temperature, I am going to do this on the last slide because I am running out of room right now.1215

I will come back to changes in temperature and also to change in vessel volume.1228

But now let's get into some problem solving with some ICE tables.1235

We will now approach chemical equilibrium from a quantitative view.1238

An ICE table allows for one to study a chemical system at three points in time.1241

Basically what are all reactant and product initial amounts?1246

That is what the I stands for.1250

During the reaction on the way to equilibrium, what are their changes in concentration/pressure?1252

That is what the C stands for.1256

Finally what are their final volumes once equilibrium has been achieved?1258

That is what the E stands for.1262

We are going to look at this right now.1263

Consider 2 water gas goes to 2H2 gas plus O2 gas.1266

K is equal to 2.4 times 10-5 at some temperature.1271

At equilibrium, the concentration of H2O gas is 0.11 molar.1275

The concentration of H2 gas is 0.019 molar; calculate O2 at equilibrium.1279

The very first thing we want to do is set up our ICE table.1285

2H2O gas goes on to form 2H2 gas plus O2 gas.1289

What I like to do, I just like to set up the letters I-C-E right underneath it.1296

We just fill in this table right now.1300

You are told that at equilibrium, the water is 0.11 molar and that the H2 is 0.019 molar.1303

We don't know what O2 is.1313

We can just call that the concentration of O2 at equilibrium.1315

We know the expression for K is equal to 2.4 times 10-5.1322

That is going to be equal to the concentration of H2 squared times1328

the concentration of O2 divided by the concentration of H2O squared.1332

That is going to be equal to 0.019 squared times the concentration of O2 at equilibrium divided by 0.11 squared.1338

When all is said and done, the concentration of O2 at equilibrium is going to be 8.0 times 10-4 molar.1349

Don't forget the units.1359

You notice that we didn't have to fill in the rest of the table1363

because we were already at halfway there to the equilibrium values.1366

This is a nice usage of the ICE table.1371

Let's go ahead and do another example though.1374

A 1 liter flask was filled with so much of SO2 and so much NO2 at some temperature.1377

After equilibrium was reached, 1.3 moles of NO was present.1382

The reaction is the following; calculate the value of Kc at this temperature.1385

The problem wants us to calculate Kc at this temperature.1390

We are going to go ahead and set up the ICE table--I, C, and E.1396

SO2 here is going to be 2.00 molar initially.1406

O2 is going to be 2.00 molar also.1411

After equilibrium was reached, 1.30 moles of NO gas was present.1415

That is what goes right here, 1.30 molar.1419

Let's fill in the rest of the table.1426

If no initial values are mentioned of SO3 and NO, it is safe to assume that they are zero.1427

Let's go ahead and do the change.1434

The SO2 is going to go down by some amount x.1438

O2 is going to go down by some amount x.1441

SO3 goes up by some amount x.1443

O is going to go up by some amount x.1452

We notice because these are all 1:1 mole ratio.1457

At equilibrium, it is just going to be the addition of the two rows added together; sum.1465

That is going to be 2.00 minus x, 2.00 minus x, x, and x.1472

But guess what? x is 1.30 molar because we were told that in the beginning; that is very nice.1479

Therefore this is 1.30 molar; this is 0.70 molar; this is 0.70 molar.1485

We know enough to calculate the value of Kc.1494

Kc is going to be 1.30 squared divided by 0.70 squared.1498

That is going to give us an answer of 3.5 for our answer.1504

This is how we use the ICE tables to help us come up with the equilibrium constant.1517

I want to quickly summarize this and then jump into sample problems.1526

Chemical equilibrium is a dynamic process where the rate of forward and reverse reactions is equal.1531

Equilibrium constant K quantifies how reactant or product favored a reaction is once it has reached equilibrium.1536

If K is between 0 and 1, we say that it is reactant favored.1543

If K is greater than 1, we say that it is product favored.1552

Le Chatelier's principle states that a system at equilibrium when disturbed will react to counteract the disturbance.1559

We already saw the concentration and partial pressures.1565

But let's go ahead and see what I left out before; this is now temperature.1573

For an exothermic reaction, if the temperature increases, it turns out we are going to shift left.1579

If the temperature decreases, we are going to shift right.1593

Basically you treat the word heat as a molecule.1599

Again if you treat the word heat as a molecule on the product side, it becomes1609

a lot more intuitive as you will see what happens when we result in the change.1616

Another thing we can disturb an equilibrium with is with the volume.1622

If the volume of the reaction vessel goes up, that means the pressure is lower.1629

That means we are going to shift to make more gases.1636

Shift toward side of reaction with more gas molecules.1640

Once again if the volume goes up, we are going to shift toward the side of the reaction with more gas molecules.1654

Those are the ways we can disrupt a system that is at equilibrium and see how it is going to counteract the stress.1662

Finally ICE tables allow for determination of the reactant and product concentrations/pressures given the equilibrium constant K.1668

That is our summary.1676

Now let's go ahead and jump into a pair of sample problems.1678

Consider 2SO3 gas goes to 2SO2 plus O2.1681

Initially 12 moles of SO3 is placed into a 3 liter flask at some temperature.1685

At equilibrium, 3 moles of SO2 is present.1689

Calculate the value of Kc at this temperature.1692

Let's go ahead and rewrite this.1696

2SO3 gas goes on to form 2SO2 gas plus O2 gas.1698

Let's go ahead and see what we can fill in, I-C-E.1707

Initially 12 moles is placed in a 3 liter flask of SO3.1712

That is going to be 4.0 molar.1716

At equilibrium, SO2 is present, 3 moles; that is going to be 1.0 molar.1721

Usually you can fill in two-thirds of the table with just one or two sentences.1728

SO2 initial is not mentioned; that is zero.1733

O2 initial is not mentioned; that is zero also.1735

We have to pay attention to the coefficents.1740

The coefficients tells us the relative amount of reactant and product that is going to be involved.1742

Really this is going to be -2x; SO2 is going to be +2x.1747

O2 is just going to be +x.1752

At equilibrium, we get 4.0 minus 2x, 1 molar, and x.1756

The nice thing about this is that because we know the 2x and the 1 molar, we can conclude that x is 0.5 molar.1763

When we do that, we are going to get here 0.5 molar.1776

Here for SO3, we are going to get 3.0 molar.1780

Great, we now have all of our molarities at equilibrium.1786

Kc is equal to the concentration of SO2 squared times the concentration1791

of O2 divided by concentration of SO3 squared, all equilibrium values.1797

This is going to be 0.5 squared times 0.5 divided by 3.0.1805

When all is said and done, this is going to give us a Kc of 0.056 for this reaction.1815

That is another usage of the ICE table.1827

Finally the last sample problem is sample problem number two.1831

At a particular temperature, K is equal to 3.75.1834

If all four gases had initial concentrations of 0.800, calculate their equilibrium concentrations.1837

Let's go ahead and set it up--I, C, and E.1843

0.800 molar initial, 0.800 molar, 0.800 molar, and 0.800 molar.1848

Because we have nonzero amounts of all reactants and products, we don't know which way we are going to shift.1858

This is a twist on things; now we have to solve for Q.1868

Q is going to tell us which way we are going to shift.1872

This is going to be equal to concentration of SO3 initial times the concentration of1875

NO initial divided by concentration of SO2 initial times the concentration of NO2 initial.1882

That is just going to be equal to 1, 1.00 which is going to be less than K.1891

Because Q is less than K, that means we are going to shift to the right.1900

Therefore the sign that goes here is going to be ?x.1907

This is ?x; this is +x; this is +x.1911

We are going to get 0.800 minus x; 0.800 minus x.1916

This is going to be 0.800 plus x.1923

This is going to be 0.800 plus x; K is equal to 3.75.1926

That is equal to 0.800 plus x squared divided by 0.800 minus x squared.1934

We can use what we call a perfect square to solve for this.1946

The square root of both sides is what we are going to do next.1951

When we go ahead and do that, we are going to get 1.93 is equal to 0.800 plus x divided by 0.800 minus x.1956

When all is said and done, x is going to be equal to 0.25 molar.1966

All we do next is simply plug it back into all of the expressions at equilibrium.1972

The concentration of SO3 at equilibrium is equal to the concentration of NO at equilibrium.1978

That is going to be 0.800 plus 0.25.1986

I will let you guys do the sum; that is going to be 1.050 molar.1992

The concentration of SO2 at equilibrium is equal to the concentration of NO2 at equilibrium.2004

That is going to be equal to 0.800 minus 0.25.2012

We are going to get 0.55 molar for these guys.2020

That is use of the ICE table when we have all four initial values present.2026

Once again you have to find which way it is going to shift.2034

You do that by evaluating Q.2036

That is our lecture from general chemistry on the principles of chemical equilibria.2039

I want to thank you for your time.2045

I will see you next time on Educator.com.2046

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is on acid base chemistry.0002

We are going to first start off as usual with our introduction followed by how to quantify acid and base strength.0007

After we learn how to quantify how acidic or how basic something is, we will then go on to0015

writing out aqueous acid base equilibria reactions which is going to become very fundamental for this section.0022

After that, we will learn something we call acidic and basic salts0028

followed by multiple aqueous equilibria which is for acids that have more than one hydrogen.0031

We will then get into some specific topics, namely Lewis acids and bases, followed by molecular structure and acidity.0038

We will wrap up the session with our summary followed by a pair of sample problems.0047

Let's go ahead and begin.0054

We are going to look at two different types of molecules in this chapter--what we call a Bronsted-Lowry acid and a Bronsted-Lowry base.0057

A Bronsted-Lowry acid basically donates a proton to water.0065

A Bronsted-Lowry base is going to accept a proton from water.0076

You see that water is involved in both of these reactions.0088

For a Bronsted-Lowry acid, you can say HA aqueous plus water goes on to form A- aqueous because it loses its proton.0092

It is going to give it to water; H2O then becomes H3O1+ aqueous.0105

This special ion is what we call the hydronium ion.0113

Anytime you see the hydronium cation, it is always indicative of acid.0119

Let's go ahead and look at the reaction for a base.0125

We can have a generic base B aqueous plus H2O liquid goes on to form...0127

The base is going to accept the proton; it becomes BH1+ aqueous.0136

H2O loses a proton; it is going to become OH1- aqueous.0141

This special guy here of course is what we call the hydroxide ion.0145

This is always indicative of aqueous base.0150

Once again hydronium cation is indicative of aqueous acid.0153

Hydroxide anion is indicative of aqueous base.0157

You see that water is involved in both of the reactions.0161

In the acid reaction, water is the base.0166

In the base reaction, water is the acid.0169

We say that water is an amphiprotic molecule meaning it can function both as an acid and as a base.0172

When we take a look at water reacting with itself, we see the dual role of water.0179

For example, one of these water molecules can lose a proton giving hydroxide.0186

Therefore the other water molecule is going to gain a proton forming hydronium.0191

What is the extent of this reaction?0198

It turns out that if we write out the equilibrium expression for this reaction, it is going to be0202

equal to the hydroxide ion concentration at equilibrium times the hydronium ion concentration at equilibrium.0207

It turns out that this value, this product is only 1.0 times 10-14 at 25 degrees Celsius.0216

We give this K a very special name; this is what we call Kw.0225

This is called the auto-dissociation constant for water.0230

It is a very handy one because if you know the hydroxide ion concentration, you can calculate hydronium and vice versa.0236

When we talk about acids and bases, there are going to be what we call strong and weak acids and bases.0247

Basically strong acids are going to donate a proton to water and fully dissociate into a proton and an anion.0253

Because they fully dissociate, we do not use an equilibrium arrow.0262

We only use an arrow pointing in the forward direction.0267

We can take a molecule HX aqueous; this is going to react with water.0273

We only use a single arrow to show full dissociation.0278

That is going to form H3O1+ aqueous and X1- aqueous.0281

It is very imperative that you know what are the strong acids.0287

For our purposes right now... you should always confirm this with your instructor.0292

But there are going to be seven strong acids that you should know.0295

HClO4, HClO3, HNO3, H2SO4, HCl, HBr, and HI.0299

Again these are the seven strong acids that fully dissociate when dissolved in water.0313

The strong bases also are going to become fully protonated.0319

They are going to fully dissociate when dissolved in water.0324

They are going to form a cation and a hydroxide.0328

The typical ones are going to be MOH or MOH2.0332

These are basically group 1 and 2 hydroxides.0340

We can take the typical one, sodium hydroxide.0349

You don't even have to show water because all it does is that0353

it is going to fully break apart into sodium ions and hydroxide anion.0356

Once again these are the strong bases and the strong acids that you should know.0362

Again check with your instructor because every instructor is going to be0368

slightly different on the ones he or she requires you to memorize.0372

We have talked about strong species, what are the weak species.0378

Basically weak species, weak acid and bases do not completely dissociate at all, do not completely dissociate at all.0381

For example, HF aqueous plus water is a weak acid.0389

We are going to use of course an equilibrium arrow just like we do for any type of weak electrolyte.0396

That is going to give us hydronium cation plus F1- aqueous.0403

That is the typical weak acid.0410

The typical weak base is going to going to be ammonia NH3 aqueous plus H2O liquid.0412

Once again equilibrium arrow.0420

That goes on to form NH41+ aqueous and hydroxide aqueous.0421

Anything related to ammonia which is what we call... this is ammonia/amines.0429

These are all going to be our typical weak bases.0439

Again these do not completely dissociate.0449

Let's go ahead and now see how to quantify acidity and basicity.0452

To quantify acid and base strength, we talk about what is known as the pH scale.0457

When we talk about the pH scale, this is on a scale of 0 to 14 where 7 is the neutral area.0464

Any pH less than 7 is acidic.0476

Any pH greater than 7 is going to be basic.0480

The equations for pH is the following.0483

pH is equal to ?log of concentration of hydronium at equilibrium.0485

pOH is equal to ?log of the concentration of hydroxide at equilibrium.0492

There is some common things that you can know that are acidic--citric juice, sodas, stomach acid.0501

Then there is some things that are basic that you should know also.0513

Any household detergents tend to be basic.0516

Milk and human blood also is slightly basic.0519

You have to also make sure that you be able to go both ways.0528

Sometimes you are asked to calculate the hydronium ion concentration at equilibrium.0532

That is going to be anti-log.0537

That is going to be log inverse of ?pH which is 10 raised to the ?pH.0538

The hydroxide ion concentration at equilibrium is equal to log inverse of ?pOH which is equal to 10 raised to the ?pOH.0548

There is going to be a relationship between pH and the pOH.0563

Basically the pH plus the pOH is going to be equal to 14.00 at 25 degrees Celsius.0566

Again you should always consult with your instructor whether or not you have to commit these to memory.0577

This is the pH scale that is typically used for quantifying acid and base strength.0589

Let's now move on to another way of quantifying acid base strength.0596

This is going to be something we are going to encounter quite a bit.0599

This is what we call the acid ionization constant Ka and pKa.0603

For any of the acids where HA reacts with water, you are going to form hydronium and A1-.0608

There is always going to be a certain extent to which this acid is going to dissociate.0624

We are going to call this a Ka.0633

This is just the same equilibrium expression that we have been working with so much already.0636

Remember it is going to be products, hydronium ion, at equilibrium times0640

A1- at equilibrium divided by the concentration of HA at equilibrium.0645

You can see that for stronger acids, the concentration of hydronium at equilibrium is going to be high.0652

Ka is going to be high also.0665

This is what we call the acid ionization constant Ka.0673

Ka is usually written in scientific notation.0678

But there is another way we can go around that, something more convenient system to use.0682

This is called pKa; pKa is just like the pH.0687

pKa is equal to ?log of Ka.0691

You see that for large Ka values, pKa is actually small.0694

Again this is going to be true for stronger acids.0703

The nice thing about pKa is that pKa, the advantage is that we don't have to use scientific notation when we compute it.0707

It is going to be a nice small number.0716

pKa advantage avoids use of scientific notation.0718

That is the acid ionization constant Ka and pKa.0730

Let's go ahead and do a sample problem now.0734

Calculate the pH of a 1.2 molar solution of acetic acid where Ka is 1.8 times 10-5.0736

Also calculate the percent ionization.0741

Anytime you have a calculation like this, this always involves the use...0746

This is always going to involve the use of ICE tables just like we learned last time.0756

Let's go ahead and write out the equation.0766

Acetic acid CH3CO2H plus H2O liquid goes on to form CH3CO21- aqueous plus hydronium aqueous.0768

When we are reading the statement, calculate the pH of a 1.2 molar solution of acetic acid,0786

that 1.2 molar, remember the problem doesn't mention equilibrium here for 1.2 molar.0790

We assume this is initial and then zero and zero.0796

This is going to be ?x, +x, and +x.0800

This is then 0.2 minus x, x, and x.0804

We are going to set it up then to the expression.0808

Ka is equal to 1.8 times 10-5 which is equal to x squared over 1.2 minus x.0810

You remember from our last session that we talked about the assumption that x is negligible.0819

We can do this for any weak acid.0827

Assume x is negligible for weak acids and bases.0829

This equation then is approximately x squared over 1.2.0840

When we solve for x, that is going to give us the hydronium ion concentration at equilibrium which is going to be 0.0046 molar.0846

When we go ahead and solve for pH, as expected, we should get an acidic pH because this is an acid after all.0857

This is going to be pH is equal to 2.33.0864

The rules for sig figs when you do the logarithm function, the number of sig figs0867

in the concentration is equal to the number of digits after the decimal in pH.0877

As you can here for 0.0046, there is two significant figures.0894

I am going to report my pH value to two digits after the decimal.0899

That is how we quantify acid base strength.0905

Another way we can quantify acid base strength is to do the other one besides acids--that is bases.0908

Instead of using Ka, we use Kb which is the base ionization constant.0917

For Kb, we are going to say B aqueous plus H2O liquid goes on to form BH+1 aqueous and hydroxide aqueous.0922

Basically Kb is going to be equal to the concentration of BH1+ at equilibrium0936

times hydroxide at equilibrium divided by the concentration of B at equilibrium.0944

You can see it is analogous to Ka.0951

Stronger bases are going to have lots of OH- at equilibrium which means Kb is going to be large.0955

Just like pKa, pKb is going to be equal to the negative log of Kb.0969

If Kb is large, this means pKb is small.0976

Once again for stronger bases, pKb is small.0984

That is how we do the base ionization constant.0998

Finally the other ways to quantify acid base strength is what we call percent ionization.1003

For acids, this is going to be equal to the hydronium ion concentration1014

at equilibrium divided by the initial concentration of the acid times 100.1022

Then for bases, this is equal to the concentration of BH1+ at equilibrium divided by the initial concentration of base times 100.1027

Basically strong acids are going to have relatively large percent ionizations near the 99 percent.1039

For weak acids and bases, it is going to be the extreme.1046

The percent ionization is actually going to be single digits, sometimes less than1050

one percent just to show you how drastic the differences can be.1054

Now that we have gone through quantifying acid base strength,1059

let's go ahead and turn our attention to writing out acid base equilibria.1062

As you can see, in order to set up the ICE table, it is very imperative that you be1066

able to write out the correct equilibria, or else the entire problem will be messed up.1070

Let's go ahead and take some practice in writing out equilibria reactions involving acids and bases.1075

For example, HNO2 aqueous, this is going to react with water liquid.1080

Because this is weak, we use an equilibrium arrow.1088

HNO2 is going to lose a proton giving nothing but NO21- aqueous.1093

H2O is going to gain it; that becomes H3O1+ aqueous.1097

Let's go ahead and look at a base, CH3NH2 aqueous plus H2O liquid.1103

This is going to be weak base; we use an equilibrium arrow.1111

The amine is going to gain a proton, CH3NH31+ aqueous.1114

H2O loses the hydrogen giving us hydroxide.1121

A couple of notes anytime you are dealing with an amine.1127

Four amines, you always attach the H to the nitrogen atom.1132

That is note number one.1141

Note number two, you see that when an acid loses a proton, you see that the charge always decreases.1143

HNO2 becomes NO21-; for each hydrogen loss, charge decreases by one.1150

You see for the base of course, for each hydrogen gained, the charge is going to increase by one.1162

That is why we go from CH3NH20 to CH3NH31+.1172

This leads us into what we call conjugate pairs.1181

This is going to become very important conceptually for the rest of our discussion.1184

Consider the following equilibrium.1188

HF aqueous plus H2O liquid goes on to form H3O1+ aqueous and F- aqueous.1192

The only difference between HF and F- is a proton.1207

The only difference between H2O and H3O1+ is a proton.1212

When you add two molecules on opposite sides of the equilibrium that1216

differ by only one hydrogen, these are what we call conjugate pairs.1222

HF and F- are conjugates.1226

Water and hydronium are also a conjugate pair.1229

HF is the Bronsted-Lowry acid, making F- its counterpart the conjugate base.1234

The reason why it is called a conjugate base is because there is a tendency for the reverse reaction to happen1244

where F- aqueous can react with water going on to form HF aqueous and hydroxide.1252

This equation is likely to happen because HF is only a weak acid.1260

What that means is that it doesn't fully dissociate.1264

That means that F- will react with water to reform at least some of the initial HF as we can see from this reaction.1267

In general, the stronger the acid, the weaker its conjugate base.1279

Something like HCl which fully dissociates means that Cl1- will not react with water1284

to reform HCl because HCl has a great tendency to remain dissociated in water.1290

In general, the stronger an acid, the weaker its conjugate base.1297

In general, the stronger a base, the weaker its conjugate acid.1300

What we see is that acid base conjugate strengths are going to be inversely related.1304

Now let's go ahead and examine salt solutions and predict if they are acidic, basic, or neutral.1314

Basically you want to go by the following three rules to help us predict1322

if a salt is acidic, basic, or neutral just going off of its formula.1326

Salts said to produce acidic aqueous solutions contain the following combinations.1330

A conjugate acid of a weak base and a conjugate base of a strong acid--something like NH4Cl aqueous.1335

Cl1- is not going to react with water to give us anything because Cl- is the conjugate of HCl.1349

We are not going to reform HCl.1357

However NH41+ will have a tendency to react with water because it is the conjugate of only a weak base.1359

Here we are going to NH3 aqueous and hydroxide aqueous.1368

Excuse me... NH3 aqueous plus hydronium aqueous; there we go.1375

You see that we form hydronium which makes sense that we expect this solution1381

therefore to be acidic when the salt is dissolved in water.1386

Let's look at basic aqueous solutions.1393

Salts that produce aqueous solutions that are basic contain a group 1 and group 21395

metal cation like Na in combination with the conjugate base of a weak acid like F.1400

Na+ is not going to react with water to form anything.1409

It is going to remain solvated; we don't have to worry about that.1417

Also F-, F- is the conjugate of HF, a weak acid, which means there will be1422

a tendency for the following reaction to occur, formation of hydroxide and HF aqueous.1429

You see that because we form hydroxide, these types of salts are expected to be basic when dissolved in water.1437

Let's now move onto neutral salt solutions containing a group 1 and 2 metal cation and the conjugate base of a strong acid.1445

For example, NaCl, we said that Na1+, this is not going to reform hydroxide when reacting with water.1454

Cl1- is not going to react with water to reform HCl.1464

In either case, both the cation and anion do not react with water to form hydroxide or hydronium.1469

Neither is going to contribute to pH.1476

This is what we call a neutral salt solution.1479

We have so far talked about monoprotic acids, those acids that only contain one hydrogen atom.1485

But what happened for acids like sulfuric acid or phosphoric acid?1491

These are what we call polyprotic acids.1496

To calculate the pH of a 1.2 molar solution of sulfuric acid which is diprotic, we want to remember one thing.1499

It is that the first dissociation is the only one contributing to pH; first dissociation only contributes to pH.1507

Once again the first dissociation only contributes to pH1521

meaning that Ka1 is going to be relatively much greater than Ka21525

which is going to be relatively much greater than Ka3.1530

For H2SO3 aqueous plus water, we are going to write out the dissociation stepwise.1534

That is very important; write out dissociation stepwise; write out steps one at a time.1545

H2SO3 aqueous plus H2O liquid goes on to form HSO31- aqueous and H3O1+ aqueous.1558

This is what we call Ka1.1569

The next one is HSO31- aqueous is going to take its turn.1572

Plus H2O liquid goes on to form H3O1+ aqueous and SO32- aqueous.1577

For this, we are going to see Ka2.1587

We are basically saying Ka1 is much greater than Ka2.1590

To complete calculate the pH, Ka1 is going to be equal to 1.2 times 10-2.1594

You can look that up which is approximately x squared over 1.2 minus x.1604

You can go ahead and use this to solve for the hydronium ion concentration which gets us pH.1610

If you look up Ka2, Ka2 is 6.6 times 10-8.1617

That is more than six orders of magnitude less than Ka1.1622

This shows how drastic a difference the first deprotonation is versus successive deprotonations.1629

Now let's take a look at the salt of a diprotic acid.1638

Calculate the pH of a 1.2 molar solution of Na2SO3.1642

Here Na2SO3, which one of these cation or anion is going to react with water?1647

We know that Na1+ is not going to react with water.1653

There is going to be no reaction; the only thing that is left is sulfite.1659

SO32- aqueous is going to react with water.1664

We just saw that SO32- is the conjugate of a weak acid.1668

Then there is going to be a tendency for this reaction to occur where we reform HSO31- aqueous and generate hydroxide.1674

We don't use Ka for this one; we use Kb.1684

But we don't have Kb; we were only given Ka.1690

But we do know that Kw is equal to Ka times Kb.1693

This is a very important equation which is equal to 1.0 times 10-14.1697

It turns out that Kb for this reaction which we call Kb1, Kb1 represents the first protonation.1704

This is going to be equal to Kw over not Ka1 but Ka2 only.1711

The reason why we use Ka2 is because we don't generate SO32-.1717

It is not formed until both protons have been removed.1726

SO32- not formed until second deprotonation which on the previous slide if you go back is associated with Ka2.1730

Again this is how you deal with a salt solution of a polyprotic acid.1745

Now let's go ahead and take a look at acids from a different perspective.1755

So far we have talked about Bronsted-Lowry acids and bases which involve a transfer of protons.1759

However we can also look at it from a more general approach.1764

This is what we call the Lewis theory of acids and bases.1768

Lewis acids accept a lone pair of electrons to form a new covalent bond.1772

Lewis bases donate a lone pair; Lewis acids accept a lone pair.1778

You want to look for metal cations that tend to be Lewis acids.1785

You want to look for elements that do not have a complete octet,1792

that do not require a complete octet which is boron, aluminum, and beryllium.1796

These do not need a full octet.1802

They have room to accommodate an extra lone pair.1806

Lewis bases however donate a lone pair.1810

You want to look for anions for visual clues.1813

Of course you have to have lone pairs to start with; look for lone pairs in the Lewis structure.1817

How do they relate to Bronsted-Lowry acids and bases?1831

You want to remember that all Bronsted-Lowry acids/bases are also Lewis acids and bases.1834

However the reverse statement is not always true; the converse is not always true.1849

Again these are what we call Lewis acids and Lewis bases.1861

We can take a look at a very interesting example.1865

Let's look at ammonia then, NH3 aqueous.1869

I am going to put its lone pair right there.1872

It is going to react with H2O liquid going on to form NH41+ aqueous and hydroxide aqueous.1874

Let me put the lone pairs on the oxygens just to highlight the point.1885

You see that the nitrogen atom has lost a lone pair making ammonia a Lewis base.1889

You see that the oxygen atom has gained a lone pair making water the Lewis acid.1899

Just like a Bronsted-Lowry acid reacts with a Bronsted-Lowry base,1906

a Lewis acid is also going to react with a Lewis base.1910

Now that we have interpreted acids and bases differently, let's go ahead and look at factors that influence acidity strictly off of molecular structure.1916

Factor number one, the effect of charge, basically acidity is higher with increasing charge.1925

For example, H2SO4 versus HSO41-, this is going to be the stronger acid.1939

Remember we said that for polyprotic acids that successive deprotonations don't really contribute to pH.1950

Successive steps are negligible to pH; look at this.1957

The reason is the following; a hydronium cation is positively charged.1966

It is easier for something with a high charge to give up something positively charged.1971

HSO4 has its -1 charge.1975

It is going to be a relatively harder time to give up that positively charged cation.1977

Once again acidity is higher with increasing charge.1983

Now, within a period/row--CH31- aqueous, NH21- aqueous, OH1- aqueous, and HF aqueous.1988

It turns out that acidity is going to increase left to right.2003

Acidity goes up left to right in a row; the reason is the following.2008

What else increases left to right in a row?--this parallels electronegativity.2018

Remember we say that Bronsted-Lowry acids, which are also Lewis acids, they donate a proton and accept a lone pair.2028

The more electronegative the atom, the more easier it is to accept a lone pair.2039

Higher EN is easier to accept a lone pair.2045

Again acidity increases left to right in a row.2055

Within a group or column is the next one we will look at--HF, HCl, HBr, and HI.2058

It is experimentally determined that acidity increases down a column.2065

This doesn't parallel electronegativity; however this does parallel atomic size of the anion.2073

Parallels atomic size of atom directly bonded to H.2081

Again that is key; it has to be directly bonded to the H.2088

What happens is when HA become A1-, A1- is gaining a lone pair.2095

Let's look at this; it starts off with three lone pairs in HA.2103

It comes with four lone pairs in A-.2107

We are going to be better off adding electron density to a larger volume.2113

Easier to add more electrons to a larger volume.2118

In other words, I- is more stable than F- which makes HI a stronger acid2129

because the dissociation of HI is much more likely to occur and to a greater extent2144

than the dissociation of HF where F- is not as stable because of its smaller size.2151

Oxo acids, for example, HNO3 versus HNO2.2158

In oxo acids, acidity increases with the number of oxygen atoms per hydrogen.2167

To answer the question why, let's go ahead and look at the Lewis structures.2184

HNO3 here; HNO2 is going to be right there.2190

Basically what is happening is we know that HNO3 is a stronger acid.2201

It is one of the seven you have memorized; HNO2 is not.2207

What makes the difference?--the only difference is one oxygen atom.2211

That explains for why HNO3 is more acidic.2214

What happens is the oxygen atoms are highly electronegative.2217

They are going to withdraw electron density away from the bond with hydrogen.2220

That makes this partial positive.2225

That is going to make the rest of the molecule partial negative.2228

This weakens the bond with the hydrogen.2233

It is going to allow for hydrogen to leave the acid much more easily.2236

Here in HNO2, the withdrawing effect is not as great.2242

The electron density is not so lopsided.2250

It is going to be a harder time for hydrogen to fall off here with fewer oxygen atoms.2255

This is what we call an electron withdrawing inductive effect.2260

Once again this is what we call an electron withdrawing inductive effect.2267

That is oxo acids.2273

Carboxylic acids, when we go ahead and look at carboxylic acids, you also go by inductive effect.2277

When you go by the inductive effect, just look for the presence of electronegative groups.2291

For example, we can take this carboxylic acid that has two fluorines versus a carboxylic acid that has no fluorines.2298

They are structurally similar.2307

You see here that with the two fluorines, we are going to get a greater withdrawing effect.2310

Again the stronger the withdrawing effect, the more easily the hydrogen is going to quote and quote fall off.2320

H falls off more easily.2329

Here in acetic acid, there is going to be a minimum withdrawing effect from just the oxygens.2336

Once again for carboxylic acids, you want to go by the inductive effect.2356

You want to look for nearby electronegative atoms.2360

Hydrated metals cations also can be acidic.2364

If we look at Fe(H2O)63+, this is aqueous.2367

This can go ahead and react with water.2375

It can actually function as a Bronsted-Lowry acid.2379

We are going to get Fe Fe(H2O)5OH plus H3O1+ aqueous.2382

How likely is this reaction to occur?2394

This reaction is more likely to occur when this charge is very high.2396

Again for hydrated metal cations, acidity goes up with metal charge.2402

Acidity goes up with metal charge.2412

For example, the Ka of Fe(H2O)63+ is going to be greater2415

than the Ka of Fe(H2O)62+ aqueous just strictly because of charge.2426

That is molecular structure and acidity.2438

Let's go ahead and summarize the section.2441

Bronsted-Lowry acid base chemistry involves a loss or gain of a proton to or from water.2443

Conjugate pairs only differ by one proton and are inversely related in terms of acidity and basicity.2449

We learned many ways of quantifying acid base strength, namely pH, pKa, and Ka and then percent ionization.2456

Finally we saw qualitatively how the structure of a molecule can have a significant impact on how acidic it can be.2466

That is our summary of the lesson.2476

Let's now jump into a pair of sample problems.2478

Calculate the pH of a 1.2 molar solution of NH3 where Kb is 1.8 times 10-5.2481

Just like the previous lecture, the first step for any equilibrium problem is to write out the actual equilibria.2489

NH3 aqueous plus H2O liquid goes on to form NH41+ aqueous and hydroxide aqueous.2496

Let's set up the problem; 1.2 molar is given to us.2510

This is going to be 0 and 0; this is ?x, +x, and +x.2512

This thing goes to 1.2 minus x at equilibrium, x, and x.2519

Kb is 1.8 times 10-5; this is going to be approximately x squared over 1.2.2524

When we go ahead and solve for x, we get the hydroxide ion concentration at equilibrium which is going to be 0.0046 molar.2533

When we solve for pH, we had better get a pH that is basic because this is ammonia after all.2545

We get 11.66; this is sample problem one.2550

Let's now move on to sample problem two--predicting if the following salt solutions are acidic, basic, or neutral.2558

Here potassium bromide, we have a group 1 cation.2563

Br1-, this is the conjugate of HBr which is a strong acid.2569

When we have this combination, we expect this salt solution to be neutral.2576

Sodium, group 1, HPO42-.2581

This is going to be the conjugate of phosphoric acid which is going to be a weak acid.2587

We expect this compound to be basic.2593

Finally lithium cyanide, this is going to be group 1 here.2597

CN is going to be the conjugate of HCN which is considered to be a weak acid.2602

We expect this compound here to also be basic when dissolved in water.2608

That is our lesson on acid base chemistry.2617

I want to thank you for your time.2620

I will see you next time on Educator.com.2621

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is on the tools of quantitative chemistry.0002

We are going to get introduced into how chemists go ahead and make measurements0010

and the basic concepts and methods of performing everyday calculations.0015

Our first point in the lesson is going to be the SI units of measurement.0023

Once we become familiar with the units of measurement,0027

we will then get into explaining how exactly good or bad your measurement was.0030

This is done by something we call percent error and also standard deviation.0039

We will then get into a discussion of what we mean by precision versus accuracy.0044

We will then get into something a little more quantifiable which is called significant figures and uncertainty.0048

Followed by the following: identifying sig figs; then using these significant figures in calculations.0054

We will then get into a very important concept which is fundamental to all of the physical sciences.0063

This is called dimensional analysis which is your use of what we call conversion factors for problem solving.0069

Finally we will wrap up the lesson with a brief summary followed by some sample problems.0078

The SI units of measurement are what we call the international system of units.0085

They are included as part of the measuring system.0090

It is used in all parts of the world except the United States.0093

The basic SI unit for mass of course is going to be the kilogram.0098

For length, it is going to be the meter.0106

For volume, it is going to be the liter.0109

Kg again stands for kilogram; lowercase m is going to be abbreviating meter.0114

Finally capital L is going to be symbolized for liter.0122

Again mass is going to be kilogram; length is going to be meter; volume is going to be liter.0128

When you make a measurement, how good or poor is it?0139

How close was a measurement to an actual or already known value?0143

This is what we call percentage error.0148

Percentage error is equal to the actual value minus the experimental value divided by the actual value.0150

Actual value is the known value.0158

The experimental value is what you actually measure in an experiment; what you measured.0163

Of course we always like to get as close as possible to the actual.0172

To illustrate this very simple equation to use, suppose you weighed an object and recorded a mass of 86.2 grams.0177

The known value is 97.9 grams; calculate the percentage error.0185

Known value is going to be actual.0189

What you recorded, this is our experimental value.0194

The percent error... let's just plug everything into the equation.0202

It is going to be equal to the actual value 97.9 grams minus the experimental value of 86.2 grams.0206

Divided by the actual value which was 97.9 grams.0216

All of this is going be multiplied by 100 to get you your percentage error.0222

The next statistic that we can also calculate is something called standard deviation.0229

Sometimes when you repeat an experiment several times, it is useful to know how successful you were in reproducing your results.0235

If you did for example ten trials of the same measurement,0243

how well were you able to get the same measurement in each of the ten trials?0248

For our purposes we want of course the standard deviation to be as low as possible.0255

Standard deviation has the following equations.0261

It is equal to the square root of all of these terms where x is equal to basically each individual score.0263

X-bar is simply the average of all the x's.0271

N is the number of values or the number of trials.0274

Summation of course means we are going to add up across all the values.0278

Let's go ahead and illustrate the use of this equation.0283

It looks like a lot but it is really not too bad.0287

You weighed an object three separate times with masses of 32.6, 31.8, 34.1 milligrams.0291

Calculate your standard deviation.0297

All I am going to do is set it up for us right now.0300

The standard deviation is going to be equal the square root of everything underneath it; summation x minus x-bar.0302

X is each score; 32 minus 6, minus the average, squared.0311

I am going to put all of this in brackets showing that we are going to add what is next.0321

Plus, 31.8 minus x-bar, squared; then plus 34.1 minus x-bar, squared; then closed brackets.0326

That is what is meant by summation x minus x-bar, squared.0349

Then divided by n minus 1 where n is going to be equal to 3 trials.0353

All of that divided by 3 minus 1.0358

All we need then to finish this problem is x-bar.0362

X-bar is just your average of the three values.0368

That is just going to be 32.6 plus 31.8 plus 34.1.0371

All of that divided by 3, giving you your average score; that is standard deviation.0377

The next item is what we call precision and accuracy.0385

Precision simply represents how well you were able to reproduce your results over several trials.0388

High levels of precision tend to support accountability and reliability.0394

Precision is going to be quantifiable by standard deviation; measured by standard deviation.0399

Again you want as low a deviation as possible; low standard deviation desirable.0410

The word accuracy however represents how well you were able to measure a value0422

in relation to an already known or published value from the literature.0427

Of course this is going to be measured by percent error.0431

Once again we want as low a percentage error as possible; low percent error is also desirable.0443

A lot of people use the terms precision and accuracy interchangeably.0454

But of course as you have just seen they are not the same.0457

One represents your ability to reproduce a certain measurement.0461

The second represents how well you were able to get to a known value.0466

The next item is very important whenever you make measurements.0473

This is called significant figures and uncertainty.0476

Consider the following two rulers; let me go ahead and draw one ruler.0480

I am going to go ahead and draw a second ruler on the bottom.0486

Now the difference is going to be their tick marks; one, two, three.0489

Then the next one is going to be the following; one and two.0495

Let's go ahead and measure a piece of wood.0510

That is this length right here measured by green.0514

I am going to do the same thing with the other ruler; just like that.0517

How would you measure the green stick with the above ruler?0529

One person could say maybe 1.4 cm; or another person could say 1.5 cm.0533

It turns out that neither person is incorrect.0541

Why?--because the 1 is what we know for sure.0545

We know that green object is longer than 1 for sure; 1 cm.0549

Really this last digit here, the .4 and the .5, is really uncertain.0554

It is the digit of uncertainty.0560

The digit of uncertainty is strictly up to you the user; up to person making the measurement.0566

Because the ruler is so poor, the tick marks are so large,0578

there is no way we can determine if it is .4 or .5.0584

It is not provided by the tick marks on the ruler at all.0587

However when we go ahead and look at this ruler down here,0593

we finally get the tick marks that is representative of the appropriate digit.0596

So you hear, one person could say 1.4.0601

Because it looks like the green mark is right on the .4, one person could even say 1.40 cm.0606

The other person could say 1.41 cm; one could say 1.42cm.0613

Someone could even say it is a little less; 1.39 cm.0619

It turns out that because of the tick marks, we can go one digit more.0623

We can provide a better measurement; better measurement due to what we call a higher level of precision.0629

Again it is a better measurement due to higher level of precision.0651

The rule of thumb is the following.0656

How far or how many digits do you know how to record a measurement?0658

You are always going to go one digit past whatever is given to you or whatever the limit is on the ruler.0663

Go one digit past what is given by the ruler or instrument.0671

For example when we go ahead and read a graduated cylinder, you always want to look at the bottom of the curve.0690

You see how this water level is slightly curved.0698

That is what you call the meniscus.0701

You always want to look at the bottom of it.0705

What is provided to us here on this graduated cylinder?0707

50 milliliters is right here; 55 milliliters is right there.0712

We know that the meniscus is approximately 53 milliliters.0726

The graduated cylinder can tell us if it is 51, 52, or 53.0735

What the rule is is we are going to go one digit past this.0741

We can say something like 53.0; we can say something like 53.1.0745

We could even go under and say 52.9 milliliters or even 52.8 milliliters.0750

Once again you are going to one digit past the last digit of certainty provided by the instrument.0757

When you make a measurement, it is important to always write it down with the correct number of these significant figures.0765

What are their significant figures?--what are the rules for identifying them?0774

The rules for sig figs are the following; all nonzero digits are significant.0779

A zero is significant when it is in between nonzero digits.0786

A zero is significant when at the end of a decimal number.0791

A zero is not significant when starting a decimal number.0795

Let's go take a look at a couple examples; all nonzero digits are significant.0799

For example 562, we have a grand total of three sig figs because none of them are zeros.0804

A zero is significant when it is in between nonzero digits.0810

Something like 501; this is three sig figs.0814

The zero counts because it is in between nonzero digits.0818

5001; both zeros count because they are in between nonzero digits.0822

How about 50010?--only the two zeros count here because they are in between nonzero digits.0829

This last zero here does not count; we only have four sig figs here.0837

A zero is significant when at the end of a decimal number; for example 500. and 500.0.0843

It turns out that in 500. all of these are significant.0853

The zeros come at the end of a decimal number.0858

Here, 500.0, these are all significant because they come at the end of a decimal number.0860

A zero is not significant when starting a decimal number.0870

For example 0.00321, the two zeros do not count here because they start a decimal.0873

We only have three significant figures here.0882

However 0.003210, these do not count at all.0884

However you see the zero here, that comes at the end of a decimal number.0891

This definitely does count; we have a grand total of four significant figures.0895

When we use significant figures in calculations, we have to learn how to incorporate the rules now.0906

For multiplication and division, the answer will have the same number of sig figs as the fewest number of sig figs present.0913

For addition and subtraction, the answer will have the same number of decimal places as the fewest number of decimal places present.0921

For example. 0.321 times 0.57, we are going to get an answer that is only two sig figs.0929

Why?--because here in 0.321, you have three sig figs.0941

Here in 0.57, you have two sig figs.0946

When we do the addition and subtraction with the same numbers, 0.321 minus 0.57,0949

you see now we go by digits after the decimal places.0956

Here there is three digits after the decimal; here there is two.0960

Our answer is going to have two digits after the decimal.0963

Finally when you have mixed operations, you never want to round until the end.0971

You want to carry all digits through.0976

Now that we have talked about significant figures, let's go ahead and discuss what we mean by dimensional analysis.0982

Dimensional analysis utilizes the following.0989

It utilizes ratios of different units that we call conversion factors to convert from one unit to another.0991

If you want to go for example from unit A to unit B, how do we go ahead and do that?0999

The general format is the following; we are going to take unit A.1007

We are going to multiply by this ratio, something over something.1011

That is going to go ahead and give me unit B.1015

Unit A goes downstairs to get cancelled.1018

Unit B goes upstairs to get carried through to the final answer.1022

This ratio right here of unit B to unit A, that is what we call your conversion factor.1027

Let's go ahead and look at a couple of examples.1038

There are 4 laps in 1 mile; how many miles are in 17 laps?1040

We are going to say 17 laps times something over something.1045

That is going to give us our answer in units of miles which is what the question is asking for.1050

I start with laps; but I want to cancel it.1056

It is going to go downstairs to get cancelled.1059

Miles goes upstairs to get carried through to the final answer.1061

The conversion factor is actually given to us in the problem because they state that 4 laps is equal to 1 mile.1065

This is then 1 mile on top and 4 laps on the bottom.1071

We are going to get 4.25 miles for your answer.1075

This is going to round to 4.3 miles.1081

Let me tell you why: 17 laps, you have two sig figs.1085

However for conversion factors, we are assuming that a conversion factor are what we call exact numbers.1090

That is they have infinite precision.1097

They have infinite significant figures; there is no uncertainty.1100

You are going to ignore conversion factors for sig fig purposes.1104

Once again you are going to ignore conversion factors for sig fig purposes.1113

Let's go ahead and look at one last example here on dimensional analysis.1121

One pound is 454 grams; how many grams are in 1.22 pounds?1125

1.22 pounds times something over something is going to give us our answer in grams.1131

I want to cancel the pounds; that goes downstairs.1141

I want to keep grams; that is going to go upstairs.1144

You are told that 1 pound is 454 grams; 1 pound on the bottom and 454 grams on top.1147

That is going to give us an answer of 553.88 grams.1155

Here we have three sig figs; our answer is going to round to 554 grams.1162

That again, that is what we call dimensional analysis.1170

It is important to really master this because we are going to be using this incredibly heavily throughout all our lessons in general chemistry.1173

One last example then; suppose a diaper cost us 35 cents.1185

You have a newborn who goes through about 14 diapers a day.1189

How much is spent on diapers in one week?1192

Suppose you have a diaper costing 25 cents; that is actually a statement already.1196

We are told that 25 cents costs us each diaper; 25 cents per diaper.1203

We are trying to multiply through and cancel units.1211

The other item we see here that has diapers is 14 diapers a day; that is another ratio.1214

14 diapers goes upstairs to get cancelled; then 1 day on the bottom.1220

Finally I want to get my answer into week or dollars per week.1226

This is going to now be 7 days on top to get cancelled, divided by 1 week on the bottom.1233

When all is said and done, you are going to get an answer of1241

24 dollars and 50 cents per week to be spent on diapers.1243

As you can see, the reason why we are doing these examples1250

that are not chemistry yet is to show you that dimensional analysis1253

which we use in chemistry and the physical sciences can actually be very easily applied to everyday life.1256

The next type of dimensional analysis deals with unit conversion.1266

You have heard of terms like centi and milli and kilo before.1271

But how do you convert between the three?1276

We are going to learn that dimensional analysis is all behind this; converting between units.1278

You should ask your chemistry instructor which units you actually have to know.1284

But let me go ahead and just point out a few.1288

Mega is 106; kilo is 103; deci is 10-1; centi is 10-2.1291

Milli means 10-3; micro is 10-6; nano is 10-9.1301

Once again please ask your instructor if you have to memorize any of these1309

if at all or if they are going to be given to you.1313

Now that we have been introduced to these prefixes and what they mean,1317

let's go ahead and see how we can use them in calculations.1322

For example, 500 milligrams is equal to how many kilograms?1326

What I always like to say is that sometimes it helps to convert to the base unit first.1330

What I mean by the base unit is that it has no prefixes.1335

In other words, let's get to the unit with no prefixes first.1341

First step is to go from milligrams to regular grams then onto kilograms.1347

This is going to be a two step process.1355

500 mg, just going to set it up, times something over something.1358

The first step is to get g; g goes upstairs; mg goes downstairs to get cancelled.1364

Once I am in g, now I can go to kg; times something over something.1369

That is going to give us our answer in units of kg.1375

You see that g is upstairs here; it is going to go downstairs to get cancelled.1378

Kg goes upstairs to get carried through to the final answer.1383

What numbers do we put in and where?--the rule is the following.1388

For the prefix multipliers like kilo and milli, you put the multiplier with the base unit; put multiplier with the base unit.1393

Once again you should put the multiplier with the base unit.1409

For milli, milli stands for 10-3.1414

That is going to go with the base unit here; 10-3 on top; 1 on the bottom.1419

Kilo stands for 103; that is going to go with the base unit.1424

1 goes to kg; 103 is going to go the g.1428

Once again you always put the multiplier with the base unit.1433

Then you get your answer in kilograms.1441

Let's go ahead and do a last one; this is centimeters to micrometers.1443

The first step is to go from centimeters to regular meters; then from regular meters on to micrometers.1447

34.1 centimeters times something over something is going to give me my answer in meters.1457

Cm goes downstairs to get cancelled.1465

M goes on top to get carried through to the final answer.1467

Once I am in meters, I can then go on and get micrometers; times something over something.1471

That is going to give me my answer in micrometers.1476

You see that m is on top; it has to go downstairs to get cancelled.1479

Then micrometers goes upstairs to get carried through to the final answer.1483

When we look up the prefix for centi, centi stands for 10-2.1488

That goes with the prefix-less unit; 10-2 on top; 1 on the bottom.1492

When we look up the multiplier for micro, it is 10-6.1497

That goes downstairs with meters; 1 on top.1503

That is going to get you your answer in units of micrometers.1507

To summarize, when you perform measurements, you want to gauge how well you are doing.1514

We can do this by percentage error calculation which again is going to tell us a little about your accuracy.1519

We can also calculate what is called the standard deviation.1527

That is going to tell us of how precise you were.1531

We also learned the concept of significant figures; significant figures is related to precision.1535

Finally we learned a very fundamental concept in all of the physical sciences which is dimensional analysis1545

which follows the same basic pattern where we can go from unit A to unit B using a conversion factor.1551

There is our summary; now let's go ahead and tackle some sample problems.1567

An intramuscular medication is given at a dosage of 5.00 milligrams per kilogram of body weight.1571

If you give 0.425 grams of medication to a patient, what is the patient's weight in pounds?1584

0.425 grams of medication is here.1592

Somehow we want to go from grams of medication to whatever the question is asking for which is pounds of body weight.1597

In addition we see in the first sentence that we have a statement here1611

that 5 milligrams of medication are given per kilogram of body weight.1615

That represents a very nice ratio.1619

5.00 mg of medication for every kilogram kg of body weight; that is our conversion factor.1622

This unit is 5 milligrams of medication; we are given 2.45 grams of medication.1637

We have to get the 0.425 grams of medication into milligrams first.1643

0.425 grams of medication times something over something is going to give us our answer in units of milligrams of medication.1648

Milligrams goes upstairs; g goes downstairs.1659

When we look up the prefix for milli, it is 10-3.1662

That goes with the prefix-less unit on the bottom; then 1 on top.1666

That is going to give us milligrams of medication which is going to be 425.1670

We can then take the 425 milligrams of medication, multiply it by something over something.1677

That is going to give us our answer in kilograms of body weight.1684

Mg is going to go on the bottom; kg is going to go on top.1698

We know that from the sentence, it is 1 kg for every 5.00 mg.1701

Finally we can then take our kilogram of body weight and we can go to pounds of body weight.1708

We do that from the conversion factor where 1 kilogram is equal to approximately 2.20 pounds.1717

We are going to take kilograms of weight, multiply it by something over something.1726

That is going to give us our answer in pounds.1732

Pounds goes upstairs; kg goes downstairs; just 2.20 divided by 1.1736

When all is said and done, we should get an answer of 187 pounds using the correct number of sig figs.1743

That is our lesson from general chemistry on quantitative tools.1752

I want to thank you for your attention.1758

I will see you next time on Educator.com.1760

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is applications of aqueous equilibria.0003

We are pretty much going to build upon what we learned in the last lesson when we talked about equilibria for acid base chemistry.0008

We are going to see how it can easily apply to other aspects of general chemistry.0015

We are going to learn how to calculate the pH of an acid base mixture.0020

We are going to learn about a very important system in chemistry which is called buffers.0024

Following buffers, we are going to learn a unique experiment called acid base titrations0030

followed by other aspects of chemical equilibria, namely what we call solubility equilibria and compensation equilibria.0037

We will always wrap up the session as always with a summary followed by a sample problem.0047

When we did equilibria before, every equilibria we have discussed has always involved a direct reaction with water.0054

For example, it has always been the acid HA plus H2O.0064

Or it has always been a base plus H2O.0069

But what happens when we react directly a Bronsted-Lowry acid with a Bronsted-Lowry base?0073

When this happens, you don't get equilibrium immediately.0079

Instead the very first step that occurs is going to be a neutralization reaction.0082

Neutralization involves complete consumption of one or both of the reactants.0088

We always get the generic reaction where we have an acid HA reacting0093

with a base aqueous going on to form HB1+ aqueous and A- aqueous.0099

Once again when we do these neutralization reactions, we use a single arrow only0108

because one or both of the reactants is going to be consumed completely.0116

After neutralization occurs, then we can apply whatever we learned last time.0126

Basically whatever remains is going to be reacting with water in a typical aqueous equilibrium.0134

In that case, it is either going to be the acid HA aqueous plus H2O liquid.0143

Or it is going to be the base reacting with H2O liquid.0153

Those are going to be our main reactions after neutralization has occurred.0163

Let's go ahead and take a look at a representative problem and jump right into a calculation.0167

What is the pH of the solution that results from adding 30.0 milliliters0173

of 0.1 molar sodium hydroxide to 45.0 milliliters of 0.100 molar acetic acid?0178

Now we have a strong base reacting with a weak acid here.0187

Step one, step one is to always write out the neutralization reaction.0194

Let's go ahead and write that out.0204

It is going to be hydroxide aqueous reacting with the acetic acid, CH3CO2H aqueous.0206

That is going to go on to form water liquid and acetate CH3CO21- aqueous.0216

We can go ahead and proceed to fill in the ICE table as usual.0226

Here hydroxide is what we want, moles.0230

That is 0.0300 liters times 0.100 molar.0233

That is going to give us 0.00300 moles of hydroxide.0240

The acetic acid we have 0.0450 liters times 0.100 molar.0246

That is going to give us 0.0045 moles of acetic acid.0257

Acetate is going to be zero.0263

Remember we said that one of these reactants or both is going to be completely consumed.0266

We can see that this is a 1:1 ratio here.0271

If we have amounts of both reactants, that is like a limiting reactant problem which we discussed a long time ago.0278

Here because it is 1:1 ratio, the reactant that is present in smaller quantity is going to be consumed entirely.0284

Here if this is I-C-E, this is then going to be -0.0300 moles.0291

This is -0.00300 moles; this is going to be +0.00300 moles.0300

The hydroxide gets completely consumed.0310

Most of the acid is going to be eaten up here and consumed or left with only 0.0015 moles.0314

This is going to be 0.00300 moles.0320

Now step two, step two, we are then going go ahead and proceed on to do...0326

Whatever is left standing is going to react with water; step two, react with water.0335

Now this is going to be like our previous lecture.0341

We are going to go ahead and select the acid, CH3CO2H aqueous.0346

That is going to react with water going on to form CH3CO21- aqueous and H3O1+ aqueous.0352

We carry everything down.0367

The final values of table one become the initial values of table two.0368

This is now 0.0015 moles; this is 0.00300 moles; this is 0.0372

This is going to be ?x, +x, and +x.0380

At equilibrium, 0.0015 moles minus x, 0.00300 moles plus x, and x.0386

The k for acetic acid is something that should be given to you or you look it up.0395

That is 1.8 times 10-5 which is relatively small.0400

All of this reduces to 0.00300 times x divided by 0.0015.0404

Because the volume is the same for each of these moles, the liters are really going to cancel out.0416

I can directly just use the moles.0424

We get an x value of 9.0 times 10-6 molar which is the hydronium ion concentration at equilibrium.0427

The pH is going to be 5.05.0438

Again always make sure your answer makes sense.0443

Here we have an acidic pH which we should have because all we have leftover is acid and no more base.0446

Again this is how you calculate the pH of an acid base mixture.0455

Let's now go ahead and examine how to control a change in pH.0460

When a solution contains both a weak acid/base and its conjugate in appropriate amounts,0467

there is going to be a resistance to small changes in pH.0474

These are solutions that we call buffers.0477

A buffer is going to contain again a weak acid HA plus its conjugate base A-.0482

Or it is going to contain a weak base B and its conjugate acid BH+.0493

These are typically going to be in the form of salts; usually in the form of a salt.0501

Anytime you see these two mentioned in the same sentence, they simultaneously occur in the same mixture.0515

We have what is called a buffer; once again buffers resist small changes in pH.0523

The reason why they work is because of the following.0531

If you disrupt the buffer by say adding acid, the solution is going to resist a drop in pH.0534

The reason is because if we add acid to the buffer, this acid is going to react with any base present.0540

Because we have both acid and base present as a buffer, this is good to go.0549

We are going to form HA aqueous and H2O liquid.0555

Or it can react with H3O1+ aqueous and base aqueous.0561

We are going to form BH plus aqueous and H2O liquid.0568

Right away you see how the buffer can resist change in pH because water is formed0574

as a product which helps to counteract5 the effect of the added acid; dilutes H3O1+ added.0581

The other way to disrupt the buffer of course is by base, adding base to the buffer.0594

The solution is going to resist a rise in pH.0598

The reason is because if we add base, say hydroxide, that will then react with possibly HA if it is an acid in solution.0601

Or the base can then go ahead and react with the conjugate acid if it is a basic buffer.0611

In either case, we see that we still form water.0619

Again it is the formation of water that helps to counteract the effect of added base; dilutes OH- added.0626

Let's go ahead and jump right into a calculation.0643

A buffer is prepared by mixing 0.50 molar of acetic acid with 0.50 molar0648

of sodium acetate where the k is given to you for acetic acid.0653

Let's go ahead and calculate the pH first.0657

Let's go ahead and write out the equilibrium.0663

CH3CO2H aqueous plus H2O liquid goes on to form CH3CO21- aqueous and hydronium aqueous.0664

This is 0.50 molar; this is 0.50 molar; this is 0.0682

This is going to be ?x, +x, +x.0688

This is going to be 0.50 minus x, 0.50 plus x, and x.0692

Ka is given to us to be 1.8 times 10-5.0698

This is then approximately 0.50 times x divided by 0.50.0702

You see that we can get an x of 1.8 times 10-5 molar.0708

The pH is going to be 4.74.0717

There is a shortcut it turns out.0724

Anytime we have a buffer, it turns out we don't need to set up an ICE table problem.0726

It turns out that we do have a nice shortcut.0732

This is what we call the Henderson Hasselbalch equation.0735

The Henderson Hasselbalch equation is the following.0753

It is pH is equal to the pKa plus the log base 10 of0756

the concentration of the base initial divided by the concentration of the acid initial.0761

This is going to be for an acidic buffer.0767

If you have a basic buffer, it is still pH is equal to pKa.0772

But this time it is going to be the log of the initial concentration0776

of base divided by the initial concentration of the conjugate acid.0778

The nice thing about this equation is that we don't need equilibrium values for the conjugate base and the acid.0785

We only care about initial values.0791

The Henderson Hasselbalch equation, you must know it is only an approximation.0793

But it does work out many times.0799

We can go ahead and plug in our values here for the previous problem we just did.0802

We are going to get pH is equal to ?log of 1.8 times 10-5.0807

That is equal to the log of 0.50 divided by 0.50.0814

When all is said and done, we get 4.74.0819

As you can see, this matches up very nicely with the previous problem where we did the ICE table.0822

Once again when you see a buffer problem, think Henderson Hasselbalch equation.0831

Next let's go ahead and see what happens to the pH when we disrupt this buffer by adding strong base.0837

Now we are going to add 0.010 moles of sodium hydroxide to exactly 1 liter of the buffer.0843

Once again because we are adding a strong acid or a base, step one is to do the neutralization step.0853

Because we are adding a strong base hydroxide, this is going to go0861

ahead and react with the weak acid that is already in solution.0866

That is going to go on to form acetate and water.0871

This is going to be 0.010 molar; this is 0.50 molar.0881

This is going to be 0.50 molar; we don't care about water.0886

Because again this is a 1:1 ratio, remember what we always said.0892

Anytime there is neutralization, either one or both of the acid and base are going to be consumed.0895

My 1:1 ratio tells me I am going to choose a smaller amount that is going to get consumed completely.0901

This is -0.010; this is -0.010 molar; this is +0.010 molar.0907

This goes to 0; this goes to 0.49 molar; this goes to 0.51 molar.0916

We can go ahead then and step two...0923

Remember we would set up ICE table like we taught you at the beginning of this session.0926

But because this is a buffer, we can use Henderson Hasselbalch directly which is always nice.0931

It saves us a lot of time.0937

pH is equal to the pKa plus the log of acetate initial divided by the initial concentration of acetic acid.0942

Remember how we said last time, the final values of table one become initial values of table two.0960

That is still going to apply.0964

This 0.51 goes right there; this 0.49 goes right there.0967

When all is said and done, we expect a pH that is going to be0972

greater than 4.74 because we added strong base but not too much greater.0977

When we do this, we get pH of 4.76.0982

As you can see, the buffer was successful in resisting this change in pH.0986

We only went up 0.02 units.0993

The next thing we are going to discuss is how to make a buffer and what we call buffer capacity.0998

Exactly 1.25 liters of a buffer solution is prepared from mixing 23.5 grams of sodium dihydrogen phosphate1003

and 15.0 milliliters of 14.7 molar phosphoric acid where the Ka is given to you.1014

Calculate the pH.1019

Because this is a buffer, again we can use Henderson Hasselbalch equation directly.1020

The pH is equal to the pKa plus the log of the conjugate base initial divided the acid initial.1025

The hardest part of this is just to figure out which one is the conjugate base and which one is the acid.1036

This is going to be equal to pKa plus the log of...1041

the conjugate is going to be the H2PO41- concentration.1045

Of course the acid is going to be the phosphoric acid initial concentration.1052

We can go ahead and again use this to calculate the pH of this buffer by inserting everything directly in.1062

This 23.5 grams of course we need to get to moles first.1073

This 14.7 molar is going to go right there.1080

Remember the final volume is exactly the same.1084

The 1.25 liters is same for both A- and HA which means the volume term is going to cancel.1089

All you need is just really the moles.1106

Here is 15 milliliters of 14.7 molar phosphoric acid.1110

Then you can convert the 23.5 grams of the sodium dihydrogen phosphate directly to moles.1114

A buffer can only do so much.1122

A butter can accommodate so much extra base or extra acid.1124

It is what we call buffer capacity.1129

A buffer is considered to be effective when the pH is within 1 unit of the pKa.1132

In other words, pH is equal to +/-1 of the pKa.1138

What that means by going from the Henderson Hasselbalch equation is that the ratio of A- to HA is less than 10.1145

Anything outside this range, the buffer will not be effective.1153

Why?--because you have too little of either the acid or base to counteract any effect from disturbing the buffer.1157

That is our brief discussion on buffer preparation and buffer capacity.1167

Following buffers is going to be now a specific experiment.1176

This is what we call an acid base titration.1180

In an acid base titration, an acid or base of known concentration is used to determine the concentration of an unknown acid or base.1183

This is known as standardization of the unknown.1192

The solution that is known is what we call the analyte.1195

Excuse me... the solution of known concentration is what we call the titrant.1201

The solution that is unknown, what we are trying to standardize, what we are1209

trying to find the concentration of, this is what we call the analyte.1213

The basic setup is going to be the following.1219

We are going to have a pretty long thin cylindrical tube.1224

This is usually 50.00 milliliter; this is what we call a buret.1229

The burets are very very precise.1236

You notice I use two sig figs after the decimal.1238

This is always two within 0.01 milliliter.1241

There is going to be an Erlenmeyer flask on the bottom.1250

Usually the buret is going to have the titrant.1253

Usually the flask is going to have the analyte.1262

With this flask with the analyte inside the flask, there is also going to be called what is called indicator.1266

Indicator solution is going to change color.1274

When it changes color, this is what we call the equivalence point, also known as the stoichiometric point.1281

At this point, we have complete neutralization.1299

When we have complete neutralization, the moles of HA acid is equal to the moles of the base.1308

You see that we can use mole to mole ratio here between acid and base to backtrack and get the concentration of the unknown.1319

But let's go ahead and go through a typical problem first.1329

Calculate the pH at the equivalence point when so much HClO is titrated with 1 molar KOH.1333

Anytime you see volume and molarity, you have heard me say this before.1342

That is always a good 1-2 combo; you are going to get moles.1347

Let's go ahead and do that.1349

0.250 liters times 0.0350 molar, that is going to give us 0.00875 moles of HClO.1350

Because we are at the equivalence point, this is also the same number1365

of moles with the KOH by definition; by definition of equivalence point.1371

Because we now have both of our amounts, let's go ahead and set up the neutralization reaction.1389

HClO plus OH1-, that goes on to form H2O liquid and ClO1- aqueous.1394

This is 0.00875 moles; this is 0.00875 moles; this is going to be 0.1406

Remember last time how when we have an acid and base neutralization reaction,1415

we said that either one or both of the reactants is going to get consumed entirely.1419

This is a good example where both of them gets consumed entirely.1424

Why?--because we have a 1:1 ratio.1427

Because this is the equivalence point, we have equal amounts.1430

This is going to go -0.00875 moles; this is going to be -0.00875 moles.1433

This is going to be +0.00875 moles.1444

This goes to 0; this goes to 0; this goes to 0.00875 moles.1448

Final values of table one become initial values of table two.1457

But this is not a buffer anymore; why?--because we have zero amount of HClO.1460

Let's go ahead and do our second setup; it is the last man standing.1465

The only thing remaining is ClO1- aqueous plus H2O liquid going on to form HClO aqueous plus OH1- aqueous.1471

When all is said and done, we are going to be using Kb.1486

We are going to use Kb to get the concentration of hydroxide.1491

From there, you can go ahead and get the pH.1496

But don't forget, the important thing is to take this moles amount and get to molarity first.1500

When you get to molarity, you need to divide by the total volume; very very important.1509

This is a typical type of question to calculate the pH at an equivalence point of a titration.1523

During a titration, something that is done that is very typical is to monitor the pH yielding what is known as a titration curve.1532

When we look at a typical titration curve, it is going to be the following: pH versus volume of titrant added.1541

We can have different types of titration curves.1556

Each one is going to be a different experiment.1558

We can have a curve that looks like this.1561

We can have a curve that looks like this.1569

In each of these cases here, the pH is going to be greater than 7 here.1575

Here the pH is going to be less than 7 here.1583

Of course the final situation is pH equal to 7.1587

Of course you can have the mirror image right here, pH equal to 7.1599

Let's go ahead and discuss what each of these means right now.1607

The first situation, here the pH is acidic.1610

The only thing you have initially is your acid.1615

This is going to be a weak acid titrated with strong base.1619

Here you see that the initial pH when no titrant has been added is basic.1629

This is going to be a weak base titrated with strong acid.1635

Let's go back one.1645

When the weak acid is titrated with a strong base, the pH is always going to be basic at the equivalence point.1647

When the weak base is titrated with strong acid, the pH is going to be acidic at the equivalence point.1655

Here pH is neutral.1662

When the pH is neutral at the equivalence point, you have one of two possibilities.1665

Either a strong acid titrated by strong base or you have a strong base titrated by strong acid.1670

A titration curve typically can be broken up into four regions.1693

I am going to choose the weak acid titrated with a strong base as our typical example.1699

Here is region one, region two; region one, region two, region three, and region four.1706

At the end of this lesson, you are going to be able to calculate the pH along any point of the titration curve.1719

You are basically in one of four regions.1725

Each region is going to have little own strategy.1727

Let's go ahead and go over them.1730

In region one, which is basically the y-intercept; in other words, zero base added.1732

The equilibria that you would use is an equilibrium from last lecture which is the1747

weak acid HA reacting with water going on to form H3O1+ aqueous and A- aqueous.1755

It is a typical ICE table setup problem; you use Ka to solve for pH.1768

Region number two, you notice that in region number two, the pH doesn't change that much.1777

It is really a plateau.1784

The pH doesn't change that much because we have a functioning buffer.1787

Region two is what we call the buffer region.1790

Again you don't need an ICE table setup to calculate the pH in the buffer region.1797

You simply use the Henderson Hasselbalch equation.1802

pH is equal to pKa plus the log of the concentration of A- initial divided by the concentration of HA initial.1806

There is special point of interest in the buffer region.1818

It is right here; that is what we call the midpoint.1821

That is what we call the midpoint of the titration.1828

At the midpoint of the titration, we have this very special relationship where pH is equal to the pKa.1831

In other words, the concentration of A- initial equals the concentration of HA initial.1836

It is called the midpoint because we are exactly halfway done with the titration in order to get to the equivalence point.1847

In other words, the moles of hydroxide added is going to be equal to half the moles of HA present.1853

Again the moles of hydroxide added is half the moles of HA present.1866

We are halfway to the stoichiometric point; halfway to equivalence point.1870

Region number three, in region number three, we get a sudden spike in pH.1885

That is because the buffer no longer is functioning.1894

We are beyond the capacity of the buffer.1897

The point of interest is of course the equivalence point.1901

We went over already how to calculate the pH at the equivalence point.1908

Remember it is going to be here.1912

Remember the acid is completely gone; the base is completely gone.1916

All you have left is the conjugate base A- reacting with water going on to form HA aqueous and hydroxide aqueous.1919

For this one, you are going to use Kb.1933

As you can see, that is why pH is going to be greater than 7.1938

Finally region number four, region number four, you have nothing but excess hydroxide added, nothing more than that.1941

That is going to be a typical pH is equal to ?log of H3O1+ equation.1957

We are going to go through a problem later on at the end of this lecture discussing that.1967

That is our basic introduction to titration curves.1973

The four regions, each require a different strategy to compute the pH.1979

Let's now take a look at another application of aqueous equilibria.1984

It is what we call solubility equilibria.1988

Remember a long time ago at the beginning of the first couple of presentations, we went over our table of solubility rules.1991

We said that some salts were relatively soluble; we said that some salts are insoluble.2000

It turns out that a really all solid salts are going to be soluble in water to a measurable degree.2006

Some of course more so than others.2014

Salts that are very soluble will have the following equilibrium lie far to the right.2016

We can go ahead and imagine an ionic solid of the formula MX solid2021

dissolving into the respective cation and anion, M+ aqueous and X- aqueous.2029

The equilibrium expression for this type of reaction would be the concentration of M+ times the concentration of X-.2041

We give this equilibrium constant a very specific name.2051

This is what we call the Ksp or the solubility product constant.2055

Basically as you can see, the larger this constant, the more ions you get.2063

In other words, the more soluble the solid is; solubility increases with Ksp.2069

Salts that are relatively insoluble such as carbonate will have the equilibrium lie to the left instead.2082

These are going to have smaller Ksp values.2090

We can use Ksp to predict if a solution will form a precipitate2099

of the ionic solid given initial amount of the cation and anion.2103

This is done by simply assessing the reaction quotient Q, something we discussed in a previous lecture.2108

We compare it to Ksp.2114

Just like Le Chatelier principle, we are going to use the same principles here.2116

If Q is less than Ksp, we have the reaction shifting to the right because we don't have enough product.2121

Because it shifts to the right, that is away from the solid.2129

Precipitation will not occur.2132

However if we find and calculate that Q is greater than Ksp,2135

the reaction will shift to the left because we have too much product.2140

Because we are going toward the solid, this is the same as saying that precipitation will occur.2144

Now that we have gone through solubility equilibria and the solubility product constant Ksp,2154

let's go ahead and examine some factors that affect solubility.2159

When a solid salt is dissolved in a solution that already contains either the cation or2164

anion of the salt, its solubility will be found to be lower than in pure water.2169

This is what we call the common ion effect.2176

Let's go ahead and look at an example.2178

The solubility of lead(II) chloride in water is 3.9 grams per liter.2181

Calculate its solubility in 0.55 molar sodium chloride solution.2186

We have solubility in water to be 3.9 grams per liter.2192

We are asked to calculate the solubility in salt water.2197

Let's go ahead and set it up.2203

Remember we must always write out the proper equilibrium.2205

In this case, because it deals with solubility, it is going to show the ionic solid dissolving water into cation and anion.2209

PbCl2 is going to go ahead and form Pb2+ aqueous plus two Cl1- aqueous.2218

You see that we do have an initial amount of sodium chloride.2229

It is 0.55 molar and zero Pb2+; this is going to be +.2233

We don't use x anymore; instead we use s as a convenience to indicate solubility.2241

This is going to be +2s; this is going to be s.2247

This is going to be 0.55 plus 2s.2250

The Ksp for the lead(II) chloride is going to be given to you.2253

Or you look it up in your textbook; that is 1.7 times 10-5.2257

That is going to be equal to the concentration of Pb2+ times the concentration of Cl1- squared.2262

That is going to be equal to s times 0.55 plus 2s squared.2269

Because Ksp is relatively small, the approximation rule still applies.2275

That is s times 0.55; we can go ahead and solve for s.2281

That is going to be equal to 2.81 times 10-5 moles per liter.2287

That is in salt water; in pure water, the solubility was 3.9 grams per liter.2298

You can easily convert grams per liter to moles per liter.2305

We get 0.015 moles per liter here.2308

As you can see, the solubility in salt water is much less than the solubility in pure water.2313

In other words, the common ion effect reduces the solubility of an ionic salt.2321

Common ion effect reduces solubility.2329

The common ion effect is one factor that affects solubility.2344

Another factor is when the solid salt contains the conjugate of a weak acid.2348

It is found that these salts are going to be more soluble when in acidic solution.2353

The typical example is going to be for example a metal carbonate, MgCO3 for example.2359

MgCO3 solid goes on to form Mg2+ aqueous plus CO32- aqueous.2366

Something that can drive the reaction to the right is addition of acid.2379

Why?--because H3O1+ aqueous will want to go ahead.2384

It is going to react with the conjugate base of carbonic acid.2390

CO32- aqueous plus H3O1+ aqueous is going to go ahead and reform part of the carbonic acid.2398

We are going to first bicarbonate aqueous and H2O liquid.2410

As you can see, because hydronium is going to react with carbonate, it reduces the amount of carbonate.2417

That is going to drive the equilibrium to the right.2423

CO32- will be consumed by acid driving the dissolution of magnesium carbonate.2427

Finally the last factor that influences solubility is of course temperature.2443

This is going to be true for all ionic solids.2449

Solubility is always enhanced with increasing temperature.2451

Basically the reason is because the thermal energy is going to become sufficient enough that2456

you can overcome the electrostatic attraction between the cation and anion in the ionic solid.2462

The last type of equilibria is what we call complexation equilibria.2473

Aqueous solutions that involve transition metal cations can also form from their aqueous cation and anion counterparts.2477

Can also form... excuse me... from their aqueous cation and anion counterparts.2488

Let's go ahead and show a representative equilibrium showing the formation of what we call a complex ion.2494

For example, Fe3+ aqueous can react with six equivalents of carbon monoxide gas.2499

That can go ahead and form Fe(CO)63+ aqueous.2512

We are going to get into these types of compounds in a later presentation.2523

But this is what we call a complex ion.2529

Basically it involves a transitional metal and what we call a ligand.2535

In this case, the ligand is going to be carbon monoxide.2543

The extent to which the complex ion forms can be described by its equilibrium constant.2547

In this case, K is equal to the concentration of Fe(CO)63+ divided by the2552

concentration of Fe3+ times the concentration of carbon monoxide raised to the sixth power.2562

Because this is involving complex ion formation, we don't just call this Keq anymore.2571

But instead this is what we call Kf or the complex ion formation constant.2576

The main difference between Ksp and Kf is that they are opposite.2584

Ksp represents a breakup of a compound.2588

Kf represents the formation of a compound from its constituent ions.2594

While Ksp values tend to be very small, Kf values actually tend to be2599

just the opposite--quite large, several orders of magnitude, much much much larger.2605

Again that is what we call complexation equilibria.2611

Let's go ahead and summarize this very important part from general chemistry.2616

Equilibrium can be applied to various systems including acid base mixtures, buffers, titrations, solubility, and complex ion formation.2620

We found a very nice shortcut that we call the Henderson Hasselbalch equation.2631

This can be used to approximate the pH of a buffer system.2636

We found that during a titration, we can monitor the pH to yield or generate what is called a titration curve.2641

Finally we also saw that solubility product constants can quantify the extent to which an ionic solid is going to be soluble in water.2650

Let's now go ahead and jump into a representative sample problem.2660

This sample problem is quite long.2665

I have chosen an acid base titration on purpose because this is where students tend to trip up.2667

I want to go through how to calculate the pH along a titration curve in each of the four regions that we pointed out.2675

Exactly 1.36 grams of trimethyl amine is going to be dissolved in 250 milliliters of water.2683

Then 23.6 milliliters of hydrochloric acid was required to reach the equivalence point.2691

Basically we are asked to calculate the following.2698

The pH at the beginning of the titration which is region one.2700

The pH at the midpoint or halfway point... sorry about that... this should read or.2707

Because we are at halfway point, this is going to be region two.2718

The pH at the equivalence point which is region three.2724

Finally the pH after 27.50 milliliters of acid was added.2728

You notice that this 27.50, that is greater than the 23.60 which means we are after the equivalence point.2734

We are out of region three; we are into region number four.2742

Let's go ahead and do this.2748

When we go ahead and calculate the pH at the beginning of the titration, remember no titrant has been added yet.2751

Initially we only have the trimethyl amine; this is a weak base.2757

Let's get the 1.36 grams into molarity right away, 1.36 grams of the trimethyl amine.2765

We are going to go ahead and divide this by its molar mass which is approximately 59 grams.2777

That is going to give us 0.023 moles of the trimethyl amine.2783

We are going to go ahead; we are going to divide this, 0.023 moles.2792

We are going to divide it by the total volume at that point which is 0.250 liters.2800

We get the initial molarity of the base to be 0.092 molar.2804

Now it is just a typical equilibrium problem with direct reaction with water.2812

Let's go ahead and write out the equilibrium then.2819

Trimethyl amine aqueous plus H2O liquid goes on to form HN1+(CH3)3 and hydroxide aqueous.2821

Let's go ahead and plug in all our values.2841

That is 0.092 molar and 0 and 0; -x, +x, and +x.2843

This is going to be 0.02 minus x, x, and x.2851

Of course we are going to use Kb to go ahead and get x.2855

X is then going to give us the concentration of OH-.2862

From there, we can go ahead and get the pH.2867

Of course we expect a basic pH greater than 7 because we only have base present in solution.2870

No titrant has been added in the form of strong acid yet.2878

That is part A; part B is very simple, the pH at the midpoint.2883

Because we can use the Henderson Hasselbalch equation, the pH at the midpoint is simply equal to pKa.2891

That is a very quick calculation there; not bad.2900

Part A is done; part B is done; region one and region two.2903

Let's go ahead and do part C now, the pH at the equivalence point, region three.2907

At region three, at the equivalence point, the moles of the base is going to be equal to the moles of the added acid.2915

Let's go ahead and get the moles of the acid then.2933

0.02360 liters times 0.974 molar, that is going to be equal to 0.0230 moles of HCl.2936

The acid is monoprotic; we don't have to worry about diprotic or triprotic.2951

Then it is a 1:1 ratio with this amine.2957

This is going to be equal therefore to 0.0230 moles of the trimethyl amine that we had initially.2960

Let's go ahead and write out the neutralization reactions.2970

That is going to be HCl aqueous plus the trimethyl amine aqueous.2973

That is going to give us HN1+(CH3)3 and then Cl1- aqueous.2983

Let's go ahead and plug in all our values--0.0230 moles, 0.0230 moles, and then 0 and 0.2992

Again this neutralization reaction, 1:1 ratio, we have the identical amounts of each reactant.3004

All the reactants are going to be consumed entirely; this is -0.0230 moles, -0.0230 moles.3010

This is going to be +0.0230 moles and +0.0230 moles.3020

This goes as 0; this goes as 0.3029

This is 0.0230 moles; this is 0.0230 moles; the neutralization step is done.3033

Now we can go ahead and proceed on to write the equilibrium with direct reaction with water.3043

You have to choose: is it going to be the conjugate acid or is it going to be Cl-?3049

We already know from a previous lecture that Cl- is the conjugate of HCl.3056

It will not react with water to reform HCl.3063

Therefore you are going to select the conjugate acid of the amine.3066

HN(CH3)31+ aqueous direct reaction with water goes on to form H3O1+ aqueous and the trimethyl amine weak base.3071

This is going to be 0.0230 moles, 0, and 0.3092

Remember final values of table one become initial values of table two.3100

This is ?x, +x, and +x; this is 0.0230 minus x, x, and x.3103

You are going to use Ka for this expression.3113

Ka is going to be equal to the following expression: x squared over 0.0230 moles minus x.3116

Again we need to get moles into molarity.3128

This is simply going to be 0.0230 moles divided by the total volume at the equivalence point which is going to be 0.2736 liters.3132

When all is said and done, you can go ahead and solve for x.3143

X will give you the hydronium ion concentration which will then get you to pH.3148

As you can see, this titration was a weak base with a strong acid.3154

We had better expect the pH to be acidic or less than 73159

as we saw in the previous slide where we introduced the graphs.3163

That is going to be the case because we only have hydronium being formed with no hydroxide.3168

That is region three.3175

Finally let's go ahead and solve how to do region four.3177

In region number four, we have nothing but excess titrant added; region four, extra HCl added in this case.3181

What you need to calculate is the amount of HCl remaining after neutralization.3194

Need to determine amount of HCl leftover after neutralization has occurred; after neutralization.3201

We know that 23.6 milliliters was required for neutralization.3222

Here we are given a total of 27.5 milliliters; we are just going to subtract.3227

That tells us that 3.9 milliliters of HCl still remains after neutralization.3234

Because this is a strong acid, we can use the pH equation directly.3243

Right now we need to get this into molarity.3248

We are going to say 0.0039 liters of the HCl.3255

We are going to multiply that by the molarity which is 0.974.3259

That is going to give us moles.3265

To get the molarity, we are going to divide by the total volume at that point.3267

The total volume at this point is actually 0.2775 liters.3271

That is going to give us our concentration of HCl which because HCl is monoprotic, gives us the concentration of H3O1+.3279

From there, we can go ahead and calculate the pH.3289

Again because this is region four of this titration curve, the pH had better be3293

much less than 7 because this is the titration of a weak base by a strong acid.3299

At this point, we have nothing but strong acid remaining.3304

Again this as you can see, each of the four different regions of3311

a titration curve require its own different strategy for solving the pH.3315

I want to thank you for your time; I will see next time on Educator.com.3321

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is on entropy and free energy.0003

This is basically going to be our second and final lecture on thermodynamics.0009

As usual we are going to start off with a brief introduction.0013

We are going to get into something very important in thermodynamics which is what we call entropy.0017

It is going to be related to spontaneous processes.0023

After we get into entropy, we are going to go through the second and third laws of thermodynamics followed by problem solving.0027

After entropy, we are going to then get into what is called Gibbs free energy followed by definition and problem solving.0038

Then we are going to go ahead and conclude with a summary and a pair of sample problems.0047

We have consistently said, you have heard me say over and over again that nature favors states of low energy.0055

In other words, reactions occur to obtain a lower energy state in the end.0063

This is also known as a downhill reaction.0069

In other words, nature tends to favor reactions that start in a high energy state and go to a low energy state.0072

That is going to be the natural tendency.0079

In this lecture, we are going to complete our discussion from thermodynamics and quantify the above statement.0081

In the end, we are going to see that nature not only favors0089

states of low energy but also states of what we call high entropy.0091

Consider the following closed reaction vessels that contain gas molecules.0099

Let me go ahead and draw a closed reaction vessel here.0104

We can make this one gas particle; the red one can be another gas particle.0109

The black one can be another gas particle; we are now going to...0115

Because these are gases, gases naturally expand; what can happen is the following.0120

What are some possible different configurations for these three gas particles if they naturally expand?0128

Situation number one is when all three gas particles are on the same side.0135

We can have a second situation when only the blue and the black are together and the red is different.0140

We can have another situation where the blue is by itself and the red and black are together.0148

You notice that these two situations here are actually the same0162

where we have two gas particles on one side versus one on the other.0169

It turns out that configuration number two actually has more what we call microstates.0177

Once again configuration number two is going to have what we call more microstates.0196

We can even do one more to show you how this is possible.0204

Now red and blue can be together with black by itself.0209

This is also considered a configuration number two.0215

Configuration number one, there is only one way to put three particles on the same side.0220

Configuration number one has one microstate.0227

Because gases expand, we are going to see that configuration number two is going to be the more probable configuration.0237

Why?--because there are a higher number of ways to do it.0246

There are a higher number of microstates to achieve, configuration number two, where one gas particle is on one side.0250

Two gas particles are on the other side.0257

We symbolize the term microstate with capital W.0260

For configuration number one, W is 1; for configuration number two, W is simply 3.0264

We can quantify that this is going to be more probable.0273

In other words, this is more likely to occur.0281

The equation we use to quantify this is following.0286

S is equal to kB times the natural log of W where S is what we call entropy.0289

I want you to think of entropy as basically disorder or chaos.0297

We see that S is directly proportional to W.0308

In other words, as W goes up, so does S.0313

What does that mean though?0318

We already said that a high value of W means it is going to be more likely to occur0321

which says that very high entropy values are going to relate to more likely processes.0326

If W large, S is large; process more likely to occur.0337

In this equation, we see that there is a proportionality constant called kB.0351

kB is what we call the Boltzmann constant.0356

That is equal to 1.38 times 10-23 joules per kelvin.0363

Because natural log of W is going to be unitless, that means the units of entropy are also joules per kelvin.0369

This entire equation is known as the Boltzmann equation.0377

S is equal to kB natural log of W.0383

The units of entropy are going to be joules per kelvin.0386

Let's go ahead and relate to entropy to what we have already discussed0393

in our first lecture from thermodynamics which involved heat flow and energy.0397

If you recall from our first thermodynamic lecture, if heat leaves the system, Q is going to be a negative value.0402

If heat enters the system, Q is going to be a positive value.0410

Entropy is going to be a state function and can be related to Q where ΔS is equal to Q over T.0415

Temperature, because you see capital T, has to be in kelvin here.0426

Again because this is a state function, what we are interested in is0433

a change in entropy just like we talked about for ΔE or ΔU.0437

If heat enters the system, Q is going to be positive which means ΔS is going to be positive.0448

Let's see if that makes sense.0460

Remember if heat enters the system, we are supplying thermal energy for molecules to absorb.0462

If they are going to absorb this thermal energy, that gives them more molecular motion such as vibrations.0470

That increases the number of states an electron can occupy.0477

Basically we are increasing the amount of disorder.0483

We are increasing the amount of chaos.0485

Yes, it does make sense that ΔS is a positive value.0488

ΔS is also related to heat flow during phase changes.0494

It is found that Q is equal to n times ΔH of the phase change.0500

This is from our first thermodynamics; from thermo I.0506

Therefore it follows that ΔS is equal to Q over T which equals to nΔH of the phase change over the temperature.0514

What does that mean then?0525

That means for endothermic processes such as melting and vaporization, endothermic means Q is going to be a positive value.0527

ΔS is positive.0540

This agrees with the previous slide where when we put energy in, everything gets a little more chaotic and a little more disorderly.0542

For example, as you go for melting, you go from a very orderly state, a solid, to a less orderly state, to a liquid.0553

When go from a liquid to a gas, a liquid is less orderly.0563

A gas is going to be the most disordered state.0567

Everything is a lot more chaotic there.0572

ΔS is going to be positive when heat enters the system.0576

Here we have the phase change backing up this point.0580

In other words then, we can summarize this very nicely.0586

If a process is likely to occur, then ΔS is going to be a positive value.0590

We say that the process is expected to be spontaneous if ΔS is positive.0596

This brings us into our second law of thermodynamics.0605

The second law of thermodynamics states that a process will be spontaneous if ΔS total is greater0609

than 0 where ΔS total is equal to ΔS of the system plus ΔS of the surroundings.0616

In other words, nature is going to favor processes that contribute to the overall entropy of the entire universe.0623

In other words, nature favors processes where the amount of entropy or disorder increases as a result of the process.0631

We see this occurring in nature quite a bit.0640

If you think of a virus, what does a virus naturally do?0643

A virus is going to naturally expand and go from cell to cell consuming the resources there and moving on and on.0648

If viruses expand, they get more disorderly, more chaotic.0658

We see this in herd migration.0662

Herd migration, flocks of animals tend to naturally expand from area to area.0667

They tend not to stay just in a single location.0678

Of course the universe itself, the universe theoretically started off as a single point.0682

It is just constantly expanding and expanding.0689

Again we see that entropy is seen in several naturally occurring examples.0694

Then you may be wondering, if ΔS positive is going to be for a spontaneous process, what does it mean when entropy is zero?0704

That brings us into the third law of thermodynamics.0715

Zero entropy really does not exist because the conditions to obtain zero entropy do not exist experimentally.0718

The third law of thermodynamics states that entropy is zero only for a perfect flawless crystal at absolute zero.0728

But absolute zero has not been obtained experimentally.0736

Only in the most extreme conditions where we have zero thermo motion, that is molecules cease to move, entropy is completely zero.0741

In other words, entropy is always going to be a nonzero quantity in practice.0750

Let's now focus on problem solving involving entropy.0758

The following are typical types of questions involving entropy.0762

Sometimes a question can ask you simply to predict the sign of ΔS given a chemical reaction.0765

This is going to be an example of a qualitative process, of a qualitative problem.0770

Basically when the process is a physical change, ΔS is going to be positive for0777

any endothermic process just like we talked about, including melting, vaporization, and sublimation.0781

When you go from a solid to a liquid, from a liquid to a gas, or from a0788

solid to a gas, ΔS of these processes are going to be expected to be positive.0793

They are opposite processes, are all exothermic which means ΔS is expected to be negative.0801

Liquid to solid, gas to liquid, and gas to solid.0807

Freezing, condensation, and deposition; again ΔS is expected to be negative.0812

Let's go on to another type of problem.0822

Another type of problem involves a quantitative problem.0828

What you want to do here is pay close attention to the physical states.0834

From a solid to a liquid to a gas, entropy increases.0838

We have already seen that.0841

What you want to keep track of is the change in physical states from reactant to product side.0843

Let's go ahead and look at an example.0850

Fe solid plus O2 gas goes on to form Fe2O3 solid.0856

We are going to need two of these, three of these, and two of these.0864

Excuse me... we will need four irons.0869

You notice that we start off with a solid and a gas.0873

But on the product side, we have no gases here; no gases on product side.0878

If we are eliminating the highest disorderly state which is a gas, we are starting off with it.0887

We are not winding with it; we are actually getting less disorderly.0894

ΔS for this process expected to be a negative value.0898

Let's go on to another type of problem.0903

If the reaction involves all gases, what you want to keep track of0906

is the net change in gas molecules from reactant to product side.0909

For example, N2 gas plus 3H2 gas goes on to form 2NH3 gas.0913

On the left side, we start off with four gas molecules.0923

On the right side, we wind up with only two gas molecules.0930

For all intents and purposes, gas molecule A is going to be the same as gas molecule B for entropy counting purposes.0935

Because we start off with four gas molecules and we only wind up with two,0945

we are actually decreasing the amount of disorder in this process.0949

ΔS is expected to be negative for this example.0953

Sometimes the problem is going to ask you to calculate ΔS0958

for what is called the system, surroundings, and total where0960

ΔS total is equal to ΔS system plus ΔS of the surroundings.0963

We have talked about system and surroundings in thermodynamics I.0967

If you recall, Q of the system equals to ?Q of the surroundings.0972

This is going to come into play.0978

If a refrigerator coolant absorbs heat from stored food items, the heat then vaporizes the coolant which boils at -27 degrees Celsius.0979

ΔH of vaporization is equal to -22.0 kilojoules per mole.0988

Calculate ΔS total when 1.471 moles of the coolant vaporizes exchanging heat with the food items that are stored at 4 degrees Celsius.0993

What we have to do is the following.1002

We know we are using this equation for sure.1005

ΔS total is equal to ΔS of the system plus ΔS of the surroundings.1007

Let's go ahead and define the system; it doesn't matter which one as the coolant.1018

Let's go ahead and define the surroundings as the food items inside the refrigerator.1024

This is then going to be equal to...1031

We know that ΔS is equal to Q of the system over T.1035

The ΔS of the surroundings is going to be equal to Q of the surroundings over T.1043

But we also know that this relationship here where Q of surroundings is equal to ?Q of the system.1050

We can then go ahead and plug everything in.1060

That is going to be equal to nΔH of vaporization over the temperature of the coolant.1062

That is going to be equal to ?nΔH of vaporization but this time over the temperature of the food.1072

Really the only difference is the temperature that the system and surroundings are stored at1080

if we assume complete transfer of heat from the system to the surroundings.1085

When all is said and done, we are going to get -0.248... don't forget the units.1092

Here this is going to be kilojoules per kelvin or -248 joules per kelvin.1098

Again you must be careful on how you define your system and surroundings.1108

Be consistent throughout the entire equation.1113

Finally another type of problem involves standard molar entropies of formations which is the1118

change in entropy when a compound is formed from its constituent elements under standard conditions.1123

Recall that we covered standard molar enthalpies of formation and used the summation equation.1131

Similarly ΔS of the reaction is going to be equal to summation nS of the products minus summation nS of the reactants.1137

These values here again just like for the enthalpy, you are going to look up in the appendix.1148

Don't forget that n is going to be the stoichiometric coefficient.1157

Unlike standard molar enthalpies of formation, known compound is going to have a standard molar enthalpy of formation.1169

Why?--because the zero is only obtained under the conditions of the third law of thermodynamics.1175

Let's go ahead and examine our... excuse me... this is very important.1182

Again you never can assume that the molar entropy of formation is a zero value ever.1191

It is only true for ΔH and, we are going to learn, for one more--what we call ΔG later.1200

Let's examine our final state function.1207

This is an introduction to Gibbs free energy; free energy is ΔG.1210

It is defined as the amount of energy available to do work.1215

We said that mother nature favors high entropy already and low energy.1220

It is that energy that we are really talking about.1225

We will see that the spontaneous processes will all have ΔG less than zero.1227

To go ahead and do this, we can do ΔG is equal to ΔH minus TΔS.1241

This is the equation that relates all three thermodynamic properties together.1258

We already said that a positive ΔS is going to be a spontaneous process.1266

We want ΔG to be negative for a spontaneous process.1279

What conditions give you ΔG negative?1287

ΔG negative is going to be spontaneous under the following conditions--for very large ΔS values and at high temperatures.1290

As you can see, if ΔS is large and temperature is large,1312

this term is going to be larger than this term giving us1318

an overall negative ΔG value; very large ΔS at high temperatures.1322

Basically ΔH is going to be a relatively negative value; ΔH, small as possible.1329

Again this is what we call Gibbs free energy.1345

Just like our previous state functions, we have standard molar free energies of formation1350

where ΔG of the reaction is equal to summation nΔG of the products minus summation nΔG of the reactants.1355

Like enthalpy, the free energies of formation are zero for all compounds in the standard state.1363

Something like O2 gas, H2 gas, N2 gas, Cl2 gas, Br2 liquid, and I2 solid just to name a couple of examples.1369

These are all going to have a ΔG of formation which is equal to ΔH of formation equal to zero.1385

But again S, the molar entropy of formation, is not going to be zero whatsoever.1397

Let's now study the temperature dependence of ΔG.1409

It is found that we already said that ΔG of the system is equal to ΔH of the system minus TΔS of the system.1414

The reason why this is important is because of the following--this term right here.1424

ΔS of the system is in this equation.1429

But the second law tells us that ΔS of the universe has to be a positive value.1433

If we go strictly by entropy, we need to know the ΔS of the total, not just of the system or the surroundings.1446

However this equation here tells us that all I have to know is ΔG of the system; that is all.1454

Let's go ahead and see what conditions are going to be right for this.1466

We can go ahead and make a chart here; ΔH, T, and ΔS.1474

Then this is going to be sign of ΔG.1481

Let's go ahead and delete temperature here--sign of ΔG, and then ΔH, ΔS.1488

Let's say ΔH is a negative value--if this is less than zero and ΔS is a negative value.1495

If ΔH is negative and ΔS is negative, it will be very difficult to get ΔG to be negative.1505

The only time ΔG can be negative is going to be at a low temperature here.1516

What if ΔH is negative and ΔS is positive?1523

If ΔH is negative and ΔS is positive, then this sign of ΔG is going to be negative at high temperature.1531

What if ΔH is positive and ΔS is positive?1547

If ΔH is positive and ΔS is positive, ΔG is going to be negative at high temperatures.1551

Finally what if ΔH is positive and ΔS is negative?1560

If ΔH is positive and ΔS is negative, ΔG will never be negative according to the equation at any temperature.1565

This is going to be greater than zero at all temperatures.1574

Again as you can see, ΔG of the system is greatly influenced by1579

not only the signs of ΔH and ΔS but also the kelvin temperature.1585

Finally let's go ahead and examine the effect of reactant and product concentration on the sign of free energy.1595

It is found that ΔG of the reaction is equal to ΔG0 plus RT natural log of Q.1602

Remember what Q is?--Q is what we call the reaction quotient.1609

This equation relates reactant and product concentration to the free energy not at1618

standard conditions which means the concentrations are not going to be 1 molar.1623

Or the pressures are not going to be 1 atm; or partial pressures not equal to 1 atm.1630

Finally it can also be shown at equilibrium that ΔG not of the reaction is equal to ?RT natural log of K1639

where this is our equilibrium constant which we have discussed so much in the previous sections.1647

This is any equilibrium constant.1654

This can be Ka, Kb, Kp, Ksp, Kf, etc; any of them.1656

Basically this equation tells us that if K is large, the reaction is product favored.1667

ΔG of the reaction is a negative value; that makes sense.1672

If K is very large, that means the reaction is highly product favored.1677

In other words, the reaction as written is going to be likely to occur.1682

If the reaction is likely to occur, ΔG of that reaction should be negative.1687

Let's go ahead and summarize our thermodynamics lecture before getting into our problems.1694

We see that nature not only favors states of low energy but also now high entropy.1701

Entropy is basically disorder or chaos.1707

It is the focus of the second and third laws of thermodynamics.1710

Basically what we have seen in this lecture is that there is a series of equations that govern entropy and free energy.1713

They allow us to calculate both of them under a given set of conditions.1722

Let's go ahead and do sample problem number one.1729

Calculate ΔS of the reaction under standard conditions for the following.1732

We have aluminum oxide reacting with 3H2 going on to form two aluminums plus three waters.1738

Anytime you see a balanced chemical equation and it asks you simply to calculate1746

ΔS not of the reaction, you are going to use the summation equation.1751

Use ΔS of the reaction is equal to summation n S molality of all products minus summation n S molality of all reactants.1755

I am just going to setup the problem for you.1772

Again you are going to look up these values in the appendix of your textbook.1775

Let's go ahead and do the products first.1779

This is going to be 2 moles of aluminum times the standard molar entropy of formation of aluminum solid.1781

Because it is summation, I am going to add... plus.1799

Let me put that in parentheses.1803

Plus 3 moles of the water vapor times its standard molar entropy.1804

This is all in brackets; that is going to be subtracted from the reactant part.1813

Here this is going to be 1 mole of Al2O3 solid times1819

its molar entropy plus 3 moles of H2 gas times its molar entropy.1830

Again you must pay attention to the physical states because the molar entropy of formation for say water vapor1848

is going to be different than from liquid water which is going to be different than from solid water.1856

Once again pay attention to the physical states.1862

When this is all said and done, you are going to get your answer1865

in of course units of joules per kelvin or kilojoules per kelvin.1868

It depends what the question specifies of course.1874

Let's go ahead and do sample problem number two.1880

The reaction N2O4 going to 2NO2 is not spontaneous under standard conditions.1883

Calculate the temperature at which the reaction becomes spontaneous.1889

What that translates to is, in other words, at what temperature does ΔG become negative?1893

Anytime you are asked to calculate a temperature at which something becomes spontaneous,1902

pretty much you always use this equation: ΔG is equal to ΔH minus TΔS.1908

They are asking you when does this change sign?1916

In other words, when does ΔG go from positive to negative?1920

What you simply is you set this whole thing equal to zero.1926

Zero is equal to ΔH minus TΔS; you solve for temperature.1931

Temperature is going to be equal to ΔH over ΔS.1936

This is all going to be standard conditions of course.1941

To get ΔH, you are going to use the summation equation.1946

To get ΔS, you are going to use the summation equation1950

which means you have to look up the formation values in the appendix.1954

Look up formation values in appendix; appendices.1959

When you do this, you are going to get a temperature value.1969

At temperatures lower than T, ΔG is going to be positive.1976

At temperatures higher than what you calculate for T, ΔG is going to be a negative value.1987

Once again anytime you are asked to calculate the temperature at which something becomes spontaneous,2006

almost always you are going to use this equation: ΔG is equal to ΔH minus TΔS.2015

Set ΔG equal to zero and solve for the temperature.2021

Finally our third sample problem, calculate ΔG of the reaction at 25 degrees Celsius2029

for the formation of 1 mole of C2H5OH gas from C2H4 gas and H2O gas.2036

The first thing we probably want to do is translate this into a balanced chemical equation.2047

The reactants are C2H4 gas; this is going to react with H2O gas.2054

That is going to form 1 mole of C2H5OH gas.2063

Let's see if everything is balanced.2069

Two carbons, six hydrogens, one oxygen; yes, this is nicely balanced.2071

You are asked to calculate ΔG of the reaction at 25 degrees Celsius.2078

Again ΔG of the reaction can simply be calculated from ΔH of the reaction minus TΔS of the reaction.2083

Again to get the ΔH, you are going to use the summation equation.2094

To get the ΔS, you are going to use the summation equation.2099

Again the kelvin temperature is already specified for you right there.2102

That gives you ΔG of the reaction; they then ask you to calculate Kp.2108

The only equation in this lecture that involves the equilibrium constant is ΔG is equal to ?RT natural log of K.2114

We can then insert this ΔG of a reaction in there.2126

K therefore is equal to 10 raised to the -ΔG0 over RT2131

where of course R is equal to 8.314 joules per k mole.2140

Always double check your work.2151

Make sure that K is a positive value because again2152

we saw in previous lectures that K can never ever be negative.2158

That is our lecture from general chemistry on thermodynamics.2165

I want to thank you for your time and attention.2169

I will see you next time on Educator.com.2171

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is electrochemistry.0003

We are going to start off with a brief introduction in electrochemistry--what it is about.0010

We are going to go ahead and go into the main reactions that are0015

at the core of electrochemistry which are what we call redox reactions.0020

A big application of electrochemistry is what we call a voltaic cell.0025

We are going to take a look at that followed by quantifying aspects in electrochemistry which deal with what we call electrochemical potentials.0028

After we introduce the potentials, we will then go ahead and get into quantitative problems0040

that include current, charge, and time, followed by a brief summary and sample problems.0046

Electrochemistry is the branch of chemistry which deals with harnessing chemical energy and converting it into mechanical energy.0054

Basically it studies the processes in which energy is released during a chemical reaction0064

so that it can be used to perform some type of work.0071

The perfect example of this is the battery which powers everyday machines including vehicles, computers, and phones.0076

Basically a battery is a chemical process that produces energy that is then converted into electricity.0084

Basically in this lecture, we are going to examine both the concepts and equations that are at the core of electrochemistry.0095

An electrochemical reaction is one that involves a complete transfer of electrons from one reactant to another.0104

The reactant which loses the electrons is said to be oxidized or undergo oxidation.0111

The nice little acronym that allows us to remember this is OIL or oxidation-is-loss.0119

The reactant that gains electrons is said to be reduced or undergo reduction.0125

The acronym for that is RIG or reduction-is-gain.0135

OILRIG, oxidation is loss; reduction is gain.0138

An oxidation reduction reaction is also known as a redox reaction for short.0143

Let's now look at how to identify each type of a reactant.0151

To identify which reactant is being oxidized and which is being reduced, it helps to0158

keep track of oxidation states which we learned in the first half of general chemistry.0164

Let's go ahead and consider the following reaction.0169

CH4 gas plus O2 gas goes on to form CO2 gas and H2O gas.0175

Let's go ahead and balance this guy.0191

We are going to need two of these and two of those.0194

This is the reaction that represents the combustion of methane.0201

We are going to now assign oxidation states to each of the elements.0207

Hydrogen is +1 of course which means overall this carbon is going to be -4.0214

Here in O2, this is a homonuclear diatomic; each oxygen is 0 here.0222

Here in CO2, oxygen is -2 each which means carbon is going to be +4.0230

Here oxygen is -2; each hydrogen is +1.0238

We see that there are two elements that experience a change in oxidation state.0243

Carbon starts off as -4; it becomes +4 in the product side.0249

Oxygen starts off as 0; it becomes -2 on the product side.0261

We see that carbon's oxidation state has increased.0269

If the oxidation number goes up, that means it has lost electrons.0275

Oxygen's oxidation state has gone down which means it has gained electrons.0284

If carbon has lost electrons, we say that it has been oxidized.0293

If oxygen gains electrons, we say that it has been reduced.0303

But it is not just the element that undergoes the process.0308

It is that entire compound.0311

To be more accurate, we say that CH4 gas was oxidized.0313

We say that O2 gas was reduced.0323

We are going to introduce two more terms here.0330

CH4 gas was oxidized; if it was oxidized, it performed the reduction.0333

Therefore CH4 gas is also known as the reducing agent.0341

O2 gas was reduced.0353

If it was reduced, it performed the oxidation which means that O2 gas is what we call the oxidizing agent.0357

You definitely want to keep track of these terms because they are somewhat confusing.0369

A reducing agent performs the reduction; but it itself is oxidized.0373

A reducing agent performs the reduction... excuse me.0379

An oxidizing agent performs the oxidation; it itself is reduced.0385

Let's now learn how to balance redox reactions.0389

Redox reactions can occur in neutral solutions and in acidic solutions and basic solutions.0397

We are going to learn how to balance redox reactions in all three of these types of medium.0407

In neutral solutions, let's go ahead and start off.0413

Cl2 gas can react with Zn2+ to go and form Cl1- aqueous plus zinc solid like that.0422

What we are going to do is we are going to balance these guys.0447

When we balance in neutral solutions, you must make sure that the0452

stoichiometry is balanced and the net charge on both sides of the equation.0456

When we go ahead and look at this, we see that we have one zinc on each side.0467

That is good to go.0472

We have two chlorines on the left and only one on the right.0473

Let's go ahead and put the two chlorines there.0476

My net balanced charge here is 2+.0478

My net charge here is going to be 2-.0482

Let's go ahead and see what the half reactions are.0492

For the half reactions, we have Cl2 gas going to become 2Cl1- aqueous.0501

That means that each chlorine is gaining an electron.0516

I am going to need two electrons on the left side.0520

Zn2+ is becoming zinc solid.0524

That means I am going to need two electrons on this side.0530

When we look at the half reactions, these are both reduction.0550

But you can't have two reduction half reactions.0558

You always have to have one oxidation; you always have to have one reduction.0561

It turns out that the chlorine one is going to be the one that reacts.0566

What is going to happen is we are going to flip this one.0577

That is going to be zinc solid going to form Zn2+ plus two electrons.0589

This immediately told me that we have done something wrong because my net charge on each side is not the same.0599

My overall reaction is going to be Cl2 gas plus zinc solid going on to form 2Cl1- aqueous and Zn2+ aqueous.0608

As you can see, now my overall charge is fixed.0624

This is 0 on the left.0628

This is 0 on the right side for our balanced redox reaction in neutral solutions.0629

Let's now jump into solutions where we have acidic or basic conditions.0637

For example, in acid, silver solid plus nitrate aqueous goes on to form NO2 gas and Ag+1 aqueous.0644

Step one is to break the unbalanced equation into your half reactions.0652

My half reaction here is going to be silver solid going to Ag1+ aqueous.0658

NO31- aqueous going on to form NO2 gas.0671

Those are our half reactions.0681

Step two, we are going to balance each of the half reactions by being careful to follow each of these following steps.0683

Balance all elements except oxygen and hydrogen by stoichiometry.0692

For the silver half reaction, we have one of each on side.0699

We are good to go on that.0703

For nitrate going to nitrogen dioxide, we have one nitrogen on each side.0705

We are good on that; part A is done.0711

Part B, we are going to balance oxygen by adding water to the side that is deficient in oxygen.0715

Here for nitrate, I have three oxygens; for nitrogen dioxide, I only have two oxygens.0721

I am going to add H2O to the side that is deficient in oxygen.0727

Let's go ahead and move on now to step number three.0737

If in acidic solution, we are going to add hydronium for0742

every missing hydrogen to the side that is deficient in H0745

and the same number of water molecules to the other side.0749

Our half reaction again was NO31- aqueous going on to form NO2 gas.0752

We added water last time; we are going to follow part C.0762

We are going to add hydronium for every missing hydrogen to the side that is deficient in hydrogen.0767

On the right side of this equation, I have the two hydrogens here.0773

I am going to add two hydroniums on the left side.0776

Remember you are adding one hydronium for every hydrogen.0780

At the same time, you are going to add water, the same number of water molecules to the other side.0784

Let me go ahead and do that in blue; plus two more H2Os.0789

If this was basic solution, we would add one water for0796

every missing hydrogen to the side that is deficient in H0799

and the same number of hydroxide to the other side.0803

What we are going to do right now is we are going to go on to part E.0807

We are going to balance the net charge by adding electrons to the side that is deficient in negative charge.0812

Let's go ahead and look at this.0818

Right now we have two hydroniums aqueous plus NO31- aqueous.0820

That goes on to form NO2 gas plus 3H2O liquid.0829

We have a net charge of +1 on the left side.0842

We have a net charge of 0 on the right side which means that0851

we have to add an electron to the left side here to get us to 0.0856

We will then multiply the half reactions by integers that will lead to cancellation of electrons.0875

In other words, each half reaction should have an identical number of electrons.0880

Ag solid going to Ag1+ aqueous.0887

It was now NO31- aqueous going to NO2 gas.0895

I believe this was three waters; this was two hydroniums aqueous.0906

Let's go ahead and fix the electrons; I am going to need one electron here.0916

I am only going to need one electron right there.0922

Because the number of electrons is the same in each reaction, we are good to go.0930

We are going to now recombine the half reactions to get the net balanced equation.0939

Be sure to cancel like terms.0943

When I add these two reactions together, the electrons cancel out.0946

We get silver solid plus 2H3O1+ aqueous plus NO31- aqueous0950

going on to form Ag1+ aqueous plus NO2 gas and 3H2O liquid.0961

This should be liquid here.0971

Let's go ahead and double check if everything is balanced.0973

On the left side, I have one silver.0977

On the right side, I have one silver; that is good to go.0979

On the left side, I have six hydrogens.0982

On the right side, I have six hydrogens.0984

On the left side, I have one nitrogen.0987

On the right side, I have one nitrogen.0989

On the left side, I have a total of five oxygens.0992

On the right side, I have a total of five oxygens.0996

Stoichiometry is good to go; let's do a final check on the charge.0999

I have a total charge of 1+ on the left side.1003

I have a total charge of 1+ on the right side.1010

Yes, this is our overall balanced redox reaction.1012

Let's go ahead now and look at how a simple battery works.1017

The first battery was referred to as a voltaic cell, also known as a galvanic cell shown below.1023

Let's go ahead and look at the redox reaction for one we already looked at.1031

This was going to be then Cl2 gas plus zinc solid going on to form 2Cl1- aqueous and Zn2+ aqueous.1040

Basically the redox reaction was the following; we can have a metal electrode here.1070

The metal electrode is going to be dipped in a salt solution that contains its ion.1091

If let's say the metal electrode was represented by A solid, the salt solution1099

could be for example A+ and maybe X1- for our salt solution.1107

Again I have another metal electrode on the right side of a different identity.1121

This could be B solid for example; this would be B+ and X1-.1125

These two electrodes are going to be connected by a conductive wire here.1133

We get a flow of electrons from A to B.1143

The flow of electrons is going this way.1147

Anytime you have a flow of electrons, that generates an electric current which can then be captured to power some type of device.1151

What we see here is that A solid is going to be losing electrons to form A1+ aqueous like so.1162

That is its half reaction.1175

B solid is going to be gaining the electrons to form B1- aqueous.1177

We need something called a salt bridge to help us complete the circuit and to balance the charge.1195

Here because we are forming a lot of A+, some type of ion has to come in and balance the positive charge.1205

Maybe some more X- comes in from the salt bridge.1218

The salt bridge can be for example A+ and X1-.1222

Because in the B side we are forming a lot of negative charge,1232

some positive cation is going to come in from the salt bridge also.1240

We see that here that this is going to be representing the oxidation half reaction.1249

Here on the right side, this is going to be the reduction half reaction.1255

We can now also look at the charges.1272

If the flow of electrons is from A to B, that means that...1275

Remember that electrons are going to go away from a negative side.1280

This is the negative charge.1284

They are going to go toward the positive side which is going to be where the electrons gravitate toward.1287

If you ever look at a battery, a battery always has a negative sign and a positive sign representing these two electrode terminals.1294

The negative electrode is what we call the anode.1303

The positive terminal is what we call the cathode.1308

A nice little trick to remember this is when you cross the t in the word cathode, it looks like a positive sign.1315

We can also have something we call cell notation.1324

Cell notation typically looks like this.1328

On the left side goes the anode; on the right side goes the cathode.1344

The single lines represent changes in physical state.1357

The double line just simply represents a salt bridge.1367

Here we can easily write in A solid going to A1+ aqueous.1372

Here we can write in B1+ aqueous going to B solid.1382

Let me go ahead and correct this then.1399

This is really B1+ aqueous plus an electron going to be solid.1402

Again this is what we call cell notations.1409

The anode half reaction is written on the left.1411

The cathode half reaction is written on the right side.1415

Another thing about the voltaic cell is that we see that the anode here is getting consumed.1421

The solid is becoming ionized.1431

We see here that the cathode, we are actually depositing more solid onto the electrode.1434

This is going to become heavier, become larger in mass.1442

Why does a battery die?1453

It is essentially because one of the terminals, the anode gets consumed to the point where no more reduction or oxidation can occur.1456

If you ever have a rechargeable battery, you know that a rechargeable battery you have to plug in.1472

What does that do?1479

When we put a rechargeable battery into the wall outlet and charge it overnight, we are basically driving the reverse reaction.1480

The reverse reaction is then replenishing the anode while consuming the cathode.1488

But that doesn't always go on forever; we cannot use a rechargeable battery infinitely.1494

It also too is going to eventually die.1500

That is because no process is 100 percent efficient.1504

You will never ever regain all of your anode.1508

It is going to become less and less efficient as time goes on.1514

Let's now go ahead and look at the quantitative part of electrochemistry.1518

All half reactions have a potential measured in volts, when you purchase for example a 1.5 volt battery.1524

This is just like a chemical reaction having a change in energy associated with it.1531

These potentials are all relative to the hydrogen reduction half reaction which is arbitrarily assigned 0 volts.1536

That is going to be 2H+ aqueous plus two electrons going on to form H2 gas.1544

This potential is what we symbolize as E standard.1555

That is going to be 0.00 volts.1559

It is very important to be able to use the following table of standard reduction potentials.1561

Here I have listed for you five half reactions from this table.1570

You see that these are all reduction half reactions.1575

On the right side are the reduction potentials; E standard values.1584

It is important that we understand how to decipher this.1590

Point number one is a large E0 means that the reduction is more likely to occur.1595

Again a large E0 is reduction more likely to occur.1606

For example, the reduction of Cl2 has a potential of +1.36 volts.1611

The reduction of Zn2+ has a potential of -0.76 volts which means that Cl2 is more likely to be reduced than Zn2+.1622

In fact if Cl2 is going to be...1638

If we combine the two, Cl2 gas plus two electrons going to 2Cl1-1644

and Zn2+ aqueous plus two electrons going on to zinc solid,1654

which one is going to be the reduction half reaction?1664

Which one is going to be the oxidation half reaction?1667

Because this half reaction for chlorine is more positive, that is going to remain reduction.1672

Zinc solid is going to become oxidized.1677

The combination of Cl2 gas plus two electrons going on to form 2Cl- and1682

zinc solid going on to form Zn2+ plus two electrons for a balanced redox reaction of1690

Cl2 gas plus zinc solid going on to form Zn2+ aqueous and 2Cl1- aqueous.1697

In other words, Cl2 gets reduced; zinc is going to be oxidized.1708

This is very important because when we read this table, we can always draw an arrow in the following direction.1720

Anything that follows this direction, you are going to have a successful redox reaction.1729

Basically we can summarize it the following way, what this line means.1738

Any element or ion can oxidize any other species...1743

Any element or ion can oxidize any other species to the bottom and right of it, to the bottom right of it.1765

For example, Cl2 can oxidize silver; it can oxidize H2.1780

It can oxidize nickel; it can oxidize zinc.1786

Cl2 itself will be reduced because its reduction potential is so positive.1790

We see therefore that when we go this direction, these are going to be very strong oxidizing agents.1798

When we go this direction on the right side, these are going to be stronger reducing agents.1812

In this table, we see that Cl2 is the strongest oxidizing agent.1822

It itself is most likely to be reduced.1827

Zinc solid is the strongest reducing agent.1830

It itself is going to be more likely to be oxidized.1832

Let's now look at calculating the potential when not at standard conditions.1837

To calculate the potential for a voltaic cell when not at standard conditions, we use what is called the Nerst equation.1843

This is just like ΔG is equal to ΔG0 minus RT natural log of Q; very very similar.1851

Once again Q is going to be the reaction quotient.1857

The temperature is in kelvin of course.1864

R is going to be 8.314 joules per k mole.1868

n is going to be the moles of electrons.1875

You get that from the balanced redox reaction; moles of electrons from balanced redox reaction.1878

F is what we call Faraday's constant.1890

Faraday's constant, you should always ask your instructor if you have to know it or not.1898

It is basically 9.65 times 104.1904

The units, I am going to draw in red, are coulombs for every mole of electron.1909

Once again coulombs per moles of electron or C over mole.1918

Again you are going to use this equation anytime you are dealing with nonstandard conditions1924

which is basically not 1 atm of pressure and not 1 molarity of concentration.1929

It can also be shown that at equilibrium, E is going to be 0.00 volts1937

which means that E0 is equal to RT over nF times the natural log of K.1943

Again this is any of the K values--Ka, Kb, Kf, Ksp, etc.1948

Again these are the two equations that are heavily used in electrochemistry.1960

Finally it can be shown that we can relate Gibbs free energy to electrochemistry.1968

The equation is ΔG0 is equal to ?nFE0.1977

This equation tells us that if E0 is large...1982

If E0 is large, that means the potential is very high which means the redox reaction is likely to occur.1988

Remember what we said if something is likely to occur>2000

If something is likely to occur, ΔG0 is negative.2003

Look, that makes perfect sense because if E is positive... we know F is positive.2008

We know n is positive making that whole side of the equation negative giving us a negative Gibbs free energy.2013

Basically if E0 is very large, that means the reaction is in the forward direction.2020

It is highly product favored which as we have seen translates to a lower energy state.2026

Let's now see how to factor in the lifetime of a battery.2032

A battery basically has three main parameters--the voltage of the cell,2038

the current that is drawn from it by the device,2043

and the total amount of time the battery can run at the specified current.2045

A simple equation relates all of these parameters.2050

n is equal to It over F where I is your current in amps.2053

Time, t is your time in seconds.2063

Of course F is your Faraday's constant, 9.65 times 104 coulombs per mole of electrons.2069

In this equation, n is not going to be a whole number.2079

n is going to be the mole of electrons that is determined stoichiometrically.2082

Let's go ahead and summarize this before we jump into our sample problems.2091

Electrochemistry we saw studies how chemical energy can be produced and converted into2095

mechanical energy which in turn can perform work such as powering an electrical device.2100

A potential is produced when there is a transfer of electrons2107

from the reducing agent to the oxidizing agent which as2110

we have seen can happen in neutral, acidic, or basic media.2114

Finally the Nerst equation allows for determination of the cell potential when not at standard conditions.2118

Sample problem number one, basically here...2128

This is a very common type of problem by the way.2132

You are given a list of half reactions here with their reduction potentials.2134

E0 is right here.2140

You are just asked to answer the following questions, all qualitative problems.2146

What is the strongest oxidizing agent?--what is the strongest reducing agent?2150

Remember that the stronger oxidizing agents are the ones that are going to be2154

most likely to be reduced which corresponds to a very high reduction potential.2158

Here the strongest oxidizing agent is Cl2 gas.2163

The strongest reducing agent therefore is going to be zinc solid.2168

Part B, list the species that silver solid can reduce if any.2173

Silver solid is not going to reduce species.2181

But instead it is going to oxidize species.2187

Silver solid is going to only... excuse me.2189

Silver solid can reduce anything to the top and left of it.2198

That is only going to be Cl2 gas; again reduction occurs to top and left.2205

List the species that zinc solid can reduce if any.2216

Zinc solid because it is such a strong oxidizing agent, it can reduce any species that is to the top and left of it.2220

That is going to be Ni2+, H+, Ag1+, and Cl2 gas.2231

Finally part D, what species is most easily oxidized?2240

The species that is most easily oxidized is going to be the strongest reducing agent.2244

Therefore this is going to be zinc solid.2254

What species is most easily reduced?2257

That is going to be the strongest oxidizing agent which is going to be Cl2 gas.2259

You see, I just want to point out a couple things.2264

That part A and part D, these are really the same question that are asked differently.2266

Again there is going to be multiple ways of phrasing, of posing the same question.2272

Sample problem number two, if a battery contains 0.355 grams of cadmium and excess2280

nickel hydroxide, how long can it operate if it draws 1.25 milliamps of current?2288

You are given the balanced redox reaction below.2294

Because this question mentions current and how long, there is only one equation that deals with that.2299

That is n is equal to It over F.2305

We know that I is amps; that is going to be 0.00125 amps.2308

We know Faraday's constant already, 96500 coulombs per mole of electrons.2316

The question is asking us to solve for time.2324

t is going to be equal to n times F over I.2327

All we need is n.2332

Remember that n is the mole of electrons that is determined stoichiometrically.2334

Let's go ahead and get it.2339

You are told that you have an excess amount of the nickel precursor.2340

The 0.355 grams of cadmium is going to be our limiting reactant here.2345

0.355 grams of cadmium times 1 mole of cadmium divided by 112.41 grams.2350

That is going to be equal to 0.00316 moles of cadmium solid.2365

The redox reaction is right here; that is cadmium zero going to cadmium hydroxide.2374

In this reaction, cadmium is 2+.2388

For cadmium to go to cadmium 2+, it has to give up two electrons.2395

My mole to mole ratio of the electrons to cadmium is a 2:1 ratio.2401

0.00316 moles of cadmium times 2 moles of electrons for every 1 mole of cadmium.2408

That is going to give me the value of n which is 0.00632 moles.2419

When all is said and done, you should get a time of 487904 seconds which is approximately 136 hours.2428

That is an awfully nice lifetime for a battery to operate, 136 hours.2441

If you look at these elements, it is nickel cadmium.2450

I am sure you have heard of batteries that are called nickel cadmium or cadmium nickel batteries2453

which are very nice batteries that help to power devices such as your smartphones and cameras.2459

That is our lecture from electrochemistry; I want to thank you for your time.2469

I will see you next time on Educator.com.2474

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is the chemistry of the transition metals.0002

We are first going to get introduced to what we mean by coordination compound and basically what they consist of.0010

After we get introduced to them, we are going to then turn our attention0017

to nomenclature which is learning the rules to name these coordination compounds.0021

Following this, we will then get into the geometries and structures that coordination compounds can commonly adapt.0027

Related to structures will be then something we call isomers.0036

Finally we are going to, in the last part of the lesson, go into0040

electronic structure of coordination compounds and basically what happens to the0045

valence electrons of a transitional metal when atoms or molecules become attached to them.0052

We are going to learn that that is called crystal field theory.0061

We will also get into what is called the spectral chemical series and the implications of the crystal field theory.0066

As always, we will finish up the lesson with a summary followed by some sample problems.0073

Coordination compounds is the term that refers to the following.0081

These are compounds that consist of a central metal atom or ion which have0085

attached a varying number of atoms or molecules via a coordinate covalent bond.0090

These atoms or molecules that are attached have their own names.0098

These are what we call ligands or ligands.0103

The term coordination compound can also be used interchangeably with the term coordination complex.0108

Basically a coordination compound can also have the central metal atom be electrically neutral or charged.0115

A case where we have a charged neutrality is when nickel has attached to it four carbonyl ligands.0122

We are going to learn the names of these shortly.0132

But here this is formerly nickel Ni0.0134

An example of a coordination compound where we have the transition metal to be charged, if it is FeCl2 and 2NH3s.0138

Here in this case, this is formerly Fe 2+.0149

What we are going to be focusing on in this session is coordination complexes where the central metal atom is actually a d-block element.0155

Let's go ahead and move on.0166

Now that we have been introduced to what we mean by a coordination compound, let's go ahead and move into how to name them.0168

This is what we call nomenclature of coordination compounds.0174

There is a lot of rules associated with this; let's get into it.0178

When writing formulas, you are first going to list the chemical formula or the chemical symbol of the metal first.0182

Followed by the ligand symbols in alphabetical order.0188

After this, if the complex is neutral overall, brackets are not needed.0194

However if the complex is an ion, you are going to enclose the entire complex in brackets followed by the charge as a subscript.0200

An example would be the following, Fe(CN)64-.0214

This tells me that the entire complex ion here has an overall 4- charge.0225

But something like this, FeCl2 followed by (NH3)2,0231

this tells me that the overall complex ion is neutral overall because you don't see brackets.0241

When the ligand is a molecule instead of an atom, sometimes the point of attachment0249

can also be emphasized in the order in which you list the letters of the molecule.0253

For example, we can say FeCl2 and (OH2)2; this is basically water.0262

We are listing the oxygen first because it is the oxygen that is0271

attached to the iron center, not the hydrogens; because point of attachment.0279

There is more rules; when naming the complex... this is always harder.0298

You are going to list the ligand names first followed by the metal name.0302

A Roman numeral in parentheses indicates the oxidation number of the central metal ion.0306

If you have a neutral ligand, neutral ligands are always going to retain their full name with the exception of the following four.0314

H2O is going to be now formally be called aqua.0324

NH3 is called ammine; CO is called carbonyl; NO is called nitrosyl.0328

Anionic ligands will also change their suffixes.0336

If your anion such as chloride ends in ?ide, it is going to become with an ?ido suffix.0340

For example, chloride is going to become chlorido.0347

If something ends in ?ate, it becomes with an ?ato suffix.0354

For example, sulfate is going to become sulfato.0359

Finally if you have something ending in ?ite, it becomes ?ito.0365

An example of this will be nitrite anion becoming nitrito.0372

Furthermore if you have more than one of the same ligand, we are going to0380

use Greek prefixes to indicate the number of each ligand, omitting mono.0384

We are going to use di- for two, tri- for three, tetra- for four, etc.0390

Next sometimes the ligand name already contains a Greek prefix.0401

If that happens, we are going to use the following system.0408

Bis- is going to be replacing di-; tris- is going to replace tri-.0412

Tetrakis- is going to replace tetra-.0418

The other scenario where you have to use these unique prefixes is when the ligand is polydentate.0421

What a polydentate ligand is that it has more than one point of attachment to the metal.0429

Once again a polydentate ligand means it has more than one point of attachment to the metal.0446

We are going to go ahead and look at more examples of those in greater detail later on in this lecture.0451

Finally the final four rows, we are going to list the ligand names in alphabetical order ignoring any prefixes.0459

If the complex ion is anionic, the suffix ?ate is added to the metal ion name.0466

An exception is iron; you don't call it ironate.0472

But instead we are going to use I believe the Latin phrase, ferrate.0475

If the complex ion is anionic, you are going to list the cation name first.0481

If the complex ion is cationic, you are going to show the anion name last.0486

Let's go ahead and look at an example.0491

Let's take the following, NH4Pt in brackets Cl3NH3.0495

In this case, my cation is ammonium.0504

My anion is the entire complex which means we don't call it platinum.0509

But we call it platinate.0517

We know that the ammonium is a 1+ overall charge.0520

We know that each chlorine is 1- each, giving me 3- overall.0527

We know that ammonia is neutral ligand overall.0535

This tells us that the platinum must have an oxidation state of 2+.0538

Now that we have established that, we can go ahead and name the compound.0546

This is going to be ammonium.0549

It is the name of the cation followed by the following.0554

We list the ligands in alphabetical order--ammonium ammine and then trichlorito-platinate(II).0556

Let's go ahead and do one more example for nomenclature.0572

This one is going to have in brackets chromium with four ammines attached and with two OH groups attached and then bromine.0579

In this case, bromine is our anion; this entire complex ion is our cation.0594

We know that amine is 0 overall; we know that hydroxo is 1- overall.0602

We know that bromine is 1- overall, telling us that chromium must be 3+ here.0608

Let's go ahead and name it; list the ligands in alphabetical order.0617

This is going to be tetra-amine followed by di...0621

Now hydroxide becomes renamed as hydroxito; then chromium(III) bromide which is the complete name.0633

That is nomenclature for coordination compounds.0648

Let's now move on to the structures of coordination compounds--what do they really look like.0653

We define the coordination number, which is also known as steric number, to be0658

the total number of points of attachment to the central metal atom or ion.0663

When I mean total points of attachment, I mean from all of the ligands.0671

The most commonly observed coordination numbers for these transition metal complexes is the coordination number 4.0675

When we have a coordination number of 4, we can get two common geometries.0684

Number one--tetrahedral; or number two--square planar.0688

In tetrahedron case, you have your central metal atom.0692

Then with the four ligands attached like so in a traditional tetrahedral format.0695

Of course for square planar, your central metal atom.0702

Now the ligands are arranged 90 degrees to each other, giving you square planar overall.0706

The other common coordination number is coordination number of 6.0716

For coordination number of 6, we are going to basically get the main geometry of octahedral geometry.0721

Here you have your central metal atom.0730

You have the four ligands forming a square base.0733

Then you have one ligand perpendicular to that; another ligand perpendicular right below.0739

We call the four ligand positions that are horizontal, that is formed as square plane, we call them equatorial positions.0746

Of course then for the ligands that are perpendicular to that plane, these are called the axial positions.0763

Again tetrahedral and square planar are for a coordination number of 4.0777

Octahedral is for a coordination number of 6.0780

Now related to structures of coordination compounds, we are now going to discuss isomers.0784

Recall that isomers are basically the following.0792

These are compounds that have the exact same formula but have different chemical and/or physical properties.0795

Specifically for the coordination number of 6, we can have a large number of different isomers.0802

The first type of isomers we are going to talk about is for coordination number 6.0810

These are what are called geometrical isomers.0815

When we have a geometrical isomer, these are two formulas that you want to look for.0818

ML4L'2 and ML3L'3 where L and L' are different ligands.0823

When we look at ML4L'2, we can have one possible configuration0842

where the four Ls that are identical to each other,0847

they form the square plane like so--L' here and L' here.0851

In the other possibility, let's switch an L' with an L; good.0860

When the L's are completely opposite to each other, 180 degrees apart, this is what we call the trans isomer.0877

However when the two Ls here and here are not 180 degrees apart, when they are0888

perpendicular to each other like this, that is what we call the cis isomer.0897

Again this is what we call trans and cis isomerism.0904

The other type of geometrical isomer that exists for a coordination number of 6 is ML3L'3.0908

In this case, we could have the following two possibilities.0916

M and then L', L', L'... L, L, and L.0922

The other possibility, let's substitute an L' for an L.0941

In the left scenario, you see that the L's actually form a plane with each other like so that bisects the octahedron.0958

This is what we call the mer isomer.0973

Mer stands for meridian such that the three identical ligands share a plane that forms a meridian through the compound.0977

The other possibility on the right is when the three ligands are actually sharing the same face of the octahedron.0990

That is what we call the fac isomer, fac being short for facial.1000

That is geometrical isomers.1009

Let's now take a look at the category of what we call structural isomers.1012

When we talk about the word structural isomers, this is a very general category.1020

Basically structural isomers are going to retain a ligand position.1032

Excuse me... they are going to change ligand position for structural isomers; change ligand position.1049

There are basically four types of structural isomers that we want to deal with.1057

One is what we call ionization isomers; two is what we call hydrate isomers.1062

Three is what we call linkage isomers; finally four is what we call coordination isomers.1072

Ionization isomers literally have a cation or an anion switch places in or out of the brackets of the coordination sphere.1087

That is, in one isomer, a certain ligand is attached to the metal.1120

In the second isomer, it is not.1125

Let's go ahead and look at an example.1127

Imagine here in brackets, cobalt, bromine, and then five amines attached, closed brackets, and then sulfate.1130

We can imagine a situation where the sulfate just substitutes for the bromine.1140

We get CO(NH3)5SO4 and then bromine.1146

My compound is still neutral overall because this is a salt after all.1155

The only difference is the location of sulfate with bromine.1162

The second type of structural isomer is what we call a hydrate isomer.1167

This simply involves ligand exchange with water.1171

Imagine chromium with three chlorines attached, coordinated to six waters.1184

We can imagine a situation where we have the waters now directly bonded to the chromium metal.1193

The three chlorines serving as counterions.1204

Linkage isomers, linkage isomers involve what are called ambidentate ligands.1210

They involve ambidentate ligands.1220

Basically the word ambidentate means the ligand can attach in more than one location to the metal.1225

Again the ligand can attach in different locations to the metal.1239

You could imagine a situation where you have transition metal; let's take a NO2 group.1247

But you could also imagine the ligand being directly attached to the oxygen instead of the nitrogen.1257

In this case, we get a pair of linkage isomers.1263

Finally we can have also a fourth type of structural isomer called coordination isomers.1267

Basically you have two metals switching location with each other.1275

An example would be a chromium with the six amines attached followed by iron with the six cyano groups attached closed brackets.1292

That will be one coordination compound.1308

But you can also have another one where you just switch places.1310

Now iron is going to get the six amines.1313

Chromium is going to get the cyanos just like that.1318

Again these are what are known as structural isomers where we actually involve a change in attachment to the metal.1326

Now that we have covered isomers, let's go ahead and take a look at how the electronic structure is in coordination compounds.1336

To examine how the d-block electrons of the central metal atom or ion are1345

arranged once ligands are attached, we use something called crystal field theory.1349

Crystal field theory treats the lone pairs on a ligand as point negative charges.1356

These negative charges influence of course how the metal d-orbitals are arranged.1362

Let's go ahead and examine the most common type of geometry which is an octahedral field and a tetrahedral field.1367

For an octahedral field, you can imagine the following.1375

Your transition metal in the middle; then the ligands are on the outside.1380

These are all treated again as point negative charges like that.1384

Here is the octahedral framework.1392

It is determined that of the five metal d-orbitals, two get raised in energy and three go down in energy.1400

It turns out that dxy, the dxz, and the dyz orbitals lie in between the ligands.1413

Because they lie in between the ligands, there is no electron-electron repulsion.1430

Their energies go down.1440

The two remaining d-orbitals, dx2-y2 and dz2, are found to lie directly pointing toward the ligand positions.1445

Because they lie directly toward the ligand positions, there is electron-electron repulsion occurring.1462

Because of this, their energies go up.1470

You can imagine the five d metal orbitals.1479

All of a sudden, once the ligands attach, you get two going up and two going downward.1482

The two metal orbitals that are together, the dz2 and the dx2-y2, these are what we call the eg orbitals.1490

The three on the bottom, the dxy, the dxz, and the dyz, these are what we call the t2g orbitals.1499

The difference in energy between the eg and the t2g levels are what we call Δo.1506

Δo is what is known as the CFSE.1512

The CFSE is what we call the crystal field splitting energy.1517

Now that we have look at an octahedral field and we know how the d orbitals are what we call split,1532

let's now take a look at electron configuration using crystal field theory.1538

Consider a vanadium(II) ion in an octahedral field.1543

Vanadium has the configuration... vanadium 2+ is going to have the configuration argon followed by 3d3.1550

We are only dealing with 3d electrons here.1560

Because this is an octahedral field, you get the following 2-3 split.1563

Just as we have learned how to fill orbitals using Aufbau principle, we are still going to follow that.1570

d3, with the first d electron, second, and the third.1577

For d3, this is the only possible ground state configuration.1582

Once again this is the only possible ground state configuration.1590

However something different happens when we consider a chromium(III) ion in an octahedral field.1598

The configuration for chromium 3+ is going to be argon 3d4.1608

Now we actually have more than one possible ground state configuration because I can do one, two, three, four.1617

But also I can have one, two, three, and four.1629

Basically is the configuration going to be t2g4?1637

Or is the configuration going to be t2g3 eg1.1644

What that depends on is the magnitude of the crystal field splitting energy.1651

It depends on Δo.1659

Basically if Δo is too large, then we are going to get the t2g4.1662

If Δo is small enough, that is negligible enough, then yes1674

it will be relatively easy to promote that last electron to the upper levels.1681

We will get the t2g3 eg1 configuration.1686

The situation where Δo is too large, this is what we call the low spin scenario.1693

The situation where Δo is very small is what we call the high spin scenario.1702

How do you determine if Δo is small or large?1713

That is strictly going to be dependent on the ligands.1717

Let's go ahead and take a look at a series of common ligands next.1722

Here we have what is known as the spectral chemical series.1727

The spectral chemical series lists commonly encountered ligands in coordination chemistry starting with the carbonyl and cyano ligands.1736

Wrapping it up on the other end of the spectrum are the halides.1750

The way to read this is the following.1754

Ligands at the leftmost end are called strong field ligands because they result in a large crystal field splitting energy.1756

As a result, with these ligands, we get low spin configurations.1764

Ligands at the rightmost end are called weak field ligands because they result in a small crystal field splitting energy.1769

As a result, we often get high spin configurations with these ligands.1778

What you have to be on the lookout for are both possible high and low spin scenarios.1783

These will arise for configurations d4, d5, d6, and d7.1792

What is the big deal?--you may be wondering; why bother with this?1800

Now let's go ahead and examine the implications.1803

This is where it gets very interesting; the implications are the following.1805

We can use crystal field splitting theory to help us predict both the color and the magnetic properties of coordination compounds.1810

Let's first compare the magnetic properties of the following.1820

Iron with six aqua ligands attached making this 2+ versus now iron with six cyano ligands attached making that 4-.1824

In this situation, this is Fe2+.1841

In this situation on the right, this is also Fe2+.1844

Both of these complexes are d6.1848

The only difference obviously then is the identity of the ligand.1853

Because here H2O, octahedral d6; here we are going to have cyano d6.1858

Let's start with the extreme.1867

Cyano is one of the first ligands you encounter.1868

It is one of the strongest field ligands.1871

Because it is strong field, that means it is going to be low spin.1874

One, two, three, four, five, six.1878

You notice that all of my valence electrons are paired.1882

For the cyano compound, we expect diamagnetism.1886

However for the aqua compound, it is determined that it is actually high spin.1893

One, two, three, four, five, six.1898

You see that we get plenty of unpaired electrons.1901

For the aqua complex, we expect paramagnetism.1904

You see how useful crystal field theory can be in explaining for the magnetic behavior of certain coordination complexes.1909

Now a brief discussion on color.1919

We want to review what is known as complimentary colors.1922

Complimentary colors is coming really from physics.1930

This tells us that the color of light that an object absorbs will be complimentary1934

to the color of light that is transmitted to our eyes, basically what you and I see.1940

You have tried to construct something called a color wheel like so.1947

Basically complimentary colors are going to be opposite to each other.1957

For example, if something is going to absorb orange color light, something is going to look blue to us.1961

How does this relate to crystal field theory?1972

Basically that crystal field splitting energy Δo is going to be proportional1976

to the energy of the photon absorbed which means that it is1984

complimentary to the energy of the photon transmitted or given off1995

or the photon that travels to our eyes, what you and I see.2010

If Δo is large, a weaker color of light is going to be given off.2016

If Δo is small, a stronger color of light is going to be given off.2034

If Δo is very large, you basically have a good bet of getting something like red, orange, or yellow.2047

If Δo is small enough, the object will most likely look violet blue or green to us or any combination thereof.2055

Once again using crystal field theory, we are able to explain for both the color and magnetic properties of coordination compounds.2067

This is very fascinating material actually.2074

To summarize, coordination compounds contain a central metal atom or ion attached to ligands which may be atoms or molecules themselves.2079

Coordination compounds are often observed with coordination numbers of 4 or 6, giving tetrahedral, square planar, and octahedral geometries.2089

Finally as we just demonstrated, crystal field theory can explain for both the color and magnetism of coordination compounds.2097

Now let's go ahead and tackle some sample problems; name the following compounds.2108

Here we see that the coordination complex is a cation and that the chlorides are the anions.2115

We know hydroxide is 1-; we know that water is neutral.2129

We know that each chloride here is 1-.2135

This tells us that in order to cancel the overall charge of the chlorides, that iron here must be 3+.2139

Let's go ahead and name them; alphabetical order.2151

We are going to get aqua first; there is five of them.2153

That is going to be penta-aqua followed by hydroxide; but hydroxide becomes hydroxido.2156

Then followed by iron; we don't use ferrate because this is a cation coordination sphere.2166

Then chloride; so pentaaqua hydroxido-iron(III) chloride.2175

Now onto the bottom one; here we have also another one.2184

Your bracket is your cation; your sulfate of course is going to be your anion.2189

We know that ammine is 0; we know that water is 0.2196

Which means that in order to balance the charges here, we have three sulfates.2200

Actually this should only be one sulfate here... my apologies.2214

We know that sulfate here is 2-; the cobalt here must be 2+ overall.2222

Let's go ahead and name this compound.2228

This is going to be triamine triaqua cobalt(II) and then sulfate.2230

That is sample problem one, focusing on nomenclature of coordination compounds.2244

The last sample problem deals with magnetism.2252

What change of magnetic properties if any can be expected when NO21- ligands2255

in an octahedral field are replaced by chloride ligands in a d6 complex?2261

When we look up the spectral chemical series, we see that the NO21- ligand is going to be a stronger field ligand than Cl1-.2266

This is going to give us a larger Δo.2279

This is going to give us the smaller Δo.2283

Remember large Δo means it is going to be low spin.2287

Small Δo means it is going to be high spin.2294

Let's go ahead and draw the possibilities out.2298

d6 low spin means that all of my electrons are paired.2301

That is going to be diamagnetic.2308

Here we have d6 high spin.2310

All of a sudden, we get at least one unpaired electron.2315

That is going to be paramagnetic.2319

To answer this question, when you replace the NO21- ligands with the chloride ligands,2321

you go from a compound that is diamagnetic to one that exhibits paramagnetism.2327

That is our lecture today on the transition elements.2335

I want to thank you for your time.2339

I will see you next time on Educator.com.2340

Hi, welcome back to Educator.com.0000

Today's lesson from general chemistry is on nuclear chemistry. Here is the lesson overview.0002

We are going to start off with a brief introduction as to exactly what we mean by nuclear reaction.0010

Then after that, we are going to go into the types of nuclear reactions, basically the different types of what we call radioactive decay.0015

We are going to learn how to write and balance nuclear equations followed by studying how fast a nuclear reaction can occur.0022

We are going to go ahead and wrap up the lesson with a summary as always followed by some sample problems.0033

Nuclear reactions are different than what we have been referring to all this whole time as a chemical reaction.0042

In a chemical reaction, chemical identity never changes.0051

Carbon remains carbon; oxygen remains oxygen.0055

But in nuclear reactions, the chemical identity of the reactant actually changes.0058

We actually change one element to something totally different.0063

In other words, we are changing the chemical composition.0067

We are changing the number of protons of the original starting material during the reaction.0072

In general, a nuclear reaction involves starting with an isotope that is relatively unstable.0078

The isotope can then undergo successive decay reactions or successive decompositions forming additional unstable isotopes or nuclei of different chemical identity.0086

We also call these daughter nuclei.0103

Because this is nuclear chemistry after all, you think of some type of energy associated with it.0109

Each decay is often associated with the release of energy in the form of what you and I commonly refer to as radiation.0119

Let's now discuss the different types of radioactive decay that can occur.0132

The first one is called alpha decay.0137

Alpha decay is the weakest type of radioactive decay.0139

In alpha decay, energy is released in the form of what we call an alpha particle.0144

An example will be the decomposition of this isotope here going to 222-86-Rn plus 4-2-He.0151

Let's go ahead and do a brief review.0177

The top number is equal to Z... excuse me, the top number is equal to A.0179

We talked about this a long time ago.0185

That is going to be the atomic mass.0187

The bottom number is of course what we call Z.0192

That is just basically your atomic number.0195

An alpha particle is essentially a helium atom.0199

Sometimes you see people write it also like that.0204

An alpha decay is the weakest type of radioactive decay.0209

You can essentially stop alpha particles in its path with simple paper.0212

A stronger type of decay, and therefore a little more dangerous, is called beta decay0221

where energy is released in the form of a beta particle.0226

A beta particle is essentially an electron.0229

Let's go ahead and look at this isotope of hydrogen which is what is called tritium.0233

That is going to decay to a helium isotope releasing also an electron.0241

An electron has 0 mass but an overall charge of -1.0247

That is how it represents an electron or also a beta particle.0252

0-(-1)-beta is sometime how it is also represented.0257

Because a beta particle is more dangerous, you require something a little tougher to stop it.0264

Something like a piece of aluminum metal is sufficient to stop a beta particle in its path.0273

However the strongest type of decay of course is going to be what we call gamma decay.0280

Gamma rays are pure energy; there is no mass.0285

All of the energy that is released is pure energy.0289

Nothing is transferred to another atom as mass.0294

The example would be an excited cobalt isotope going to a lower state cobalt isotope and releasing just a ridiculous amount of energy.0298

A gamma ray is often referred to as 0-0-γ or just γ for short.0312

Again this is pure energy.0321

Of course you need something a lot more dense to stop gamma radiation.0322

This is what is coming out of nuclear facilities after all.0327

Something like concrete or something like lead is going to be the only suitable objects.0331

There are other types of particles of varying energy that can also be released.0341

One type of decay is what we call positron emission.0347

A positron is essentially the positive equivalent of an electron.0353

That is going to be represented as Β1+ or as 0-1-e.0357

You can also have nuclear decay that releases a proton; proton emission.0366

Of course proton emission is basically just your hydrogen.0376

That is going to be 1-1-p.0380

Finally you can also have the release of neutrons which is going to be represented as 1-0-n.0385

Have a mass of 1 and a relative charge of 0.0396

Now that we know the different types of radioactive decay, let's go ahead and learn how to express nuclear reactions.0401

Just how we have seen this entire session on general chemistry, a chemical equation expresses a chemical reaction.0408

Similarly we are going to have what is called a nuclear equation to express a nuclear reaction.0417

The same principles hold though, especially the conservation of mass and the conservation of charge.0422

Basically the sum of the superscripts must equal each other on both sides of the equation.0433

The sum of the subscripts must equal also each other on both sides of the equation.0453

It is essentially we are balancing the equation essentially.0463

Let's just go ahead and take a look at three examples here.0467

We can have polonium-84 going to an alpha particle plus blank.0471

Let's go ahead and fill in what we can.0483

This subscript here has to be 82; subscript here has to be 207.0485

When you look up element 82 on the periodic table, you get lead.0491

Let's go ahead and do another one.0496

We can have sodium-11 undergoing decay to form magnesium-12 plus blank.0498

The subscript here has to be -1; superscript here is going to be 0.0509

That is going to be therefore an electron.0519

Excuse me... this is a typo here; let's make this 22; there we go.0522

Finally we can have calcium-20 combining this time with an electron.0529

When that happens, we are going to get 41 on top on the0540

right side and 19 on the bottom on the right side too.0543

That is element 19 which is potassium; it is relatively straightforward.0547

Nothing too difficult really when writing and balancing a nuclear equation.0552

Let's now go ahead and examine how fast a nuclear reaction can occur, basically the rate of nuclear decay.0559

The rate at which an unstable isotope decays is a matter of kinetics.0569

If you recall, kinetics as we saw was either zero, first, or second order which are the common ones.0574

It turns out that all radioactive decay processes follow first order kinetics.0582

Let's go ahead and refresh our memories and exactly what we mean by first order.0589

The integrated rate law for first order kinetics was the following.0594

The natural log of the concentration of A at some time t divided by the initial concentration of A is equal to -kt.0599

The half-life is equal to natural log of 2 over k.0612

If you all can recall what we mean by half-life, let's plot the amount on the y-axis.0620

On the x-axis, let's plot time.0630

Let's say this dot represents the initial concentration of A right here.0632

As it gets consumed, it is going to follow the following profile.0640

At some point, we are going to reach half of this initial amount.0652

The time that is required to reach half the initial amount, that is what we call the half-life.0658

We can come up with a nice equation in terms of mass where the mass0666

at time t is equal to the initial mass times e raised to the ?kt.0672

That is an equation we are going to be using when we do some sample problems.0680

Once again radioactive decay processes follow first order kinetics.0686

To summarize this very short session, nuclear reactions involve a change of chemical0693

composition and can be accompanied by substantial amounts of energy which is why0698

when nuclear reactors melt down, that is why they are so tragic.0704

Number two, the energy release can be in the form of what we call alpha, beta, and gamma rays,0709

with gamma rays of course being the most harmful and penetrating.0715

Writing and balancing a nuclear equation follows our most fundamental laws of conservation of mass and charge.0719

Finally radioactive decay processes follow first order kinetics.0727

Let's go ahead and tackle some sample problems.0734

Complete the following nuclear equations.0737

14-7-n plus something is going to go on to form oxygen-8 and a proton.0740

Let's go ahead and look at this.0752

Here my subscript is going to be 2; my superscript is 4.0754

That essentially is going to be my alpha particle.0758

Let's go ahead and do a second example.0763

Here we can have something plus a neutron going on to form Bk-97 plus an electron.0765

My subscript here is 96; my superscript is 248.0780

This is going to be a curium, Cm.0785

Another one can be something plus a neutron going on to form 244-96-Cm plus a gamma ray plus blank.0790

This should be americium-95... excuse me about that.0812

Now on the right side here, this is going to be 0 on top, -1 on bottom, and an electron.0816

Finally the last example is going to be carbon-6 reacting with a neutron.0823

That is going to go on to form something plus a gamma ray plus carbon-6.0835

I'm sorry, excuse me... the question mark is the carbon-6 to balance out everything.0851

That is just some straightforward examples on balancing nuclear equations.0856

Now finally onto sample problem two.0862

Potassium-40 can be used to date materials because it is presumed to have existed at the formation of the earth.0865

If three-fifths of the original K-40 exists in a rock, how old is the rock?0872

Three-fifths is the fraction remaining.0878

Once again three-fifths is the fraction remaining of K-40 in the rock.0887

We can look up the half-life of potassium-40.0900

The half-life of potassium-40 is 1.26 times 109 years.0905

The rate constant is going to be equal to natural log of 2 over the half-life.0915

We are going to get 5.50 times 10-10 reciprocal years.0922

We can set up and use the equation M at some time t is equal to M0 times e raised to the ?kt.0932

Three-fifths is equal to the M0... which is what we call 1 because at the initial point,0941

we have 1 if we are dealing with fractions... times e raised to the ?kt.0953

Solving for t, we are going to get 9.29 times 108 years.0959

Once again when using fractions, don't forget, when using fractions,0967

the initial amount can always be represented as simply 1; 1.0.0976

That is our lecture on nuclear chemistry.0993

I will see you next time on Educator.com.0997

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is on atoms, molecules, and ions.0003

We are going to go ahead and look at the lesson overview right now.0009

We are going to get a brief introduction into atomic structure which is more of a historical background.0012

Following this, we will get into what we mean by isotopes.0019

We will get introduced to the periodic table in this lesson followed by naming compounds.0022

Basically ionic compounds, those that are made from metals and nonmetals.0028

Polyatomic ions, hydrates, and also what we mean by molecular compounds.0036

We are then going to get associated with the concept of the mole0042

and how that can be used to determine what is called a percentage composition0047

which is going to ultimately lead to the finale of this lesson0053

which is on determining what is known as the empirical and molecular formulas.0057

By the beginning of the twentieth century, physicists had already known the following.0065

Number one, that opposite charges attracted and that like charges repelled.0071

Number two, a subatomic particle existed that was negatively charged which had been called the electron; commonly abbreviated e-.0075

One of the first accepted models of the atom was called the plum pudding model.0086

What the plum pudding model was the following.0090

It was essentially a sphere of positive charge.0092

Embedded in the sphere of positive charge were the subatomic particles, the electrons.0099

These were completely stationary if you will.0107

But there was no accounting for any other type of subatomic particle.0113

Once again, this is called the plum pudding model.0119

A sphere of positive charge that contained electrons scattered throughout.0122

We now get into one of the earliest theories that was accepted to explain for atomic structure.0129

This was called John Dalton's atomic theory.0138

John Dalton's atomic theory contained the following premises.0141

Number one was that matter was composed of individual units called atoms.0146

Number two, that these atoms were the smallest unit possible and that they were indivisible.0171

That is we could not break them down any further; atoms were indivisible.0177

Number three, the third premise of John Dalton's atomic theory was that atoms of one element were all identical to each other.0188

Number four, atoms of one element are going to be different than atoms of another element.0213

That is what we reference as chemical identity.0218

Atoms of one element different than atoms of another element.0224

Finally the fifth and final premise was that when elements combine, they can combine in small whole number ratios to form compounds.0239

Atoms combine in small whole number ratios to form compounds.0250

This fifth premise here is going to be looked at again.0268

That is what we call the law of multiple proportions.0273

I am going to go ahead and put an asterisk next to0284

each of the premises which we actually find not be so true today.0286

Atoms were indivisible; that is not true because we can divide an atom in a process called nuclear fission.0292

So premise two is not completely accurate.0298

The next one is atoms of one element are all identical to each other.0302

We are going to talk about this--is that not all atoms of the same element are identical.0307

Because atoms can exist in different forms which we call isotopes.0312

That is going to be coming on later on.0317

Following John Dalton's atomic theory was one of the first major experiments on atomic structure.0323

This is what we call Ernest Rutherford's gold foil experiment.0331

What Ernest Rutherford did was the following.0334

He shot a beam of high energy particles that were positively charged.0337

These positively charged particles were what we call alpha rays.0343

This was shot at a very thin piece of gold foil.0348

What was the expected result?--the expected result was the following.0355

That all of the alpha rays should have gone straight through.0360

Why?--because the positive sphere was mostly empty space.0366

And that the positive ray should have been attracted to the electrons that are embedded in the sphere.0370

This is expected result due to plum pudding.0377

However the actual result was the following.0387

From the gold foil, yes the majority of the rays did indeed go straight through.0390

However what was determined was that some rays actually got deflected at various angles.0397

You would actually have a couple rays get deflected completely backward.0405

This was highly unusual because this was not predicted by the plum pudding model.0411

Instead what came about was the following implication; an implication was the following.0416

That matter and atoms contained a dense center of positive charge.0424

Because it is positive charge that was making the alpha rays deflect.0441

This positive charge is what we today refer to as the nucleus.0449

The nucleus is at the center of all atoms and is positively charged.0456

To resummarize the implications of the gold foil experiment was that all atoms0464

have a dense center that is positively charged which we call the nucleus.0469

Inside the nucleus, two subatomic particles exist.0474

The proton is positively charged; the neutron has zero charge.0477

Number three, the nucleus is responsible for the mass of the atom; that is what gives an atom weight.0482

Finally electrons, which were already known to have existed, exist outside the nucleus and have negligible weight.0489

Let's go ahead and summarize these subatomic particles--neutron, proton, and electron.0499

Please make a note that these relative masses and the relative charges are indeed in relation to each other.0510

A neutron is going to be assigned a mass of 1 and a charge of 0.0520

A proton is assigned a mass of 1 and a charge of +1.0524

An electron has a negligible mass, a mass of 0, and a charge of -1.0529

Of course, we symbolize the neutron lower-case n, proton p, and the electron is going to be e.0535

One fallacy in John Dalton's atomic theory was that atoms of the same element are all identical to each other.0547

But we are going to learn that atoms of the same element can exist in different forms; these are called isotopes.0555

Isotopes are atoms of the same element; they differ only in their number of neutrons.0563

Number of protons is going to be equal to the number of electrons; this is because of charged neutrality.0570

But the only difference is the number of neutrons; number of n differs.0577

We are going to introduce something called isotope format.0586

For example, carbon has two isotopes, carbon-12 and carbon-13.0589

Anytime you see the number immediately following the hyphen, this is what we call the mass number of the isotope.0597

The mass number of the isotope is simply equal to the number of neutrons plus the number of protons.0607

When we write an isotope in a specific format, it is always going to have the following design.0618

Element symbol, X; A on top and Z on the bottom as a subscript.0628

X is simply your element symbol; A is your mass number; Z is simply your atomic number.0634

The atomic number is strictly equal to the number of protons of the element.0648

A mathematical relation follows that if you take the difference of A minus Z, you are going to get the following.0659

The number of neutrons plus the number of protons and then minus the number of protons; as you can see protons cancels.0668

When you take the difference of A minus Z, you simply get the number of neutrons.0677

For example, if you take 12, 6 and carbon.0683

We have 12 is equal to the number of neutrons plus the number of protons.0689

6 is on the bottom; that is the number of protons which is also equal to the number of electrons.0698

Therefore A minus Z is equal to 6; that is equal to the number of neutrons.0704

So carbon-12 has six protons, six neutrons, and six electrons.0712

However if you do the same analysis for carbon-13, we are going to get six protons, seven neutrons, and six electrons.0721

As you can see, isotopes strictly differ by the number of neutrons present inside the nucleus.0730

Now we get introduced to the periodic table.0739

The periodic table is essentially organized into different rows and columns.0742

Each column is also referred to as a group; each row is referred to as a period.0749

Typically there is always a staircase that is embedded in all of our periodic tables.0766

Anything to the left side of the staircase is what we call a metal.0776

Anything to the right side of the staircase is what we call a nonmetal.0785

When we look at the periodic table, you see always the following.0797

You see a number on top; the number of top is what we call the element number.0802

The element number is actually the atomic number, the number of protons.0810

The number on the bottom is the atomic mass of the element.0816

Finally right in the middle is your element symbol.0823

We are going to talk now more about different areas of the periodic table.0830

We talked about metals; metals have the following characteristics.0839

They appear lustrous which is shiny; they are good conductors of heat and of electricity.0843

They are malleable which means we can form dinged sheets out of them.0850

And they are ductile which means we can pull them into fine wires.0856

Nonmetals are going to be the opposite.0866

They are going to be electrically and thermally insulated which is what we mean by poor conductivity.0869

In addition, they are brittle which means that if we take a hammer and hit them, they are going to crumble.0878

Instead are not too malleable; so will crumble.0885

But right in the middle of the border between metals and nonmetals, that is what we find semimetals.0891

Semimetals are also known as metalloids.0900

They border that staircase that we drew previously; border the staircase.0904

Because they are called semimetals or metalloids, they are going to have properties of both.0915

They tend to be good conductors of electricity but poor conductors of heat; they tend to be malleable.0923

The typical examples are of course silicon and germanium.0928

We see these two being used in a lot of electrical devices and what we call semiconductors.0938

What we are going to analyze next are some of the major columns of the periodic table.0960

Group 1 is what we call the alkali metals such as lithium and sodium and potassium, etc.0966

Group 1 metals tend to have the following characteristics.0974

That is they are moisture sensitive which means they have a tendency to react with water.0977

They can be explosive when they react with water.0982

Group 2 are what we call the alkaline earth metals such as calcium and magnesium and barium.0986

These are harder than group 1 and less moisture sensitive.0995

Group 6 are the halogens which are fluorine, chlorine, bromine, and iodine.1001

All of these occur naturally in nature diatomically which means they occur in pairs.1007

When someone says fluorine, it is really F2, Cl2, Br2, and I2; the halogens are incredibly reactive.1015

Finally group number 8, the rightmost column of the periodic table are what we call the noble gases such as helium, argon, and neon.1028

Unlike the group 7s, these are going to be monoatomic; that is they occur just as individual units.1039

They are called noble gas species because they are relatively inert and have a smaller tendency to react.1045

Now that we have been introduced to the periodic tables and the elements that comprise them, let's now talk about different types of compounds.1057

The first type of compound we are going to study are called ionic compounds.1066

Ionic compounds are formed between a metal and a nonmetal or they contain what are known as polyatomic ions.1069

Because of the strong electrostatic attractions that exist between opposite charges, ionic compounds tend to have very high melting points.1078

Below are a couple of these polyatomic ions; polyatomic could be the following.1087

Something like CO32- which is called carbonate or SO42- which is called sulfate.1094

NO31- which is called nitrate; you see why these are called polyatomic ions.1107

Because it is several atoms coming together to form one entity; those are what we call polyatomic.1111

That is different than monoatomic ions of course which are just formed from one unit.1120

Something like Cl-, O2-, Al3+, etc.; polyatomic versus monoatomic.1124

It turns out that we can actually use the periodic table to predict the ionic charge for main group elements.1135

Main group elements are going to be group 1 and group 2, then group 3 to 6, and then 7 and 8.1141

It turns out that all of the group 1's tend to form a 1+ charge.1160

All of the group 2's tend to form a 2+ charge; group 3's tend to form a 3+ charge.1167

We are going to skip 4 and 5 because they tend not to form ionic compounds to such a large degree.1174

Column 6, we are going to get a 2- charge; column 7 is going to be a 1- charge.1183

Group 8, we say no charge because group 8 are the noble gases and they tend not to react with other species.1188

You noticed also very importantly that the positive charges, which are what are called cations, these are going to be formed by metals.1196

The negative charges are what are called anions; these are all nonmetals.1208

I am sorry; I am leaving out group 5.1219

Group 5 is of course... this is going to be a 3- charge.1222

So you have your cations and then your anions.1226

Once again cations are going to be formed by metals and anions are going to formed by nonmetals.1229

Now we have to learn how to name ionic compounds, such as NaCl, such as BaF2.1237

We need to now go over the systematic procedure for doing so.1247

You are going to name the cation first followed by the anion and the anion will end in ?ide.1253

Something like NaCl; the cation is simply going to be sodium.1259

Cl is usually chlorine; but now we are going to become chloride.1267

So the correct name of NaCl is simply sodium chloride; very simple.1275

You do not use prefixes mono, di, tri, etc, to denote subscripts in a formula.1284

Something like for example BaF2; let's go ahead and name it.1289

That is going to be barium followed by fluoride; you do not say difluoride.1294

No, no difluoride, for example; again we ignore the prefixes.1302

For certain transition metals, you are going to use a Roman numeral in parentheses to indicate the metal charge.1310

If we want Fe2+, we are going to denote it as iron(II); Fe3+ cation is going to be Fe(III).1317

Again you should definitely ask your instructor as to which transition metals he or she1331

would like you to know for sure where you have to use a parentheses to indicate charge.1336

Finally you can also have a hydrate; a hydrate is when you have water attached.1343

For example, CoCl2·5H2O, we are going to use the prefixes followed by the word hydrate.1350

Five using the Greek prefix would be penta; we would just say that this is the pentahydrate compound of CoCl2.1360

Let's go ahead and put everything together and do a couple of examples.1372

Al2O3, FeCl3, CuI2·3H2O; we are going to name the cation first.1378

Al2O3, that is going to be aluminum followed by oxygen becoming oxide.1394

FeCl3, Fe is one of those we need to know.1402

Iron, something in parentheses, and then chlorine becomes chloride.1406

Now we have to come up with the charge; each chlorine is a 1- charge.1411

There is three of them, giving me a 3- overall.1416

For iron, iron is going to be 3+ here, giving us iron(III) chloride in the end.1422

Don't forget for the ionic compounds, the net charge should equal zero to maintain charge neutrality.1431

Right here we are going to have copper.1442

Copper is one of those where you need a Roman numeral for, followed by iodide.1444

Then we have three waters which becomes trihydrate.1451

Here iodine is a 1- charge; there is two of them giving me a -2 overall.1457

Copper must be a +2 to balance it; I am going to indicate that in the parentheses as Roman numeral.1464

Now we have to do the other way.1474

Before I gave you the formula and you were asked to provide the name.1476

Now let's go ahead and write the formula given the name.1480

Let's go ahead and look at the following examples; barium phosphide; ammonium phosphate.1485

Barium phosphide; barium is Ba; phosphide is phosphorus which is P.1511

What I would like to do is I would like to put the charges up in the top just for my own purposes.1518

Barium is 2+; phosphorus is 3-; what I always tell my students to do is then I use the crossing rule.1524

Basically if you cross the charges, they become the subscripts for each of the other elements.1530

Ba is going to get subscript of 3; phosphorus is going to get a subscript of 2.1537

You can convince yourself to make sure that the charges cancel.1543

Three of the bariums times 2+ plus two of the phosphorus times 3-, yes they do cancel to zero.1547

Next one is ammonium phosphate; ammonium is NH4; phosphate is PO4.1556

Ammonium is a 1+ charge; phosphate is a 3- charge; let's go ahead and cross them again.1564

When we cross them, we are going to get (NH4)3PO4 with a 1.1572

Let's go ahead and make sure that they all work out.1582

Three of the ammoniums times a +1 charge plus one of the phosphates and a 3- charge; yes it does cancel to zero.1585

That is some nice sample problems on naming ionic compounds.1596

The other type of compound we are going to learn about now is called a molecular compound.1603

Molecular compounds are not a metal and nonmetal but instead they are composed from two nonmetals.1608

Something like CO2; something like CO; and of course H2O itself.1613

To name a molecular compound, the rules are going to be very similar with one main exception.1620

We are going to name the first element first by its entire name; then the second element has a ?ide suffix.1627

This time we do pay attention to the subscripts.1636

We are going to use the Greek prefixes, mono, di, tri, tetra, etc, to determine what the subscript is.1639

However you omit mono for the first nonmetal.1647

Something like CO2, name the first element first which is carbon.1651

The second element is oxygen becoming oxide.1656

There is two of them which is how we come up with carbon dioxide.1660

Something like N2O5; there is two nitrogens which is di.1665

The first element gets the full name; so dinitrogen; then five oxides or just pentoxide.1670

So actually naming a molecular compound is always easier than naming an ionic compound.1679

Why?--because you don't have to worry about charge.1685

Now onto what we call the concept of the mole.1691

When you go to the supermarket and you purchase eggs, they are always in either a half dozen or a dozen.1697

It is very hard and very seldom that you are going to purchase just an individual egg.1704

It is convenience; it is convenience for us to refer to eggs as a dozen.1708

For chemistry, we have the equivalent of our dozen.1716

The mole is our chemists' dozen for convenience.1721

When you go into lab, you don't measure just the weight of for example one atom.1730

You don't deal with an individual atom; it is not practical.1735

What the mole is all about is that it goes from the microscale to a macroscale; it is more practical.1739

Basically our dozen which is what we call the mole... the mole is abbreviated simply m-o-l.1755

It is equal to the following: 6.022 times 1023 units of whatever you are measuring.1763

This is equivalent of saying one dozen is equal to twelve units of whatever you are measuring.1774

Again our dozen, the chemists' dozen is what we call the mole.1782

This mole is probably... this number you have to memorize I am pretty sure.1786

6.022 times 1023 is what we call Avogadro's number; Avogadro's number.1792

The mole we are going to find is a central unit; it can connect the following quantities.1803

The mole, we can equal and calculate individual units; the mole we can also get grams.1809

When we go from mole to units, we are going to multiply this way.1823

By Avogadro's number, 6.022 times 1023 units for every one mole.1830

But when we go the opposite direction, we are going to divide.1838

We are going to divide by Avogadro's number.1842

Mole can also be used to go to mass.1854

When we go to mass, we can actually multiply the mole by a rate conversion factor of gram for every one mole.1858

When we go the opposite direction, we are going to divide by one mole.1871

We are going to divide by grams; so we can see that mole is truly a central unit.1879

The conversion factor of gram per mole, that is what we call the molar mass.1892

This is equal to the grams that one mole of a substance weighs.1903

We are going to come back to this quite heavily.1920

We are going to do a lot of example problems utilizing the mole.1924

Where do we get this molar mass from then?--on the periodic table.1929

On the periodic table, atomic masses... remember it is the number below the element symbol... are also called molar masses.1935

When you look at C-6, 12.101 on the periodic table, this bottom number here it the molar mass of carbon.1943

One mole of carbon weighs exactly 12.01 grams; that is how you write it out.1953

That is going to be a conversion factor.1960

How many moles are in 25.7 grams of sodium?1962

You say 25.7 grams of sodium times something over something.1967

That is going to give us moles of sodium.1973

Remembering our dimensional analysis, now grams of sodium is going to go downstairs.1978

Moles of sodium is going to go upstairs.1984

You just look up the value; that is going to be 22.99 grams of sodium for every one mole of sodium.1988

Giving you your answer in units of moles of sodium.1994

Let's now go ahead and utilize Avogadro's number.1998

How many atoms are in 1.2 moles of carbon?2002

1.2 moles of carbon times something over something is going to give us our answer in units of atoms of carbon.2008

To get cancelled, moles of carbons goes downstairs; then atoms of carbon goes upstairs.2021

We know from our flow chart with the mole that this is involving Avogadro's number.2029

One mole will go downstairs and 6.022 times 1023 goes upstairs to give our answer in units of atoms of carbon.2034

Whenever you see the word atoms or molecules, that is pretty much a good give away2046

that you are going to be using Avogadro's number as your conversion factor.2053

Again always keep in mind that; especially if the word moles is mentioned too.2060

Now that we have been introduced to what the mole can help us get,2068

we are going to find something else that the mole can be used to calculate,2073

what is called molar mass of not only an atom but also of a compound.2079

Let me say that again; the mole can be used to calculate the molar mass of a compound.2084

The molar mass of an element is already provided to us on the periodic table.2088

Basically the molar mass is simply the net sum of all molar masses of each element.2093

The molar mass of carbon dioxide is simply equal to2111

the molar mass of carbon plus the molar mass of oxygen2114

plus the molar mass of oxygen because we have two oxygens.2122

That is going to be equal to 12.01 grams plus 16.00 grams plus 16.00 grams.2126

Giving us a grand total of 44.01 grams; one mole of CO2 weighs exactly 44.01 grams.2135

Having said that, we can also use molar mass as a conversion factor for a compound.2149

How many grams are in 1.2 moles of carbon dioxide?2154

1.2 moles of CO2 times something over something is going to give us our answer in units of grams of CO2.2158

The mole is going to go downstairs; the grams is going to go upstairs.2169

Right there in the previous problem, we calculated this conversion factor or molar mass of CO2.2175

Anytime you want to relate mass and moles, it is always molar mass to be your conversion factor.2183

It is going to be 1 on the bottom; it is going to be 44.01 on top.2190

Giving us our answer in units of grams of CO2.2195

The next application of the mole is what we call percentage composition.2206

Percentage composition is given to us to be the following.2211

Percent composition of an element tells us basically the parts of the element divided by total parts times 100.2216

It basically tells us your relative amount that a specific element makes up of the compound.2236

This equation that we actually use is going to then be the total mass of the element2244

divided by the molar mass of the compound; all of that times 100.2256

Let's go ahead and answer this question.2269

How many grams of carbon dioxide are contained in 65.1 grams of CO2?2270

This problem is strictly wanting to know how much carbon is going to be contained in so much CO2.2276

Let's go ahead and first calculate the percentage composition of carbon in CO2.2283

We are going to say 12.01 grams because there is only one carbon.2294

Divided by the entire molar mass of CO2 which is 44.01 grams; times 100.2299

That is going to give us 27.3 percent.2306

Carbon dioxide is only 27.3 percent carbon; the majority is oxygen.2309

That tells me that any sample I hold of CO2 in my hand, 27.3 percent of that is carbon.2317

We are going to use percentage as our conversion factor.2324

We are going to say .273 times the 65.1 grams of CO2 is going to give me my grams of carbon.2328

We are going to get 17.8 grams as our answer.2339

Once again anytime you see a problem that relates the amount of an element2345

contained in a certain amount of compound, you want to use percentage composition as your tool.2351

Finally we are now going to see how we can determine what is known as an empirical and molecular formula.2360

An empirical formula is the simplest whole number ratio of a formula.2367

For example, CO2, this is an exact 1:2 ratio of carbon to oxygen.2371

That is my smallest ratio; I cannot go smaller than that.2378

However let's go ahead and look at one that is maybe C2H6.2382

In this example, I have a ratio of 2:6 of carbon to hydrogen.2388

But you notice that I can divide; I can divide by a lowest common multiple which is 2.2395

I am going to get 1:3 ratio instead; therefore 1:3 is actually my smallest ratio.2402

That is going to be CH3 which is my empirical formula.2410

What we are going to do is we are going to determine how to calculate the empirical formula.2415

The empirical formula is determined from elemental analysis.2422

Basically you have an unknown sample and you put it through...2428

You perform what is called an elemental analysis test on it.2434

After you perform this analysis, you are going to get a printout2440

of the percentage that a certain element makes up in your unknown.2445

You are going to get x percent of for example carbon.2450

You are going to get y percent of for example oxygen.2454

You are going to get z percent for example of hydrogen.2457

These percentages again are only to be used for empirical formulas.2461

They may or may not necessarily give you the actual formula; only the relative amounts are reported.2466

If you want to get the actual formula, which is known as the molecular formula, you can determine it.2473

But only if the molar mass is already provided for you.2478

Let's go ahead and look at a typical example.2486

A compound used as an additive for gasoline to help prevent engine knock has the following percent composition by mass.2490

71.65 percent chlorine; 24.27 percent carbon; 4.07 percent hydrogen.2497

The molar mass is known to be 98.96 grams per mole.2505

Determine both the empirical and molecular formulas.2509

The very first step is to get your percentages into grams; get percentage to grams.2516

After we get everything into grams, then we can go to our central unit from the chapter which is moles.2527

Again whenever you are in doubt, get to moles.2533

The nice thing about percentage is that percentage is always made out of 100.2536

If we assume 100 gram of compound, the percentages automatically become grams.2541

We have 71.65 grams of chlorine, 24.27 grams of carbon, and 4.07 of hydrogen.2548

Again assuming 100 gram of compound.2560

Now we are going to go ahead and get to moles.2572

This is something we know how to do by now.2575

When we go ahead and get to moles, we are going to get 2.02 moles of chlorine.2577

We are going to get 2.02 moles of carbon.2584

We are going to get 4.04 moles of hydrogen.2587

Step one is done; now step two.2596

Step two, we are going to take our moles and we are going to divide.2601

We are going to divide by the smallest moles present; divide by smallest moles present; everything.2604

We are going to get a nice whole number that is going to become our subscripts in our empirical formula.2617

The result is the empirical formula subscript.2626

When we take chlorine and divide it by the smallest number which is 2.02, we are going to get 1.2642

Carbon divided by 2.02, get 1.2648

For hydrogen, 4.04 divided by 2.02, we are going to get 2.2651

Giving us our empirical formula of CH2Cl.2655

What happens if you don't get a perfect whole number?--what if you get 1.5, 1.5, 2?2665

Dalton's law of multiple proportions tells us that we cannot combine elements in other than a small whole number ratio.2684

Basically any time you have a decimal number, we are just going to then2694

multiply everything by the same factor to get everything in whole numbers.2697

When we do that, we can multiply 1.5 by 2 and everything by 2.2702

Giving us nothing but whole numbers; 3, 3, and 4.2707

If that happened, then those numbers would become the empirical; we would get C3H4Cl3.2711

To get the molecular formula, you simply do the following.2722

You take the molar mass of the molecular formula which is given to you.2726

You divide it by the molar mass of your empirical formula.2733

That is always going to give you a nice whole number.2738

When we do this, we are going to get 98.96 divided by our molar mass of our empirical formula which is CH2Cl.2741

We actually get 48.468; when we do this, we get approximately 2.2752

You take that whole number and you multiply the subscripts of the empirical formula by this.2758

I get a molecular formula of C2H4Cl2 which is the formula of the actual compound.2764

You can always double check your work.2773

You can add up the molar mass of C2H4Cl2.2774

We are going to get very close to 98.96.2778

Let's go ahead and summarize.2783

Atoms are composed of a central nucleus that is positively charged.2786

Protons and neutrons reside within the nucleus; electrons are outside.2790

We saw that isotopes are the same element; differ only by their number of neutrons.2796

We have went through some specific rules for naming ionic and molecular compounds.2801

We also saw that the mole is a central unit that allows for conversion between number of atoms and molecules and for mass.2806

Finally the mole is a central unit that is required for several types of different problems2814

Including percentage composition and the empirical and molecular formula problems.2822

Let's go ahead and look at some sample problems.2831

You have 1 milligram of lithium metal reacting with molecular fluorine gas.2834

The resulting fluoride salt has a mass of 7.3 mg.2838

Determine the empirical formula of lithium fluoride; you have 1.00 mg of lithium.2842

After it reacts with the molecular fluorine gas which is F2, you get 3.73 mg of your lithium fluoride.2850

If I start with only 1 mg of lithium and I wind up with 3.73 mg of compound, why did my mass get heavier?2865

The mass got heavier because it reacted with fluorine; fluorine came on board.2874

We can actually determine the mass of fluorine.2879

It is just the difference, 3.73 mg minus 1.00 mg.2882

That is going to give us 2.73 mg which is the mass of fluorine reacted.2888

We got all of our masses now; now let's get everything into moles.2902

2.73 mg of fluorine is going to become 1.44 times 10-4 moles of fluorine.2908

1.00 mg of lithium is going to become 1.44 times 10-4 moles of lithium.2919

Now that we have all of our moles, we can divide by the smallest number present.2930

For this problem, they are identical.2934

When we divide everything, we are going to get Li1 and F1 for our empirical formula; or just LiF.2937

This answer actually makes sense because we know that lithium is a 1+ charge and that fluorine is a 1- charge.2947

Indeed the charges do balance each other out.2954

Now moving onto sample problem two.2962

How many atoms of carbon are present in 2.67 kg of C6H6?2964

In this case, we are asked about an element within a certain compound.2970

That sounds a lot like percentage composition.2976

Let's go ahead and calculate the percentage composition of carbon in C6H6.2978

The percentage composition is equal to the total mass of carbon...2984

There is six of them by the molar mass 12.01.2988

Divided by the total mass of C6H6 which is 6 by 12.01 plus the 6 hydrogens by 1.008.2991

All of that times 100; when we do this, we get 92.3 percent.3001

So C6H6 is mostly carbon; 92 percent carbon.3007

Remember what we did last time.3013

We are going to take our percentage composition 0.923.3015

We are going to multiply it by the total mass 2.67 kg.3019

That is going to give us the 2.46 kg of carbon in this specific sample.3023

The question is asking for atoms of carbon though.3031

Somehow we have to go from kilograms to atoms.3035

Remember anytime you see the word atoms or molecules, it is going to be via an Avogadro number.3040

Our first step is to go from kg just to regular g.3047

Then from g to the central unit which is moles.3052

Then of course moles onto atoms; we can go ahead and do this.3056

We are going to say 2.46 kg of carbon times something over something.3063

G goes on top; kg on the bottom; that is going to be 103 kg over 1 kg.3071

Now onto moles; times something over something; g goes downstairs to get cancelled.3078

Moles goes upstairs to get carried through the final answer; that is molar mass.3084

You look up the molar mass of carbon which is 12.01 grams for every one mole.3090

Then finally times something over something, giving us our answer in atoms of carbon.3095

That is Avogadro's number; mole on the bottom; Avogadro's on top which is atoms here.3102

That is 1 mole on the bottom and 6.022 times 1023 atoms on top.3110

You should get an answer of 1.23 times 1026 atoms of carbon.3116

I want to thank you for your attention.3128

This concludes our lecture on atoms, molecules, and ions.3130

Thank you for using Educator.com.3135

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is on chemical reactions; let's go over the outline.0002

We are first going to talk about something very fundamental to all of the physical sciences.0011

That is the law of conservation of mass.0017

We are going to see that in order to obey the law of conservation of mass,0020

anytime we have a chemical equation, we have to make sure that it is completely balanced.0024

After discussing this, we will then go on to something we call chemical equilibrium.0031

Because chemical equilibrium is going to be very applicable to our discussion of reactions in solution.0037

The solutions that we are specifically talking about are electrolyte solutions, nonelectrolyte solutions, and etc.0048

After talking about electrolytes and nonelectrolytes, we are then going to dive into a couple of different reactions that can occur in aqueous solutions.0058

The first one is what we call single replacement reactions.0067

The second one is what we call double replacement reactions.0072

The third one is neutralization reactions.0076

The fourth and final one is going to be reactions that produce unstable compounds0078

that are going to further decompose and form other compounds we cannot initially predict.0084

When this section is done, we are going to be able to predict0092

the compounds that are formed from any of these types of reactions.0095

We will wrap up the session with of course our summary followed by a pair of sample problems.0101

Let's go ahead and start.0107

The law of conservation of mass is fundamental to all of the physical sciences.0110

We always will be obeying this in chemical reactions.0114

Basically the law of conservation of mass tells us the following: that matter can neither be created nor destroyed.0118

In other words, what you have, you are going to have permanently.0126

It is coming from other sources.0131

You are never going to truly eliminate it or get rid of it.0134

But matter as we know can be transformed into different states, etc.0140

If we were to translate this into a chemical reaction before and after, the law of conservation of mass basically tells us the following.0146

That the total mass before a reaction is going to be equal to the total mass after a reaction.0153

In other words, the total mass of what you start with equals to the total mass of what you end up with.0162

Let's go ahead and look at a typical way of representing a chemical reaction.0170

Basically the general format of a chemical reaction is the following.0175

We have the reactants on the left side of the arrow.0180

We have the products on the right side of the arrow.0184

We can go ahead and verbalize this the following way.0186

That A combines with B to form C and D; or A and B yields C and D.0190

Or A and B mix C and D; whatever you are more comfortable with.0197

Basically if we were to apply the law of conservation of mass, this is telling us the following.0204

That the total mass combined of A and B will be equal to the total mass combined of C and D.0209

In order to obey the law of conservation of mass, we have to do what is called balancing a chemical equation.0222

You are going to be using this from now on pretty much in every lesson.0230

This is one of the most fundamental parts of general chemistry that you want to learn to master early on.0235

When we balance a chemical reaction, let's go ahead and look at the following.0242

N2 plus H2 goes on to form NH3.0246

We want to look at the number of elements on both sides of the equation.0252

For example, here we notice that there are a total of two nitrogens on the left side.0255

But there is only one nitrogen on the right side.0262

In addition on the reactant side there are two hydrogens.0266

There are three hydrogens on the right side.0271

We say that this is an unbalanced chemical reaction.0274

We never want to leave a chemical equation unbalanced.0280

The way to go about this is the following.0285

We are going to put what are called coefficients right in front of each of the molecules to go ahead and balance this.0288

The coefficients are usually going to be whole numbers.0301

I want to give you a couple tips.0304

Whenever the reactants are all homonuclear, it doesn't matter which element you try to balance first.0306

What I mean by homonuclear is that it is basically a compound composed of just one element; one element type only.0312

Examples of homonuclear would be N2, H2, O2, Cl2, etc.0326

Those again are what we call homonuclear.0335

A lot of the time, it is trial and error; so let's go ahead and start.0339

N2 plus H2 goes on to form NH3.0343

What I would like to do, I would like to maybe balance nitrogen first.0349

Again it doesn't matter what order you do it in.0352

I am going to go ahead and put a 2 in front of NH3.0356

Then I ask myself: I have now two nitrogens on the left.0361

I have now two nitrogens on the right; the last thing to fix is hydrogen.0364

Now I have two hydrogens on the left; I have six hydrogens on the right.0370

Basically we ask ourselves what multiplied by two is going to give us six.0376

The answer is going to be three.0382

My balanced chemical equation is going to be N2 plus 3H2 goes on to form 2NH3.0384

Always, this is very important, always double check your work to make sure that you have done the balancing correct.0393

As you can see, we have two nitrogens on both sides of the equation.0397

And we have a total of six hydrogens on both sides of the equation.0401

Before we move on, another thing I just want to point out is the following.0409

When you have the coefficient and a subscript, we just multiply them together to get the actual number.0413

That is how we get six hydrogens there; coefficient multiplied by the subscript.0419

Let's go ahead and now move on to another chemical reaction where the reactants are not both homonuclear.0432

For example, CH4 plus O2 goes on to form CO2 and water.0440

We have CH4 which is going to be what we call heteronuclear.0449

That is the compound is made up of more than one element.0453

My advice I usually give is the following: you always save the homonuclear reactant for the last step.0457

That is going to make your life a lot easier.0469

Let's go ahead and try to balance this.0471

Again it is CH4 plus O2 goes on to form CO2 and H2O.0474

I am going to save oxygen for last because it is part of a homonuclear compound.0481

I am left with either carbon or hydrogen to start with.0485

Again it doesn't matter for those; let's just go ahead and pick carbon.0488

Carbon, we have one on each side of the equation.0493

We have one in the CH4 and we have one in the carbon dioxide; so carbon is already balanced.0496

The only thing remaining is hydrogen now to work with before we tackle oxygen.0502

On the left side of the equation, I have a total of four hydrogens.0507

On the right side of the equation, I have a total of two hydrogens.0510

I am going to put 2 right in front of water.0515

Now my hydrogens are balanced; I have four on both sides of the equation.0518

Remember multiply the coefficient by the subscript, two times two gives us four.0522

Now we can go ahead and tackle oxygen.0527

I have a total of two oxygens on the left.0529

I have a total now of two oxygens in carbon dioxide.0534

I have a total of two oxygens in water; remember to carry through the coefficient.0538

So actually I have four oxygens total on the right side of the reaction.0545

I am going to put a 2 right in front of O2.0553

That gives me a total now of four oxygens on both sides of the reaction.0556

That is it; that is our balanced chemical reaction.0560

Another piece of advice I usually give students is the following.0565

Balancing chemical reactions can be very frustrating.0569

You don't always get it right the first time; a lot of it is trial and error.0572

If going by one element the first time doesn't work, try another element until you get it.0577

But remember it always can be balanced.0582

It is a fundamental law of nature; it always can work.0585

Right now, I want to introduce you a chemical reaction where sometimes the coefficients0593

that we are going to use initially are not going to be all whole numbers.0599

Instead the coefficients that we are going to be using are going to be fractional.0603

Remember, remember what we have to do.0609

It is going to be the coefficient times the subscript.0613

That is going to give us the total number of that element in the compound; the total number of element in compound.0620

Let's go ahead and balance this; C2H6 plus O2 goes on to CO2 and water.0633

I have two carbons on the left side; I only have one carbon on the right side.0639

I am going to go ahead and put a 2 in front of CO2; now my carbons are fixed.0645

Remember that I am leaving oxygen for last because oxygen is part of a homonuclear compound.0649

What we have to do now is fix hydrogens.0655

I have six hydrogens on the left; I have two on the right side.0658

Then we ask ourselves: what coefficent, what number times two is going to give me six?0663

That is going to be a three; so now my hydrogens are fixed; six on each side.0669

Now we have the following C2H6 plus O2 goes on to form 2Cl2 and 3H2O.0675

Let's go ahead and look at the oxygen count right now.0685

I have two oxygens on the left side.0688

Here on the right side, I have a total of four oxygens in CO2.0691

I have a total of three oxygens in water.0695

That means I have a grand total of seven oxygens on the right side and only have two on the left side.0700

This is when we have a case of an even-odd number.0710

If you look at the previous examples, we have always had an even number on one side and an even number on the other side.0713

In that case, it is always convenient just to use a whole number.0720

But when you have an even-odd pair like this, there is going to be a little trick we are going to do.0724

Basically you ask yourself the following.0729

You see that I am shy of oxygens on the left side.0732

We have to ask ourselves: what coefficient times two is going to give me seven?0736

Basically 2 goes down stairs; 7 goes upstairs; it is very simple.0743

When you have an even-odd pair, you will always use a fractional coefficient.0750

Basically the bottom number is going to be the subscript from the homonuclear compound.0757

The top number is always going to be the greater number; what you are trying to get the homonuclear compound up to.0771

Once again when you have an even-odd pair, use fractional coefficients.0785

The fractional coefficient I am going to use is 7/2.0789

Remember a few sessions back, we talked about Dalton's atomic theory.0796

Part of his theory was that elements combine in small whole number ratios to form compounds.0802

What that means is the following: we never want to leave a chemical equation in a fractional form.0808

In other words, what we are going to do right now, we are going to eliminate the fraction.0816

We are going to be left with nothing but whole numbers.0819

The fraction we have 7/2; if we want to get rid of a fraction, we simply divide by its denominator.0825

But I just can't multiply oxygen by 2; if I change one element, I have to do it proportionally to everything.0833

My grand and final answer is going to be the following: it is going to be 2C2H6 plus 7O2.0842

That is going to go on and form four carbon dioxides and six waters.0858

This gives us our balanced chemical equation with no fractions.0877

Let's just go ahead and do one final check, make sure we have done it right.0880

Let's go ahead and see if we have the same number of elements on each side.0883

I have 4 carbons total here; 4 carbons total here; carbons are good to go.0889

I have 12 hydrogens here; I have 12 hydrogens here; hydrogens are good to go.0895

Finally I have 14 oxygens here.0903

I have 8 oxygens in CO2 and 6 oxygens in the water; giving me a grand total of 14 oxygens.0907

As you can see, this method does work.0920

Again the final thought is to never leave a coefficient as a fraction.0923

Now that we have learned how to balance chemical equations, let's move on to the next topic of the session.0932

This is what we call chemical equilibrium.0939

When an ionic compound fully dissociates... what I mean by fully dissociates is that it completely breaks up.0944

It forms individual cations and anions.0951

Whenever this happens, we use a single arrow in the forward direction.0954

For example, hydrochloric acid is going to be fully dissociating when dissolved in water.0959

That is going to break up into its respective cation and anion.0966

But not all ionic compounds do this.0972

A lot of ionic compounds really incompletely dissociate which means they don't fully break up into cation and anion.0976

When this occurs, we use a new type of arrow.0985

It looks like a pair of opposite arrows.0991

This is what we call an equilibrium arrow.0995

What the equilibrium tells us is that the reaction is reversible.1004

In other words... let's look at HF, as hydrofluoric acid breaks apart.1019

In other words, as we move in the forward direction, H+ and F- instantaneously form.1024

However look at the reverse arrow; let's go ahead and translate that.1032

What that is telling us is that H+ and F- are going to somehow find each other again.1037

They are going to recombine and reform the original hydrofluoric acid molecule.1042

This is what we call dynamic equilibrium.1048

That is it is a constant process; it is not static whatsoever.1052

Basically both the forward and the reverse reactions happen simultaneously.1057

In other words, as soon as a compound dissociates, the ions are going to recombine into the original compound.1063

Again you are going to call this an equilibrium arrow.1072

We are going to be using this quite heavily in the latter half of this general chemistry class.1075

Now we are going to talk about specific compounds when they dissolve in water.1085

They are called electrolytes and nonelectrolytes.1091

When you hear the word electrolytes, it obviously sounds like electricity or electron.1094

Basically electrolytes are going to be ionic compounds that form ions when dissolved in water.1101

For example, we can go ahead and take a salt, table salt, sodium chloride.1107

Sodium chloride is going to dissociate and form the respective ions; sodium cation and chloride anion.1112

The following is going to be a fundamental requirement for a solution to be conductive.1122

Basically a solution has to have free ions in order to conduct an electric current.1128

That is why they call them electrolytes.1133

Strong electrolytes completely dissociate.1138

For strong electrolytes which completely dissociate, we are going to use a regular single arrow in the forward direction.1143

Weak electrolytes incompletely dissociate.1151

Remember going back to the previous slide, for ionic compounds that incompletely dissociate, we use an equilibrium arrow.1155

Finally there are compounds that when dissolved in water, they do not form ions whatsoever.1162

These are going to be nonelectrolytes; these are going to be typically molecular compounds.1171

For example, if you take ordinary glucose for example, C6H12O6; you go ahead and dissolve it in water.1176

You don't form individual ions; it just becomes aqueous.1190

You see that there are no ions whatsoever.1197

Glucose is going to be a nonelectrolyte solution.1200

Now that we have gotten those vocabulary terms out of the way,1205

what we are going to do right now, we are going to study specific types of aqueous reactions.1208

We are going to learn how to predict the products.1215

The first type of aqueous reaction is what we call a single replacement reaction; also known as single displacement.1218

A single replacement reaction is always going to have the following general format.1228

A... this is usually going to be a metal... plus B and C.1233

Remember this is going to be the metal; this is going to be the nonmetal; usually.1240

Or the cation and the anion goes on to form AC and B.1247

You notice that A has come in and essentially displaced B from its bond with C.1252

We go on to form AC and B; this is called a single replacement.1259

Let's go ahead and look at a couple of examples.1263

You take lithium metal and you react it with copper(II) chloride.1266

Lithium is going to come in and knock out copper; we form lithium chloride instead.1273

Copper is also going to come out as copper solid; however let's reverse the process.1279

If we take copper solid and go on and react it with lithium chloride, we get no reaction at all.1286

Let's go ahead and see why this is going to occur and how you are going to be able to predict this.1295

All of the experiments have been done already.1301

The results are summarized in what is called an activity series.1303

Basically an activity series is the following.1307

You see that lithium is right at the first element.1310

We call lithium a very active metal.1314

It can come in and knock out essentially any other element.1318

Let's look at sodium; the way to use this activity series is the following.1325

Sodium can knock out any element below it, but it cannot knock out any element above it.1329

For example, sodium will not be able to knock out calcium.1335

But it will be able to displace magnesium in a single replacement reaction.1340

Depending on your instructor, you may or may not have to memorize this activity series; definitely be on the lookout for that.1346

Another very common type of single replacement reaction occurs between the following: a metal and acid.1353

For example, when zinc solid is going to get dissolved in hydrochloric acid, we get the formation zinc chloride and hydrogen gas.1363

Essentially zinc has come in and knocked out hydrogen.1373

Let's go ahead and see if we can find hydrogen on this activity series.1377

We know that zinc is going to be right here and right below it is going to be H2.1381

Yes, this activity series does predict that zinc is going to come in and displace hydrogen in a single replacement reaction.1389

Here is the thing; when hydrogen gets displaced, it always forms hydrogen gas.1398

When displaced in a single replacement reaction, it is going to form H2 gas.1405

This is how we use an activity series again for single replacement reactions.1417

The next type of reaction we want to talk about is going to be called a double replacement reaction.1422

It is also known as double displacement.1428

Basically instead of one quote and quote knockout, two occur.1432

We are going to have two swaps; I always like to show brackets with this.1437

Basically A is now going to hook up with D and now C is going to hook up with B to form AD and CB.1443

Please note how the cation always hooks up with the other anion, etc.1452

An example of this is going to be the reaction between sodium chloride and lead(II) nitrate.1459

Sodium is the cation; it is going to hook up now with the other anion which is nitrate.1465

Lead, Pb2+, is going to hook up with the other anion chloride.1471

We get sodium nitrate and lead(II) chloride.1474

Please note... this is very important when you do this... that the charges never change, especially for the transition metals.1479

Remember when we went back to nomenclature and you learned that transition metals can have different oxidation numbers.1488

Here this is going to be Pb2+; there should be really a 2 there; sorry about that.1496

Here in the product side, it is going to remain Pb2+.1505

Please do not ever change a metal's charge in a double displacement reaction.1509

Pb2+ is going to remain Pb2+; that is why it is PbCl2 and not just PbCl for example.1523

The next type of equation though, something very related to this, is the following.1532

That is called a net ionic equation.1537

A net ionic equation is usually written to show the formation of the precipitate.1539

A precipitate is just a fancy name for the solid that is formed in a reaction.1544

Basically a net ionic equation shows the formation of the precipitate, in this case, lead(II) chloride, from its constituent ions.1550

That is basically going to be Pb2+ combining with two chlorides.1558

It is always aqueous plus aqueous goes on to form the solid.1563

Now that we have covered net ionic equations, how do you predict that a compound is going to be solid?1570

Or how do you predict if it is going to remain aqueous in these reactions?1579

This is something you may or may not have to memorize also.1584

So please be on the lookout for that from your instructor.1586

This is what we call the solubility rules for ionic compounds.1591

Basically this table is divided into two areas; first the compounds that are usually soluble.1595

What soluble means is that the compound is going to remain broken up into ions and dissolved.1602

Something that is soluble is going to be essentially aq or aqueous in the chemical reaction.1608

You are going to find ammonium salts, group 1 salts, nitrates, etc.1616

What is important in this solubility table is to pay attention to the exceptions.1623

For example, acetates are usually soluble; for example, sodium acetate will be aqueous in water.1627

However there are exceptions such as silver, Ag1+, and aluminum, Al3+.1635

While sodium acetate is going to be aqueous, silver acetate and aluminum acetate are exceptions.1640

They are going to be solid; that is they are going to be insoluble.1647

So you always have to pay attention to the exceptions.1654

The same goes for the other side of this table; compounds are usually insoluble.1657

They are going to be remaining intact and not break up relatively speaking.1662

They are going to remain as solids; so carbonates, phosphates, etc, are going to be solids.1667

However there are exceptions.1673

Something like magnesium carbonate is going to be mostly a solid in water.1676

But an exception is ammonium; ammonium carbonate is going to remain aqueous in water.1681

Let's go ahead and apply another type of double displacement reaction.1688

That is what we call a neutralization reaction.1697

A neutralization reaction always has the generic formula.1700

An acid is going to react with a base.1703

It is always going to form a salt, water, and it is always going to give off heat.1706

Anytime you mix an acid and a base, it is going to feel warm to the touch.1713

A typical reaction would be between hydrochloric acid and sodium hydroxide.1720

Neutralization reactions are going to be typically double displacement reactions.1725

Hydrogen is going to hook up with hydroxide.1729

Chloride is going to hook up with sodium.1733

Let's do the salt; the salt is going to be sodium chloride.1736

From the table of solubility rules which we just covered, this type of chloride is going to be soluble which is aqueous.1742

You see that when hydrogen combines with hydroxide, that is essentially going to form water.1750

That is going to be a liquid.1757

There we have it; a double displacement reaction in the form of what we call a neutralization reaction.1759

Let's do one last example; that is going to be between sulfuric acid and potassium hydroxide.1765

Sulfate is going to hook up with potassium.1772

Again hydrogen is going to hook up with hydroxide.1775

This time the salt is going to be potassium sulfate.1778

I need two potassiums to balance sulfate's 2- charge.1786

Look at the table of solubility rules; that is going to be aqueous.1790

Again I am going to always form water; let's go ahead and balance it.1794

I am going to need two potassium hydroxides to go ahead and react with the sulfuric acid.1799

I am also going to form two waters as a result.1808

Remember always balance your chemical reactions.1811

We saw that a neutralization reaction is a type of double displacement reaction.1817

There are a few exceptions where using the table of solubility rules1823

and simply going by a generic format for double replacement does not work.1829

These three reactions here are going to form compounds initially that are unstable.1835

These unstable compounds are going to in turn produce additional products which we just can't predict.1842

We are going to now introduce these to you.1850

The first type, number one, is going to be the reaction between a metal carbonate...1852

This is usually going to be a group 1 metal... reacting with an acid, HA.1858

Usually A is going to be the following: Cl, Br, iodine, nitrate, and also can be sulfate.1868

Sulfuric acid, nitric acid, hydrochloric acid, hydrobromic acid, hydroiodic acid, etc.1885

These are what we call the strong acids.1894

Again depending on your instructor, you may have to know a number of different ones.1897

So please definitely be on the lookout for that.1902

When we do the double displacement format, we are going to get H2CO3, carbonic acid, in combination with an HA aqueous.1905

It doesn't stop there; what happens is the following; that carbonic acid is actually unstable in water.1915

It is going to decompose to form CO2 and liquid water; that is one example.1920

Again a metal carbonate reacting with a strong acid goes on to form carbon dioxide gas and liquid water.1928

Similar to metal carbonate reacting with acid, we are also going to see metal sulfites reacting with acid.1937

Very similar, we are going to form instead of H2CO3, we are going to get H2SO3, sulfuric acid, in combination with the salt.1945

Sulfuric acid like carbonic acid is unstable in water.1955

It is going to form not CO2 but SO2 in combination with water.1959

Once again it is going to be a group 1 metal sulfite1965

reacting with strong acid to give you sulfur dioxide gas and liquid water.1968

The third and final example that is an exception is going to be the following.1975

It is going to be an ammonium salt where X is usually a halide.1980

It is going to be the ammonium salt reacting with a metal hydroxide.1988

Again this is usually going to be a group 1 hydroxide.1994

When you do the double displacement, the M is going to hook up with X.1998

Ammonium is going to hook up with hydroxide.2003

It turns out that ammonium hydroxide is unstable in water.2006

It is going to decompose and form aqueous ammonia and liquid water.2011

Once again the third exception, something we just can't predict, is the following.2017

It is going to be an ammonium halide salt in combination with a group 1 metal hydroxide.2021

You are going to form ammonia, NH3, in combination with liquid water.2029

That pretty much covers solutions that are occurring in aqueous environments.2036

Let's go ahead and summarize what we learned.2043

We started off talking about the law of conservation of mass.2046

Remember a big implication of the law of conservation of mass is that we always have to balance any chemical equation.2049

We went on to discuss electrolytes.2059

We learned that strong and weak electrolytes are going to each dissociate to different extents when dissolved in water.2061

We learned several rules; we saw an activity series; we learned the table of solubility rules.2071

We learned the three exceptions that we can use to help us predict the product2079

of any reaction that occurs in aqueous solutions, especially single replacement, double replacement, and neutralization reactions.2085

Let's go ahead and tackle some sample problems.2097

Here I have presented to you the reactants.2100

I want you to come up with the products and of course always balance the chemical equation.2104

Let's go ahead and look at this.2110

We have a zinc carbonate, that is going to be a metal carbonate, reacting with a strong acid, H2SO4.2113

That is one of the exceptions; what is going to happen is the following.2123

We are going to get zinc sulfate.2127

That is going to be aqueous; use your table of solubility rules.2134

This is going to form not H2CO3 but H2O liquid and CO2 gas.2138

Remember H2CO3 is going to be one of those unstable compounds in water.2145

H2CO3 we saw breaks up into liquid water and CO2 gas; let's go on.2153

Next one, ammonium bromide reacting with lead(II) acetate.2163

In this case, lead is going to now hook up with bromine.2172

Ammonium is now going to hook up with acetate.2178

This is yet again another double displacement reaction.2182

When we go ahead and draw the products, let's go ahead and do ammonium now with acetate, NH4C2H3O2, and lead(II) bromide.2186

Remember Pb2+ remains Pb2+; let's not forget the physical states.2201

Ammonium acetate is going to be aqueous.2208

Lead(II) bromide is going to be the solid.2211

Most importantly always balance your chemical equation.2214

I am going to need two of these and I am going to need two of these for my balanced chemical equation.2217

Next one, potassium nitrate aqueous reacting with copper(II) chloride.2226

This is going to be another example of a double displacement reaction.2234

Nitrate is going to hook up with copper.2241

Potassium is going to hook up with chlorine.2244

We are going to get KCl; when you look up the table of solubility rules, that is going to be aqueous.2249

Forming copper(II) nitrate; when you look up the table of solubility rules, that is also going to be aqueous.2258

This is an example of a double displacement reaction that forms no precipitate which is not too uncommon.2267

Final example of sample problem one, lithium sulfate reacting with silver nitrate.2275

Let's go ahead and do the double displacement.2282

Lithium is now going to hook up with nitrate.2285

Sulfate is going to now hook up with silver.2288

When we go ahead and do this, we are going to get lithium nitrate.2294

That is going to be aqueous when you use the table of solubility rules; and silver sulfate.2300

Silver sulfate is going to also be aqueous when you look up the table of solubility rules.2309

We have to balance this; this is going to be two of these and two of those.2318

I forgot, for the third example, the potassium nitrate reacting with copper chloride, I left it unbalanced.2325

Shame on me; again let's go ahead and fix that.2330

We are going to need two of these and two of those.2333

I hope this sample problem one was really testing your ability to not only predict the products2339

but also the physical states in addition of course to balancing.2345

The final set of sample problems involves the following.2351

I have seen a lot of tests.2355

A lot of the times, you have to translate from a sentence or a phrase into a chemical equation.2359

This is really testing your ability of nomenclature also.2365

So this is a nice cumulative type of problem.2368

Solid zinc reacts with aqueous lead(II) nitrate to form aqueous zinc(II) nitrate and solid lead.2373

Let's go ahead and translate it.2382

Zinc solid reacts with lead(II) nitrate going on to form zinc(II) nitrate aqueous and solid lead.2385

Let's go ahead and balance this; we are going to need two zinc nitrates.2407

No, it is going to be zinc(II) nitrate; I'm sorry; this is going to be two of these.2413

I am going to erase that coefficient; very sorry about that.2417

My balanced chemical equation is good to go.2423

This is an example of a single replacement reaction.2426

If you go ahead and look up the activity series, zinc will come in and knock out lead.2430

Second example, solid iron reacts with molecular chlorine gas to form solid iron(III) chloride.2437

Solid iron reacts with molecular chlorine; remember what we mean by molecular chlorine?2446

That is going to be Cl2 gas not just Cl.2453

That is going to go ahead and form solid iron(III) chloride, FeCl3 solid.2457

Let's go ahead and balance this.2465

Iron is good to go already; I have one on each side.2466

But I have two chlorines on the left and I have three chlorines on the right side.2469

We can go ahead and simply put a 3 in front of the Cl2 and a 2 in front of the FeCl3.2474

Finally I am going to need 2 irons.2482

This is again an example of an even-odd pair where you could have used fractional coefficients; that would have worked too. 2485

This is an example of what we call a combination reaction where we have the form A plus B goes on to from AB.2493

Once again a combination reaction.2505

Last example, we have solid barium peroxide forming solid barium oxide and molecular oxygen gas.2508

Let's go ahead and write this and balance it; we have solid barium peroxide.2516

You are told that this forms solid barium oxide in combination with molecular oxygen gas.2529

Remember molecular oxygen; very important; that is going to be O2.2536

My bariums are good to go; the only thing we have to balance are the oxygens.2542

I have two oxygens on the left; I have three on the right.2548

Let's go ahead and put a 2 in front of BaO.2551

That is going to give me four oxygens on the right side now and two bariums.2557

All I have to do is put a 2 in front of there.2561

So I have two bariums on both sides of the equation.2564

I have four oxygens on both sides of the equation.2567

This is what we call a decomposition reaction where we have one compound AB essentially dissociating into A and B.2570

One compound forming multiple products.2585

This is a nice cumulative exercise where it really tests your ability,2589

your knowledge of the nomenclature, and of course to balance a chemical equation.2594

Thank you very much for your time.2600

I will see you next time on Educator.com.2601

Hi, welcome back to Educator.com.0000

Today we are going to be talking about additional chemical reactions; especially those that happen in aqueous solution.0003

Let's go ahead and go over the lesson overview; we are going to cover a few more areas.0013

We are going to build upon what we discussed last time on chemical reactions.0019

The very first type of reactions that we are going talk about are going to be involving what we call acids and bases.0024

That is going to be our first really general category.0035

We are going to be then jumping into another category of acids and bases.0040

These are going to be nontraditional ones; something that we couldn't look at just by formula; I will build on that.0046

After that, we are going to then go on to what we call redox reactions.0058

We are going to learn how to identify when redox reactions occur and exactly what is occurring during the reaction.0064

Finally we go on to our summary and sample problems.0071

We are first going to tackle acids.0078

The first interpretation of what an acid and a base is is called the Arrhenius definition.0081

The Arrhenius definition tells us the following.0090

That acids are going to dissolve in water and dissociate into a proton and anion.0093

For example, we can take hydrochloric acid; this is going to dissociate in water.0100

This is going to form H+ which we call proton and its anion, Cl1-.0109

This is very familiar to what we talked about in the previous session on the first part of chemical reactions.0119

If you recall, we talked about what was labeled an electrolyte, strong and weak.0127

This is pretty much an electrolyte; the Arrhenius acids essentially behave as the electrolytes.0135

We can go ahead and look at another example.0144

HNO3 aqueous can also break up into H+ aqueous and NO31- aqueous.0147

Just want to do a reminder for strong electrolytes.0162

Remember we are going to use a single arrow pointing in the forward direction0165

to show we get full dissociation--that we completely break up into ions.0168

However there are also weak electrolytes.0175

There is going to be Arrhenius acids that are going to be weak electrolytes.0179

For example, hydrofluoric acid is the typical example.0183

If everybody tried to recall, remember this was called an equilibrium arrow.0187

This is going to form H+ aqueous and F- aqueous.0193

Now that we have covered the Arrhenius definition of an acid,0201

let's go ahead and cover the Arrhenius definition of a base.0204

According to Arrhenius, bases dissolve in water and dissociate into a cation and hydroxide.0208

In other words, Arrhenius bases are also electrolytes.0217

The typical example will be sodium hydroxide.0221

Sodium hydroxide is going to be a strong electrolyte; we use a single arrow.0228

That is going to dissociate and break up into its cation, Na1+, and hydroxide, OH1-.0233

There is also going to be other types of Arrhenius bases that don't completely dissociate.0243

Again we are going to use a double arrow for them.0249

We can imagine maybe the following; magnesium hydroxide aqueous.0253

This one again is not going to completely dissociate; we use an equilibrium arrow.0259

That is going to give us Mg2+ aqueous and two hydroxides aqueous.0264

The Arrhenius interpretation of an acid and a base is relatively straightforward.0273

Once again they simply are ionic compounds that break up into their respective cations and anions when in aqueous solution.0278

The Arrhenius definition is fine for most purposes.0289

However what we are going to cover now is called the Bronsted-Lowry definition.0292

The Bronsted-Lowry definition, the main difference is the following.0297

It always takes to in count the rule of water.0301

Let's first start off with the definition of acids.0306

Acids dissolve in water; what they do is the following.0309

They are going to donate a proton to water.0313

The first example that we talked about was hydrochloric acid.0318

Let's go ahead and talk about that once again.0321

This time interpreting it from the Bronsted-Lowry perspective; HCl aqueous.0324

This time we are going to follow the definition.0330

Instead of just dissociating into cation and anion, the acid is going to react with water.0333

It is going to donate a proton to water.0339

So HCl aqueous plus H2O liquid goes to the following.0342

If HCl loses its hydrogen, all that is left is going to be Cl1- aqueous.0347

If H2O is accepting the hydrogen from HCl, H2O becomes the following; H3O1+ aqueous.0356

There is a couple of things I want to point out here.0368

Number one, we give H3O1+ a unique name.0370

This is what we call the hydronium cation.0376

The hydronium cation, anytime you see it, it is always indicative that an acid is present in water; acid in water.0382

Basically a Bronsted-Lowry acid will always form hydronium in water; Bronsted-Lowry acids form H3O1+.0395

This is the Bronsted-Lowry interpretation of what an acid is.0408

Let's go ahead and look at another example.0414

Let's go ahead and look at HClO4; HClO4 aqueous.0421

Once again this is going to go ahead and react with water liquid.0428

That goes on to form ClO41- aqueous plus H3O1+ aqueous.0433

Always remember everyone that you always want to pay attention to balancing; don't forget.0444

You always make sure your equation is balanced; not only the coefficients but also the charges.0449

My overall charge on the reactant side is 0 and 0 here.0454

Overall here it is -1 and +1 on the right side giving me a net sum of 0 too. 0461

By breaking up into hydronium and the anion, we are going to get the balanced chemical equation also.0471

There is a couple of different types of acids that we want to talk about right now.0480

Right now when you just see one hydrogen in the formula of the acid, that is what we call monoprotic acids.0485

But of course there are going to be many other types of acids that contain more than one hydrogen.0497

These are what we call polyprotic acids.0504

The typical example of a polyprotic acid would be sulfuric acid, H2SO4 aqueous.0510

H2SO4 specifically, this is called a diprotic acid; the prefix di meaning there is two of them.0520

Let's go ahead and react this with water; here is the rule of thumb.0527

For polyprotic acids, anytime you write them reacting with water, you never take off both of the hydrogens at once.0535

You always do one deprotonation at a time; for polyprotic acids, again one deprotonation at a time.0544

That means the following; I am not going to get just sulfate.0563

I am going to get just one proton is going to be taken off.0568

That is going to be HSO41- aqueous; again we always form H3O1+ aqueous.0571

The rule of thumb again, don't forget, it is going to be one deprotonation at a time.0586

Let's go ahead and take HSO4- and deprotonate it one more time.0597

That is going to be HSO41- aqueous plus H2O liquid.0603

We have another rule of thumb that we also have to go over.0610

For polyprotic acids, only the first step is going to be significantly strong.0614

Everything else is going to be considered a weak step.0621

In other words, HSO41- will be functioning as a weak electrolyte.0624

We are going to use therefore an equilibrium arrow.0635

We are going to get now sulfate plus H3O1+ aqueous.0639

This is the basic definition of Bronsted-Lowry acids.0648

Once again a couple rules to remember.0654

That Bronsted-Lowry acids are going to react with water and are going to donate a proton to water.0657

If we are dealing with polyprotic acids, you only write their deprotonations one step at a time.0662

Finally for the polyprotic acids, only the first step is considered a strong electrolyte step.0671

Everything else after that is going to be considered weak.0679

You may be wondering: how do I know what are the strong electrolytes for acids?0683

How do I know which ones are the weak electrolytes for acids?0688

We are going to go over that right now.0691

Basically the strong electrolytes are what we call strong acids.0693

Traditionally there are seven strong acids that are normally taught to a freshman level chemistry course.0700

Your instructor may give you more or less; so definitely pay attention to him or her.0706

But the basic seven that are always strong are the following.0712

Hydrochloric acid, HBr, HI, HNO3, H2SO4, HClO4, and HClO3.0716

Pretty much it is pretty safe to assume that any acid that is not on this list you can assume to be weak.0734

Again these are the seven strong acids that I would recommend definitely committing to memory.0742

You will always use a single arrow for them.0747

But remember, make a note, for sulfuric acid which is what we just looked at.0750

That is the polyprotic acid; only the first step is strong.0756

Now that we have talked about the Bronsted-Lowry definition of an acid,0762

let's go ahead and talk about how Bronsted-Lowry defines a base.0766

Bases dissolve in water and what they do, instead of donating a proton, they accept a proton from water.0773

The typical example of a Bronsted-Lowry base is going to be ammonia.0779

Ammonia... this is definitely something you should just have it memorized... is NH3.0785

Ammonia is going to react with water, H2O liquid.0790

It turns out that ammonia is going to be what we call a weak base.0794

Weak bases are going to be weak electrolytes; we are going to use an equilibrium arrow for them.0800

NH3 aqueous plus H2O liquid, it is going to accept the proton from water.0806

NH3 is going to become... remember your polyatomic ions... NH41+ aqueous.0812

H2O is going to lose a hydrogen this time forming hydroxide.0818

We see it pretty clear cut; Bronsted-Lowry acids always generate hydronium.0826

And Bronsted-Lowry bases are always going to generate hydroxide when reacting with water.0831

What are some other typical bases, Bronsted-Lowry bases, that you want to be on the lookout for?0840

Pretty much anything that is related to ammonia is going to be a weak base.0847

Specifically the molecules that I am talking about that are related to ammonia, these are going to be called amines.0861

Amines contain nitrogen and usually with carbon and hydrogen.0871

NH2R would be an example where R is a molecule that contains hydrogen and carbon; what we call a hydrocarbon.0890

Not only NH2R but NHR2; not only NHR2 but NR3.0904

These are the again generic formulas for amines; also Bronsted-Lowry bases.0913

Just by looking at the name, you can usually tell if it is going to be a base or not.0922

A lot of the organic bases that are also biomolecules, they all end in the same suffix.0926

What you want to look for is a ?ine suffix.0935

You actually know a lot already of examples that are weak Bronsted-Lowry bases that are also bimolecules.0942

For example, here is one of my favorite.0951

Of course it is going to be caffeine; you see this caffeine ends in ?ine.0956

Adrenaline is another one that ends in ?ine.0961

A lot of the drugs also end in ?ine; they are also bases and amines.0968

For example, morphine, cocaine, and heroine; the list goes on and on.0975

I am sure you will encounter these in some of your biology classes.0983

But just by looking at the name again, the ?ine suffix usually indicates if something is an amine.0986

What are the strong bases and what are the weak bases?0998

Pretty much the strong bases are going to be group 1 metal hydroxides and some group 2.1002

Sodium hydroxide, potassium hydroxide, lithium hydroxide, etc; those are the group 1 metal hydroxides.1023

Some of the group 2 metal hydroxides that are also considered strong are going to be calcium hydroxide and magnesium hydroxide.1032

Pretty much it is safe to assume... again you should check with your instructor about this.1045

It is safe to assume that anything that is not on this list of strong bases can be considered to be weak.1050

Especially ammonia and its derivatives which we call amines.1057

Now that we have talked about the Bronsted-Lowry interpretation of acids and bases...1065

We saw that they have one thing in common; that they all react with water.1071

But in one situation, that is with acids, water is accepting the hydrogen.1076

In the second situation, that is with bases, water is actually donating the hydrogen.1083

Let's take a look more closely at that.1089

Water has the unique ability of functioning as both a Bronsted-Lowry acid and also as a Bronsted-Lowry base.1093

This is what we call the amphiprotic nature of water.1099

Because of this, let's go ahead and further scrutinize water reacting with itself.1105

H2O liquid plus H2O liquid; this is going to be a weak process.1111

If we focus just on maybe the left water molecule... let's call that the acid.1124

Let's call the other water molecule the base.1135

If the acid is going to function as an acid, it is going to donate its hydrogen.1143

But if it donates a hydrogen, it becomes hydroxide.1149

The second water molecule which we are labeling as the base is going to accept the hydrogen.1157

All of a sudden it becomes H3O1+.1163

This process is known as the autoionization of water.1168

It is physically representative of what happens in normal water.1172

The water that you get out of your tap, it is not just H2O.1178

There is a little bit of hydroxide and there is a little bit of hydronium.1184

But the majority is clearly going to be water.1187

Again this is what we call the autoionization of water.1192

When we go ahead and look at quantitative acid-base chemistry,1200

down the line, this is going to come into play a lot more.1204

Now that we have covered the traditional acids and bases, Arrhenius and Bronsted-Lowry,1210

we are now going to look at some more acids and bases1218

that are going break the trend from Bronsted-Lowry and Arrhenius.1224

So far we learned that acids can be identified as basically having one or more hydrogens.1229

Just by looking at the formula; the hydrogens are usually at the beginning.1235

Bases were pretty easy to spot; they either contained hydroxide or they were amines related to ammonia.1240

There are molecules that fit neither profile.1249

Specifically we are going to now look at oxides.1254

Basically oxides are anything that contain oxygen.1257

Certain oxides of metals and nonmetals, they contain neither hydrogen or hydroxide.1260

However they can still react with water to produce hydronium or hydroxide.1266

What we are going to do now, we are going to spend some time on going over1275

what are called the acidic oxides and what are called the basic oxides.1279

Certain oxides of nonmetals... remember certain oxides of nonmetals are actually acidic.1286

That is they will dissolve in water to form hydronium.1294

Here I present to you three examples.1301

They are going to be the oxides of carbon, of sulfur, and of nitrogen.1305

Let's first look at the oxide of carbon, carbon dioxide.1315

Carbon dioxide can go ahead and react with water to form H2CO3; this is called carbonic acid.1318

Carbonic acid remember is a compound we learned before; it is a relatively unstable product.1329

It can go on to react with water to form bicarbonate and hydronium.1337

Essentially carbonic acid is going to function as a Bronsted-Lowry acid.1343

This process is very interesting; this is actually what is happening in your body.1351

If you take in too much carbon dioxide, the carbon dioxide is going to react with the water in our bodies.1356

It is going to form H2CO3; H2CO3 in turn is going to form hydroxide.1363

If we get too much carbon dioxide in our body, that is actually make our blood and etc more acidic.1367

That is going to be very very dangerous.1378

The next oxide is that of sulfur; that is sulfur dioxide.1381

Sulfur dioxide is a slightly different reaction.1387

Sulfur dioxide is going to react with molecular oxygen, O2, to form sulfur trioxide.1390

Sulfur trioxide is going to go ahead and react with water to form sulfuric acid.1397

You know sulfuric acid is one of the seven strong acids that we covered.1402

That is definitely going to go on to form hydronium.1408

Finally the last oxide is that of nitrogen; that is going to be nitrogen dioxide.1415

Nitrogen dioxide is going to go ahead and react with water.1421

We are going to go ahead and form a strong acid, nitric acid, and a weak acid, nitrous acid.1424

Again both of them are going to go on and form hydronium.1431

Again it is the oxides of carbon, sulfur, and nitrogen that are going to react with water somehow to give us hydronium.1436

That is what we call acidic oxides.1447

Now that we have covered acidic oxides, let's go on to basic oxides.1451

Basic oxides are going to be oxides of certain metals1455

because they are going to go on and react with water and to form hydroxide.1459

The typical examples are going to be group 1 and group 2 oxides; group 1 and group 2 oxides.1465

Here I want to present to you calcium oxide.1477

Calcium oxide is going to react with water to go ahead and form calcium hydroxide.1480

Remember calcium hydroxide was one of those relatively strong bases that we talked about.1486

That is going to go ahead and form Ca2+ and two hydroxides.1494

The other example is the group 1 oxides.1499

Sodium oxide is going to go ahead and react with water to form sodium hydroxide.1502

We saw that sodium hydroxide was one of those strong bases also.1506

It is going to go ahead and go on to form Na1+ and hydroxide.1511

We have learned the Arrhenius interpretation of an acid and base.1517

We have learned the Bronsted-Lowry interpretation.1520

Now we have learned that acids and bases will also include oxides of metals and nonmetals.1523

We now move on to the second half of this presentation on chemical reactions.1535

This type of chemical reaction is what we call oxidation-reduction reactions; which are also known as redox reactions.1542

In an oxidation-reduction reaction, basically one reactant is going to get oxidized and a second reactant is going to get reduced.1551

The most basic, the simplest interpretation of an oxidation-reduction reaction is one that involves the loss or gain of oxygen atoms.1564

That is pretty easy to remember; when you hear the word oxidation, it pretty much sounds like oxygen.1576

Let's go ahead and look at the following example.1583

We are going to take iron(III) oxide; we are going to react it with carbon monoxide.1585

We are going to form iron solid and carbon dioxide gas.1592

You see that one compound has gained oxygens and one compound has completely lost them.1597

Fe2O3 has lost oxygens; CO has gained oxygens.1605

Now to build upon our terminology, if a reactant gains oxygens, we say that it has become oxidized.1619

Therefore carbon monoxide became carbon dioxide; it was oxidized.1632

If a reactant is going to lose oxygen, we say that it has become reduced.1642

In this example, it is the iron(III) oxide that has become reduced.1648

Two more vocabulary terms; the oxidized reactant is what we call the reducing agent.1656

The reduced reactant is what we call the oxidizing agent.1663

Let's go ahead and move on then.1672

Again the simplest interpretation of a redox reaction is involving a loss or gain of oxygens.1677

Another interpretation involves a transfer of electrons.1683

The reducing agent is going to be the reactant that loses electrons while the oxidizing agent gains electrons.1691

This is very important that electron transfer is always from reduced agent to oxidizing agent in a redox reaction.1699

Let's go ahead and take a look at the following example.1714

Zinc solid plus Pb2+ is going to form is going to form Zn2+ and lead solid.1717

Look what happened here; zinc solid became Zn2+; Pb2+ became lead solid.1726

In other words, zinc gave up electrons and became a cation; Pb2+ gained electrons to get reduced.1737

In other words, look at the charge.1752

Zinc started off with a charge of 0 and it became 2+.1753

Lead 2+ started out here on the reactant side and it became 0.1760

We can easily tie this in to the activity series that we talked about in chemical reactions from the previous section.1769

Basically good reducing agents include those at the top of the activity series that are above hydrogen.1779

Basically the higher on the list, the better its reducing ability.1788

Now that we saw that we can interpret a redox reaction as a transfer of electrons, let's go ahead and talk about another one.1796

This is what we call a change in oxidation number.1806

This is going to be the less obvious one.1811

It is easy to see the change in number of oxygen atoms.1815

It is easy to see a change in ionic charge.1819

So this one we are going to spend a little more time on, takes a little getting used to.1824

Let's go ahead and identify and define what we mean by an oxidation number first.1829

Basically an oxidation number is going to be a pseudo charge.1833

It has no physical meaning and don't think of it as an ionic charge.1839

But it is a convenient way for us to keep track of electrons.1842

What you want to remember is the following.1849

During the redox reaction, an oxidation number is going to change for the compound that is involved.1851

If the oxidation number has increased, we say that the compound the element was a part of was oxidized.1859

If the oxidation number has gone down, then reduction has occurred.1868

There is a couple rules that we have to know in order to do this.1879

This is in order of priority; please remember that.1885

Rule number one: for all homonuclear species including monatomic, diatomic, and polyatomic, the oxidation number is 0 for each element.1891

For example, if you just see sodium solid, that is going to be a neutral atom sodium; its oxidation number is 0.1906

Let's look at some diatomic homonuclear species.1919

That would be Cl2 gas, Br2 liquid, I2 solid, N2 gas, H2 gas, and F2 gas.1923

These are what we call homonuclear diatomic compounds.1941

Each element in the compound has an oxidation number of 0.1946

There are some homonuclear polyatomic species; for example, sulfur.1953

The natural form of sulfur is actually S8 solid.1960

In S8 solid, all eight sulfurs, each have an oxidation number of 0.1966

That is the first rule; not too bad to remember.1975

The second rule is actually very straightforward too.1979

It tells us that for monatomic ions, the oxidation number is identical to the charge.1982

When we look at Na1+ aqueous, N3- aqueous, and O2- aqueous, these are what we mean by monatomic ions.1989

The ionic charge is equal to the oxidation state.2004

Next rule, this deals with fluorine.2010

Fluorine is the first element that gets very high priority.2013

Fluorine is going to be having an oxidation number of -1 when part of a heteronuclear compound.2017

For example, if you look at the molecule HF, this is a heteronuclear compound.2025

The fluorine here is going to have an oxidation number of -1; pretty straightforward.2034

The next element which gets a pretty big priority is going to be oxygen.2040

Oxygen is going to have an oxidation number of -2 which is very very common.2045

For example, in water, oxygen has an oxidation number of -2.2051

However there is going to be one main exception for oxygen; this is in what we call peroxides.2063

Remember peroxides we had you memorize for polyatomic ions; an example would be Na2O2.2069

In this case, for sodium peroxide, each oxygen is -1 oxidation number.2090

Once again peroxides is a big exception for oxygen.2103

Going on, after fluorine and oxygen comes the remainder of the halogens.2110

The halogens, chlorine, bromine, and iodine, are also pretty high on the list of rules here.2117

They are going to get oxidation numbers of -1.2127

Coming up is hydrogen; hydrogen is almost always a +1 oxidation number when part of a heteronuclear compound.2133

For example, if you look at HF which we just dealt with.2143

Heteronuclear compound so hydrogen is going to get a +1 oxidation number.2147

There is an exception however; this is when hydrogen is going to be in a compound that contains a metal.2151

When it is directly bonded to a metal; for example, NaH.2159

In this case, hydrogen is going to get an oxidation number of -1.2165

These types of compounds are what we call metal hydrides.2171

When hydrogen is part of a compound with a metal, what we call metal hydrides.2176

The final rule is going to be something very familiar to us.2183

Remember when we learned how to name ionic compounds.2189

We said that the overall charge, the net charge, must be equal to the overall charge of the compound itself.2193

Along a parallel track, the net sum of all oxidation numbers in the compound must add up to the overall charge.2200

This is very familiar; let's look at those examples we just talked about; HF.2212

We said that hydrogen is going to get +1 and fluorine is going to get -1.2220

HF is neutral overall so we get +1 and -1 adding up to 0.2227

If you look at NaH, Na is going to get a +1 charge.2232

Remember this is an ionic compound so we go by group 1; we just go by the charge.2241

Hydrogen we said is going to get -1; once again +1 and -1 giving us 0.2246

Let's do one final look; that is going to be H2O.2253

What get priority?--is it hydrogen or oxygen?2256

Going back to your rules, oxygen is going to get the priority.2260

That is going to have a -2 oxidation state.2263

Hydrogen here we said is going to get a +1 oxidation state when part of a heteronuclear compound.2266

However how many hydrogens are there?--there is two of them.2274

When we do the arithmetic, we are going to get +2 plus -2 adding up to 0 again.2278

This is again very familiar territory where the net sum of the oxidation numbers must equal to the overall charge of the compound.2290

Now that we have talked about the rules...2302

Unfortunately, these are rules you are just going to have to commit time to memorizing.2306

Let's go ahead and apply them to do some problems.2310

We are going to now assign oxidation numbers to all elements in each of these compounds.2315

Ammonia, nitrogen and three hydrogens; if you look at your list, hydrogen is going to get the priority.2324

Each hydrogen is +1; there is three of them giving me +3 overall.2334

Ammonia is neutral; that means nitrogen has to be high or low enough of an oxidation number to balance the +3 from hydrogen.2341

Therefore nitrogen is going to be -3 oxidation number.2352

Next one is a barium chloride.2356

We recognize this as a metal and a nonmetal; this is an ionic compound.2359

Barium is column 2; that is going to be +2.2364

Each chlorine, group 7, is -1; there is two of them.2368

2 plus -2 is going to give us 0; so we are good to go.2373

C6H12O6, this is a molecular compound, nothing but nonmetals.2379

This is actually the formula for glucose; let's take it one by one.2384

Oxygen gets priority; each oxygen is -2; there is six of them giving me -12 overall.2391

After oxygen, hydrogen is going to get the next priority.2401

There is +1 oxidation number times 12 hydrogens giving me +12 overall.2406

What does that mean for carbon then?2418

If this whole molecule is neutral, that means carbon has an oxidation number of 0 in this compound.2423

When I add all the numbers together, 0 times 6 added to +12 added to -12, I add up to 0.2431

Finally, oxygen difluoride, this is a molecular compound; oxygen difluoride.2440

Which one is going to get priority?--fluorine is going to get priority.2449

Each fluorine is -1; there is two of them giving me -2 overall.2452

That means the oxygen has to be +2 to balance the -2 oxidation state.2459

Now that we know how to assign oxidation numbers to elements within a compound, let's apply this now to the following chemical reaction.2469

We have CH4 plus 2O2 goes on to form carbon dioxide and water.2482

Let's go ahead and assign oxidation numbers to each element.2489

And see if there is any changes from the reactant side to the product side.2492

Carbon is in CH4; hydrogen is going to get priority.2498

The hydrogen is going to be +1; there is four of them which means carbon has to be -4.2504

Molecular oxygen is a homonuclear diatomic so each oxygen is going to be 0.2517

Carbon dioxide; in carbon dioxide, oxygen gets priority; each oxygen is -2.2525

There is two of them giving me -4 overall which means carbon has to be +4.2533

Finally water; oxygen gets priority; it is going to be -2.2540

Therefore each hydrogen in water must be +1 oxidation number to balance the -2 overall.2550

Have there been any elements in this problem that have undergone changes from reactant side to product side?2560

Let's go ahead and see; carbon changed from a -4 to +4.2565

In other words, carbon's oxidation number increased.2571

In addition, oxygen's oxidation number decreased from 0 to -2.2579

Carbon went up from -4 to 4.2592

Therefore, remember what we said for compounds that have an element whose oxidation number has gone up.2598

That compound we say has been oxidized.2606

So CH4 has been oxidized; it is the reducing agent.2610

Molecular oxygen has been reduced; it is the oxidizing agent.2626

This is how we, as you can see, use the concept of oxidation numbers2643

to help us determine exactly what is going on in a redox reaction,2650

what compound is being oxidized, and what compound is being reduced.2654

Let me go ahead and summarize this lesson on chemical reactions.2660

We covered many interpretations of what we mean by an acid and base.2665

The first one was called an Arrhenius interpretation.2672

The second one was a Bronsted-Lowry interpretation.2676

Right after Bronsted-Lowry, we also covered the common strong acids and bases that you will just have to memorize.2682

There is no other way of going about it.2690

In addition, we saw that acids and bases include oxides.2693

What is interesting about these oxides is that they contain neither hydrogen nor hydroxide in their formulas.2698

Yet they still react with water to form hydronium or hydroxide.2705

Finally the last type of chemical reaction we covered were called redox reactions.2711

Redox reactions basically involve a transfer of electrons.2716

We saw that we can use the concept of oxidation numbers2720

to help us keep track of the oxidizing agent and of the reducing agent.2725

Now that we have covered all this material, let's go ahead and tackle some sample problems.2734

This is again a very traditional type of problem.2739

Write out the reaction for nitric acid reacting with water.2742

Let's go ahead and do that; HNO3 aqueous plus H2O liquid.2748

You have to try to think back: is nitric acid one of the seven strong acids?2760

The answer is absolutely yes.2765

Because of that, we are not going to use an equilibrium arrow.2767

But we are going to use a single arrow in the forward direction.2770

That is going to go ahead and form hydronium of course and nitrate aqueous.2774

The second part of the question is the following: is this redox?2782

If this is a redox reaction, we have to have a transfer of electrons involved.2787

In order to determine that, we have to see if any element's oxidation number2794

has changed from the reactant side going to the product side.2801

Let's go ahead and assign oxidation numbers to all elements in this chemical reaction.2806

In nitric acid, oxygen is going to get priority.2814

That is going to be -2 by three.2817

Hydrogen is going to get the next one which +1.2822

Therefore in order to have everything add up to the overall charge which is 0, nitrogen has to be +5.2826

Let's go ahead and look at water; water is something we have done before.2836

Here oxygen is -2; here each hydrogen is +1.2840

That adds up to 0; so we are good to go on that.2847

Let's go ahead and do hydronium; oxygen is -2.2850

Hydrogen, when it is part of a heteronuclear compound, it is going to be +1; there is three of them.2856

Let's see if this adds up to the overall charge of hydronium which is just +1.2862

I have +3 and -2; I do get a sum of +1; it definitely works out.2867

The last one is nitrate; nitrate, we have -2 for oxygen times three of them.2875

Each nitrogen therefore has to be what?2882

You see that nitrate is -1 overall charge.2886

If I already have -6 from oxygen, nitrogen has to be +5.2891

Now that we have assigned all oxidation numbers, let's see if anything has changed.2898

Nitrogen remains +5, oxygen remains a -1, and all hydrogens remain +1.2904

Not a single oxidation number has changed; therefore this reaction is not redox.2912

Again this reaction is not redox.2923

The final two questions are focusing on redox reactions.2929

It is basically having you identify the oxidizing agent and the reducing agent.2936

That involves first assigning oxidation numbers once again; let's go ahead and do that.2941

We have silicon solid; that is a monoatomic homonuclear compound; so that is neutral, 0.2946

Two Cl2s, molecular chlorine is a homonuclear diatomic; both of them are 0s.2957

Now we reached silicon tetrachloride.2964

Once again chloride is going to be getting the priority here, -1; there is four of them.2968

The overall molecule is neutral; therefore silicon must be +4.2973

We have seen that silicon goes from 0 to +4 and that chlorine goes from 0 to -1.2980

This is also how you tell if you are doing it right.2992

Because when you do a redox reaction, you have to have the following combination always.2995

Exactly one element is going to have an increase in oxidation number and one is going to go down.3000

In this case, that is what we exactly have.3006

We say therefore that silicon, because it experienced an increase in oxidation number, was oxidized.3009

Molecular chlorine, each chlorine experienced a decrease so molecular chlorine was reduced.3019

That makes silicon the reducing agent and that makes molecular chlorine the oxidizing agent.3029

The last problem is a little longer chemical equation.3043

But again I want to present this to you because it doesn't make a difference how long the chemical reaction is.3047

You are always going to have one being oxidized and one being reduced.3052

Just follow the rules, assign the oxidation numbers, and you will be fine all the time.3056

We have sodium iodide; we recognize that immediately as metal and nonmetal; that is going to be ionic.3062

Therefore iodine is going to be -1; sodium is going to be +1.3068

Let's go ahead and look at sulfuric acid.3073

Each hydrogen is +1; there is two of them; the overall molecule is neutral.3083

So far I have 2 and -8 giving me -6 overall.3090

Which means sulfur must be +6 to help me balance out everything to 0.3095

Next one is manganese(IV) oxide.3102

Manganese(IV) oxide we recognize as being ionic also; metal and nonmetal.3105

Therefore each oxygen is -2; so manganese here is going to be +4 overall.3109

Moving on, sodium sulfate.3118

We recognize that as also being ionic; it is a metal and nonmetal.3121

Each sodium is going to be +1; there is two of them.3127

Each oxygen is -2; there is four of them.3132

Everything has to equal up to the overall charge which 0.3138

I have +2 and -8 overall which is going to be -6; that means sulfur has to be +6.3141

Next compound, manganese(II) sulfate.3152

Once again this is cation and anion so we just go by ionic charges.3157

Each oxygen is -2 overall; there is four of them.3164

Sulfur is going to be +6; how did I get that?3172

We know from our polyatomic table of ions that you memorized, sulfate is -2.3177

If oxygen is already -8, that means sulfur has to be +6 to give me a -2 overall charge.3183

Therefore manganese... we know that sulfate is -2 so manganese has to be +2.3194

Two more compounds; each iodine is going to be 0; that is homonuclear diatomic.3203

Finally in water, each oxygen is -2 and here each hydrogen is +1; there is two of them.3209

It looks like a lot of material, but let's go ahead and see if anything has changed.3218

Sodium, has sodium changed?--the answer is no.3225

Sodium started off +1; in the product side, it has remained +1; so sodium is not involved here.3230

The next one is iodine; iodine started off -1 and iodine became 0.3242

You see the oxidation number of iodine has changed.3250

In fact we say that I- has become oxidized; therefore it is the reducing agent.3255

We have one more left; we have to now find which element or compound became reduced.3273

That is going to be, looking here, that is going to be manganese.3281

Manganese started off +4 and manganese became +2; look, it went down.3287

We would be in trouble if we had two elements go up or if we had two elements go down.3296

Remember you always have one of each for a redox reaction to occur.3301

In this case, manganese Mn4+ was reduced, making it the oxidizing agent.3307

We are going to talk about redox reactions quantitatively down the line later on in this class.3323

We will come back to this before you know it.3331

Thank you for using Educator.com again; I will see you all next time.3337

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be another...0003

It is one of those central topics that you are going to be using for the rest of your general chemistry career.0007

That is going to be the concept of stoichiometry.0014

This is going to be the first of two lectures on stoichiometry.0017

Let's go over the lesson overview.0024

What is very core to stoichiometry problems is what is called a mole to mole ratio.0027

This is what we are going to introduce right from the start.0033

After we introduce the mole to mole ratio, we are going to go through0038

some traditional types of problems and sample calculations you can encounter.0041

Moreover the mole to mole ratio can be used to solve0048

a very common type of problem which is called the mass to mass conversion.0053

Right after the mass to mass conversion, we are going to encounter0060

another unique type of problem which is called a limiting reactant.0064

In addition to that, we are going to be discussing what is called a percent yield.0069

I am going to not only show you how to do it.0073

But more importantly, how do you know when you are dealing with a limiting reactant problem?0075

Because it involves a different strategy of its own.0079

We are going to finish up with a brief summary as always.0083

Then we will go ahead and tackle some sample problems together.0087

What exactly is a mole to mole ratio?0094

In chemical formulas, that is the first place where we are going to look.0098

In chemical formulas, the subscripts can actually be interpreted as the following.0102

The moles of that element for every one mole of compound.0108

For example, in one mole of water, there are two moles of hydrogen and there is going to be one mole of oxygen.0113

We literally can translate the subscripts in terms of moles.0128

That is going to make calculations a lot easier to interpret later on.0135

We not only have to deal with one mole of a compound.0141

For example, let's go ahead and look at the next question.0144

It is the same question: how many moles are there of each element?0147

But not in one mole of water but this time in two moles of water.0150

Remember the concept of dimensional analysis.0156

We saw that dimensional analysis used what was called a conversion factor.0160

In this whole lesson, conversion factors are key.0166

We are going to be doing dimensional analysis so much in this chapter.0175

But the nice news about this is that you already know how to do dimensional analysis.0182

You already know how to use conversion factors.0186

It is the same mathematical tools that we were introduced to several lectures ago.0189

Basically 2.6 moles of water is given to me; 2.6 moles of water.0195

Let's go ahead and get the moles of hydrogen.0204

2.6 moles of water times something over something, that is going to give me my moles of hydrogen.0208

Let's go ahead and enter the conversion factor.0218

I want moles of water to cancel; so moles of water goes downstairs.0220

I want moles of hydrogen to go upstairs to get carried through to the final answer.0224

Where do I get the numbers for the conversion factor?0231

You get it from the subscripts in the chemical formula.0233

For every one mole of water, there is going to be two moles of hydrogen.0237

We get an answer of 5.6 moles of hydrogen; that is it; it is that simple.0242

Let's go ahead and finish up the problem and now calculate the moles of oxygen that are contained in 2.6 moles of water.0250

2.6 moles of water times something over something equals to the moles of oxygen.0258

We are going to be using the same tools as before.0272

Moles of water is going to get cancelled; that goes downstairs.0276

Moles of oxygen now is going to the numerator.0280

That is going to get carried through to the final answer.0286

My mole to mole ratio is just 1:1, very conveniently.0289

In 2.6 moles of water, there are 2.6 moles of oxygen.0294

What we just did was we used and we came up with mole to mole ratios from a chemical formula.0300

Mole to mole ratios from a chemical formula.0311

The nice thing about mole to mole ratios is that they easily carry over from the chemical formula to a balanced chemical equation.0317

For example, in the following equation, nitrogen plus three hydrogens goes on to form two ammonias.0329

The coefficients in the balanced chemical equation, they are actually interpreted and treated as moles; we treat the coefficients as moles.0341

For example, let's go ahead and translate this.0354

One mole of N2 requires three moles of H2 for this reaction to occur.0357

Next line, we can expect two moles of ammonia to form from one mole of N2.0366

Finally we can expect two moles of ammonia to form from three moles of H2; from three moles of H2.0372

That should be H2; my apologies.0380

It is that simple; again we use the coefficients in terms of moles now.0386

This now leads us to our definition of stoichiometry.0394

If you look at everything in this phrase here, every sentence is relating one compound from the chemical reaction to another one.0399

Anytime you relate any two elements or molecules in a chemical reaction, this is what is called stoichiometry.0410

Once again this is what is called stoichiometry.0419

Another name to be specific for these coefficients, sometimes you will hear this, these are also called stoichiometric coefficients.0423

Again it is just a fancy name for what you already know how to do.0435

That is to use whole numbers to balance a chemical reaction.0439

Let's go ahead and now apply this to some problems.0447

We are going to take the following, the same reaction, the ammonia formation.0450

The question is the following: how many moles of ammonia can form if you have 3.1 moles of H2?0456

I want you to get into habit; what type of problem is this?0465

This is stoichiometry because we are asked to relate one compound of the chemical reaction to another.0467

In other words, we are asked to relate ammonia with H2.0475

We are going to use dimensional analysis again to do our problem.0481

3.1 moles of H2 times something over something is going to give me the moles of ammonia that we can expect to form.0485

The moles of hydrogen goes downstairs to get cancelled.0500

The moles NH3 goes upstairs to get carried through to the final answer.0508

What is going to make the numbers for the conversion factor?--the coefficients; that is it.0515

When I look here, it is basically a 3:2 ratio of hydrogen to ammonia.0523

We put the coefficient in front of ammonia there and you put the coefficient in front of H2; just like that.0533

When all is said and done, we are going to get an answer of 2.1 moles of ammonia.0542

Again all we are doing is we are using the coefficients in a balanced chemical equation directly into our conversion factor.0550

It is that simple; let's go on to another illustration of this.0557

Same reaction, now how many moles of hydrogen gas are required to react with 6.4 moles of nitrogen gas?0563

Once again this is stoichiometry all the way through because we are asked to relate one compound of the balanced chemical equation with another.0572

6.4 moles of nitrogen times something over something is going to give me the moles of H2 required.0583

Moles of N2 is going to go downstairs to get cancelled.0600

Moles of H2 is going to go upstairs to get carried through to the final answer.0606

Once again where do we get the numbers from?--we get it from the coefficients.0611

When I go ahead and look at the balanced chemical equation... that should be a three; this should be a two.0618

It is going to be three moles of H2 for every one mole of N2.0629

3 goes in front of the moles of H2 and 1 goes in front of the moles of N2.0636

When all is said and done, we are going to get 19.2 moles of H2 required.0644

Once again what we just did was we were able to derive a conversion factor from the coefficients in the balanced chemical equation.0655

It is always more practical to work not in terms of moles but in grams.0668

Because when you go into the laboratory, that is what you are actually weighing.0672

So what we are going to do now is a mass to mass conversion.0675

What we previously did was we converted from moles of one compound to moles of another.0680

All we are going to do next is now convert from mass of one compound to mass of another.0686

That is usually of course going to be grams.0692

So we need a conversion factor that relates moles to grams; but we already know that.0700

Do you remember what the conversion factor is called that relates moles of one entity to grams of the same entity?0705

It is called molar mass; once again we are using nothing new.0715

We are using material that we have learned previously.0720

Remember, let's just go ahead and do a refresher.0724

If we start with the grams of one compound and we want to go to moles of that compound.0727

Grams goes downstairs and 1 mole goes upstairs; so we are essentially going to divide by molar mass.0736

If I start with moles of the compound and I want to go to grams of the compound.0744

1 mole goes downstairs and grams goes upstairs; I am essentially multiplying by molar mass.0750

Let's go ahead and apply this to a question now.0759

I am going to now work with the same identical question.0764

Except that you notice that I changed the units of moles to grams now.0768

How many grams of ammonia can form if you have 3.1 grams of H2?0773

Before what we did was we did moles of A to moles of B.0781

But now we want to go from grams of A to grams of B.0790

All we are going to do, we are going to retain this very important step.0794

All we are going to do is put two steps on the outside now.0799

We are going to start with grams of A and then go to moles of A.0803

Once we are in moles of A, we are going to go to moles of B.0808

Then once we are in moles of B, we are going to finish up the problem and go on to grams of B.0811

We already know how to do this.0820

This is just a mole to mole ratio from the balanced chemical equation; that is what we just did.0822

If I want to go from grams of A to moles of A, I am going to divide by molar mass of A.0829

Here if I want to go from moles of B to grams of B, I am going to multiply by molar mass of B.0838

We are looking essentially at a minimum of three steps, three conversion factors.0844

Let's go ahead and do this; 3.1 grams of H2, the first step is to go to moles of H2.0850

Times 1 mole of H2 divided by roughly 2.016 grams of H2; that gets me as the moles of H2.0861

Now from moles of H2, I am going to go on to the next step to go to moles of ammonia.0870

That is going to be moles of NH3 on top and moles of H2 on the bottom.0876

That is going to be a 2:3 ratio; finally this gets me into moles of NH3 after everything cancels.0883

Now I want to go to grams of NH3; times something over something gives me grams of ammonia.0891

1 mole of ammonia goes downstairs and grams of ammonia goes upstairs.0898

This is going to be roughly 17 grams for the molar mass.0904

After following this three step process, we are going to get an answer of 17.4 grams of ammonia.0910

What we just did again is called a mass to mass conversion.0924

Grams of one compound to grams of another; it is a basic three step process.0929

Let's go ahead and do one more problem.0936

How many grams of hydrogen gas are required to react with 6.4 grams of nitrogen gas?0939

It is the same repetitive machinery; we are just going to go through the work now.0949

6.4 grams of N2, the first step is to go from grams of A to moles of A.0956

Times something over something; grams of N2 goes downstairs; 1 mole of N2 goes upstairs.0965

The molar mass of molecular nitrogen is going to be roughly 28 grams of N2.0973

After I am in moles of A, I then go on to moles of B.0980

Times something over something; moles of H2 on top; moles of N2 on the bottom.0983

Remember I get this mole to mole ratio from the balanced chemical equation, the coefficient.0992

That is going to be 3 moles of H2 on top and 1 mole of N2 on the bottom.0996

I am now in units of moles of B; let's go ahead and finish up now and go to grams of B.1000

Times something over something is going to give me grams of H2 required.1005

1 mole of H2 on the bottom; grams of H2 on top.1012

The molar mass of molecular hydrogen is roughly 2.016.1016

When all is said and done, we are going to get an answer of 1.4 grams of hydrogen gas that is required.1021

Remember it is a basic three step process.1033

You go from grams of A to moles of A; moles of A to moles of B.1036

Then on to moles of B to grams of B; this is called a mass to mass conversion.1043

Because we are working with mass, that does not mean we have to end in grams.1053

Any of the mass units, you should be able to get.1058

Let me go ahead and redraw what we just did.1062

What we did was we went from grams of A to moles of A.1065

Then to moles of B; then to grams of B.1071

But remember we know the prefixes for a multiplier, such as kilo, such as milli; this is always important.1075

From grams of A, you can also work with milligrams, kilograms, etc.1086

Similarly for the mass of the product, you can also work with milligrams, kilograms, etc.1094

This is a nice cumulative problem; it really requires you to know those prefixes1102

that you are undoubtedly going to have to memorize; especially milli and kilo.1107

Let's go ahead and do an example.1113

How many milligrams of ammonia can form if you have 1.2 kilograms of H2?1116

We are just going to follow this basic flow chart.1121

1.2 kilograms of H2, we want to go to grams of A or grams of H2.1128

Times grams of H2 for every 1 kilogram of H2.1136

Remember what the prefix multiplier is for kilo?--it is 103.1140

Remember you always put the multiplier with the prefix-less unit; times.1144

Now I want to go from grams of A to moles of A.1152

Times moles of H2 for every gram of H2.1156

That is going to be roughly 2.016 for the molar mass of H2.1162

Now I am in moles of H2, moles of A.1167

Now I want to go to moles of B which is going to be moles of ammonia.1170

Times moles of ammonia for every mole of H2.1174

Remember I get the mole to mole ratio from the balanced chemical equation.1180

That is going to be 2 here and 3 there.1183

Now I can go from moles of ammonia to grams of ammonia.1187

Times roughly 17 grams of NH3 for every one mole of NH3.1191

The question is asking for milligrams; so now I am going to multiply this by 1 mg over 10-3 grams.1198

When all is said and done, I am going to get 6.7 times 106 milligrams of ammonia.1212

That is expected to form from 1.2 kilograms of H2.1219

Just be on the lookout for that again.1225

Just because the most common unit of mass we work with is grams, it doesn't mean we have to stop there or start there.1229

It is any of the mass units we can work with using stoichiometry and mole to mole ratios.1235

We now move on to a very specific type of stoichiometry problem.1244

This is what is called the limiting reactant and percent yields.1248

What we have done is the following before.1253

How many grams of ammonia can form if you have 3.1 grams of H2?1256

You notice one thing is that only one reactant amount is specified; that is of hydrogen gas.1260

When you encounter these types of problems where only one reactant amount is specified, you are making several assumptions.1268

First, you have enough of the other reactant.1276

Second, there are no side reactions or unexpected occurrences.1280

Finally, third, say you are doing this in a lab.1286

You are assuming that everything goes perfect; there are no experimental errors at all.1290

If all goes well, you expect the maximum amount possible of product to form.1297

This is what we call the theoretical yield.1306

In practice, you never ever get the theoretical yield because errors happen all the time.1312

There are going to be inefficiencies in the system; you may get side reactions, etc.1318

Again the theoretical yield is never obtained; in practice never obtained.1324

Remember there is no such thing as a 100 percent efficient process.1335

Let's look at the following question.1346

How many grams of ammonia can form if you have 3.1 grams of H2 and 3.1 grams of N2?1348

This is the first time we have encounter this type of problem, when both reactant amounts is specified.1355

This is what we call a limiting reactant problem.1361

Because one of these reactant amounts is going to dictate, is going to limit how much product we can get.1364

To solve a limiting reactant problem, we are going to be using what is called the smaller amount reacted.1371

Let me go ahead and demonstrate.1377

We are going to first determine the amount of product that can form from each of the reactants; let's go ahead and do both.1380

3.1 grams of H2 times 1 mole of H2 divided by 2.016 grams of H2.1389

Times 2 moles of ammonia for every 3 moles of H2.1401

That is going to give me 1.03 moles of ammonia expected; remember that is the key word, expected.1415

What we are going to do now, we are going to repeat the process.1426

But for the other reactant amount, the 3.1 grams of nitrogen gas.1428

3.1 grams of N2 times 1 mole of N2 divided by roughly 28 grams of N2.1432

Now times 2 moles of ammonia for every 1 mole of N2.1441

That is going to give me 0.22 moles of NH3 expected.1449

This is again called the smaller amount method for the following reason.1459

The route that forms the smaller amount of product is going to be what is called the limiting reactant.1464

It is going to be what dictates the amount of product formed.1476

We are not going to get the 1.03 moles of ammonia expected.1481

But instead we expect the smaller amount, the 0.22 moles of NH3.1484

Because of this, we conclude that N2 is a limiting reactant in this example.1489

Also H2 therefore is something we have plenty of; we have more than enough.1504

The nitrogen gas, the limiting reactant, we don't have enough of it; that is why it is limiting.1511

That is why it is going to dictate how much product we are going to get.1516

We say that H2 is in excess; we have too much of it and not enough of N2.1519

The percent yield, the percent yield is the following.1528

This is what is called the actual yield divided by the theoretical yield times 100 percent.1534

The actual yield is what you physically get in the lab; this is from practice.1547

You are usually told this in the question; given.1558

The theoretical yield is what you always calculate from a mole to mole ratio.1562

In other words, it is the 0.22 moles; from a calculation.1568

In practice of course we want the percent yield to be as close to 100 percent as possible.1577

But that is just never ever going to happen.1581

What your percent yield tells you is pretty much how well of an experiment1586

you were able to pull off doing this type of synthesis problem.1593

What we just did was called the smaller amount method.1602

It is used to solve when two different reactants are given.1604

Another common type of question you can encounter is the following.1611

How many grams of the excess reactant is going to remain?1614

In other words, how much H2 is going to be left over?1620

This is another type of common question you can be asked.1630

I have seen it asked many many times which is why I want to bring it up.1633

Basically what we are going to do is we are going to once again use stoichiometry mole to mole ratios.1637

Let's start with what we know for sure; we know for sure we are going to consume all of the limiting reactant.1646

We don't have enough of it, the 3.1 grams of N2.1650

Basically we are going to go from 3.1 grams of N2 to how much grams of H2 required.1655

Once we get the grams of H2 required, we simply subtract that from the amount that we have to get the amount left over.1667

Grams of H2 initially have minus grams of H2 required is going to give you the grams of H2 remaining.1675

Let's go ahead and demonstrate; 3.1 grams of N2 times 1 mole of N2 divided by roughly 28 grams of N2.1693

Now we go to moles of H2; 3 moles of H2 for every 1 moles of N2.1707

Now we go to grams of H2; 2.016 grams of H2 for every 1 mole of H2.1715

That is going to give us 0.67 grams of H2.1723

What this number is, this 0.67 grams, this is the amount of H21727

that is going to react with the 3.1 grams of N2... will react.1731

But the question is asking for now how much is going to react or how much is going to remain.1738

Now 3.1 grams of H2 minus 0.67 grams of H2.1743

That is going to say 2.43 grams of H2 left over.1750

Once again we are not reinventing the wheel at all today.1757

We are simply using tools that we know already, dimensional analysis.1760

We are really using the concept of the mole to mole ratio incredibly heavily.1765

Let's go ahead and summarize before we get into the sample problems.1776

The mole to mole ratio is incredibly fundamental; I cannot underscore this enough.1780

We derive the mole to mole ratio from two sources today.1786

Number one was the chemical formula; number two was from a balanced chemical equation.1789

That is why you always have to make sure to balance your chemical equation.1794

Always make sure it is balanced if not done so for you already.1798

After introducing the mole to mole ratio, we then define stoichiometry.1802

Once again stoichiometry refers to relating the amounts of one compound to that of another in a chemical formula or a reaction.1807

Finally we introduced a very specific commonly asked stoichiometry problem, the limiting reactant.1821

How do you know you are doing a limiting reactant?1828

Because you are going to be specified two reactant amounts.1830

That is our summary for this first presentation of stoichiometry.1836

Let's now spend some time on doing some calculations and representative problems.1841

How many grams of carbon are in 2.2 kilograms of carbon dioxide?1848

How many grams of carbon are in so much carbon dioxide?1854

You don't see any chemical equation or there is no mention of any chemical equation.1858

This is using a mole to mole ratio from a chemical formula... mole to mole ratio from a chemical formula.1863

For carbon dioxide, it is basically 1 mole of carbon dioxide contains 1 mole of carbon.1878

And 1 mole of carbon dioxide contains 2 moles of oxygen.1886

So 1 mole of CO2 for every 1 mole of carbon.1890

And 1 mole of CO2 for every 2 moles of oxygen.1894

Remember we use the subscripts directly as moles from a chemical formula.1900

Now that we have our conversion factors that we can possibly use, let's go ahead and solve using dimensional analysis.1906

2.2 kilograms of carbon dioxide, remember we want to get this to grams first.1912

Times 103 grams of CO2 for every 1 kilogram of CO2.1919

Now that we are in grams of CO2... remember grams of A.1927

We now go to moles of A or moles of CO2; times 1 mole of CO2.1930

The molar mass of carbon dioxide is roughly 44 grams of CO2.1937

Now we are in moles of A, moles of carbon dioxide.1943

We want to go to moles of B which is going to be the moles of carbon.1950

That is going to be 1 mole of carbon for every 1 mole of CO2.1954

Finally moles of B goes to grams of B; times 12.01 grams of carbon for every 1 mole of carbon.1960

When all is said and done, we are going to get 2.6 times 104 grams of carbon.1971

Again this is a representative problem of using a mole to mole ratio directly from a chemical formula.1979

Right away, sample problem two, you see a chemical equation immediately.1989

You know this is going to be using a mole to mole ratio from the coefficients of the balanced chemical equation.1994

The very first thing you want to do is to balance or make sure it is balanced.2001

This is an ordinary combustion problem that we have looked at before.2005

We are going to need in the end two waters and two oxygens.2009

Now the question: how many milligrams of carbon dioxide can form from so many kilograms of CH4 gas?2017

You see milligrams and you see kilograms; therefore what type of problem is this?2027

That is right; this is a mass to mass conversion.2032

In addition, do you see amounts of both reactants specified?2039

We don't; we only see the amount of one reactant specified, the CH4.2045

So we know that this is not limiting reactant at all... not limiting reactant.2052

Let's go ahead and jump into the problem now.2063

23.1 kilograms of CH4, again we want to get to grams first.2068

Times 103 grams of CH4 for every 1 kilogram of CH4; that gets us into grams of A.2074

Now from grams of A, I want to go to moles of A.2083

Times 1 mole of CH4 for every... molar mass of CH4 is roughly 16 grams of CH4.2086

That gives me moles of A; now from moles of A, I want to go to moles of B.2096

That is going to be carbon dioxide; it is conveniently a 1:1 ratio.2102

1 mole of CO2 for every 1 mole of CH4; moles of B.2111

Now from moles of B, I want to go to grams of B.2120

Carbon dioxide is roughly 44 grams for every 1 mole.2125

The question is asking for milligrams of carbon dioxide; times 1 milligram divided by 10-3 grams.2129

After we do this calculation, we are going to get 6.3 times 107 milligrams of CO2.2143

That is expected to form from 23.1 kilograms of CH4.2150

Again this is a mole to mole ratio from the balanced equation and not a limiting reactant problem.2162

Sample problem number three; once you see a chemical equation, make sure it is balanced.2174

Again it is going to be two of these and two of those.2183

How many milligrams of carbon dioxide can form when 23.1 kilograms of CH4 are combined with 23.1 kilograms of oxygen gas?2189

First of all, you see that both of the reactant amounts are specified.2205

You know right away, 100 percent, there is no question, it is limiting reactant.2211

Because it is a limiting reactant problem, we are going to use our method2219

that we call the smaller quantity approach or the smaller quantity method... smaller quantity method.2223

Remember we are going to now determine the theoretical amount of product that can form from each reactant and compare.2235

The smaller one is going to be the answer.2242

Now the 23.1 kilograms of CH4; we already did that.2246

We got roughly 107 milligrams of CO2; that is just directly from the previous example.2259

Now we need to determine how much CO2 is going to form from 23.1 kilograms of oxygen gas.2271

23.1 kilograms of oxygen, I want to get to grams first.2280

Times 103 grams for every 1 kilogram of oxygen; that is grams of A.2288

Now from grams of A, I go to moles of A.2296

Times 1 mole of oxygen for every... the molar mass of molecular oxygen is going to be 32 grams of O2.2298

That is moles of A; now from moles of A, I go to moles of B.2314

That is going to be 1 mole of CO2 for every 2 moles of oxygen.2318

Remember I am getting this from the balanced chemical equation.2324

Now from this I want to go ahead and get the grams of CO2.2329

Again I want to get the grams of CO2.2342

Now I am going to multiply this by 44 grams of carbon dioxide for every 1 mole of CO2.2345

When all is said and done, we are going to get approximately 16,000 grams of CO2.2355

I apologize; I don't have that number in front of me.2363

But it is going to be roughly that number; roughly 16,000 grams of CO2.2365

What is important is that this is going to be smaller than the 107 milligrams of CO2 gotten from the other answer.2374

Just go ahead and compare; 16,000 grams of CO2 is going to be...2390

That is going to be 16,000,000 milligrams of CO2 or 1.6 times 107 milligrams of CO2.2396

From the previous answer, this was I believe 6.3 times 107 milligrams.2409

The 1.6 times 107 milligrams of CO2 is going to be our actual answer.2418

We conclude that molecular oxygen is the limiting reactant.2429

CH4 in this case is going to be in excess.2440

We don't have enough oxygen; that is why it is limiting.2448

But we have plenty of CH4.2450

What that means is that we can calculate the amount of CH4 that is going to remain.2454

We started with 23.1 kilograms initially of CH4.2461

We are going to be using roughly 16,000 grams of CH4 required; roughly.2471

That means 23,100 grams minus 16,000 grams tells me approximately 5,000 grams of CH4, the excess amount, will remain unused.2489

What we just did again for sample problem three was a limiting reactant problem.2511

We not only determined the limiting reactant.2516

We also determined how much of the excess reactant is going to remain.2519

Thank you for using Educator.com; it was great to see you guys again.2525

I will see you all next time.2529

Hi, welcome back to Educator.com.0000

Today's lesson in general chemistry is going to be our second lecture on stoichiometry.0003

Let's go ahead and look at the lesson overview.0011

Last time we laid out a pretty good foundation for how to tackle stoichiometry problems.0014

We are just going to now build upon that foundation and apply stoichiometry to more specific cases.0022

These cases include the following; we are going first talk about solutions and their concentrations.0029

We are going to work with a lot of terminology today such as molarity and dilutions.0038

After discussing molarity and dilutions, the next specific application of stoichiometry is0046

going to be to double replacement precipitation reactions, acid-base neutralization, and finally0053

something we have not talked about yet which is called acid-base standardization via a titration.0062

After that we will go ahead and do our summary followed by our sample problems.0067

Moving on now, molarity.0077

We tend to think of a solution in chemistry as composed of two items--typically a solute and a solvent.0079

The solute is going to be present in the smaller quantity.0089

The solvent is going to be present in the major amount.0095

Of course for general chemistry lecture and for laboratory, the typical solvent of course is going to be water.0100

We always like to express a solution's concentration--basically how much solute we have relative to solvent.0111

There is a lot of different units we can use.0119

But the most common one we are going to use is called molarity.0121

The equation for molarity is defined as the following.0125

Molarity is equal to moles of solute for every liter of solution.0128

From this equation of course, we should always be very comfortable with solving for a single variable.0136

We can solve for liter of solution.0144

That is going to be basically moles of solute over the molarity.0148

Of course we can also solve for moles of solute.0157

Moles of solute is just going to be equal to molarity times liters of solution.0160

Once again you want to definitely have these equations committed to memory, especially for general chemistry.0172

Let's go ahead and jump right into a calculation.0181

How many grams of potassium bromide are needed to make 350 milliliters of a 0.67 molar KBr solution?0186

When we verbalize capital M which stands for molarity, this is verbalized as molar.0196

This is the language; this is the terminology that we use; 0.67 molar potassium bromide solution.0203

This is a first example.0210

We are just going to plug everything we know into the equation for molarity and solve for the unknown.0212

The molarity is given to us; that is 0.67 molarity.0217

That is equal to moles of solute divided by liters of solution.0223

We don't know what the moles of the solute is, the moles of KBr.0234

That is going to be what we are trying to find.0238

The liters of solution, you are actually told that in the problem.0241

It is here; it is 350 milliliters, which is 0.350 liters.0243

When we go ahead and solve for moles of potassium bromide, we get 0.2345 moles of KBr.0251

Of course you see that the question is not asking for moles, but the question is asking for grams.0262

Do you recall how to go from moles to grams?0268

What is the conversion factor called?--it is molar mass.0271

0.2345 moles of KBr times something over something.0276

That is going to give us the grams of KBr.0284

The molar mass is approximately 119 grams over 1 mole.0289

We are going to get an answer of approximately 28 grams of KBr.0294

We are really playing on the concept of the mole very heavily.0301

Once again what we learned last time in stoichiometry is going to serve as our foundation for the rest of this lecture.0306

So very important that we have those fundamentals down; let's go ahead.0312

What we just did was we used the molarity equation directly, moles over liters.0319

We just solve for the unknown.0323

But there is another way of using the molarity equation.0325

You see that the equation for molarity is basically a ratio of two units.0329

If you recall, I have said this many times before, that when you have a ratio of two units,0333

you can get a conversion factor that can be used in dimensional analysis.0338

Let's go ahead and use this approach now.0345

You see this is the exact same problem as before.0347

How many moles of KBr are in 350 milliliters of a 0.67 molar potassium bromide solution?--and how many grams is this?0350

Remember how we tackled dimensional analysis?--you always start with the unknown.0359

What is unknown?--what is the only thing that is given to us that is not part of a ratio yet?0366

350 milliliters is by itself; that looks pretty good.0372

0.67 molar, remember this is a ratio of two units.0374

This is really 0.67 moles for every 1 liter.0380

Or this is going to be 1 liter over 0.67 moles.0386

Remember this is how we use molarity as a conversion factor.0392

It is going to be moles over liters or liter over mole.0395

Let's start with what is given to us that is not part of a ratio.0401

That is going to be 350 milliliters.0404

You see that in the question for molarity, the volume is actually liters.0409

What do you think we have to do to 350 milliliters?0414

We have to get it to liters.0417

That is going to be 0.350 liters times something over something.0418

Let's go ahead and use our conversion factor.0425

We want liters downstairs; we want moles upstairs.0428

That is going to be 0.67 moles for every 1 liter.0432

That is going to give us our answer of 0.2345 moles of KBr.0437

From there you can go ahead and get the 28 grams of KBr on your own.0445

Again all this is is this is the exact same