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Lecture Comments (15)

1 answer

Last reply by: Professor Franklin Ow
Wed May 7, 2014 6:44 PM

Post by Edgar Suarez on May 5 at 09:30:30 PM

I have a question, you said that milk was basic, but i thought that it was an acid, because of its lactic acid.

1 answer

Last reply by: Professor Franklin Ow
Sat Apr 26, 2014 5:17 PM

Post by Anthony Mendoza on April 25 at 02:00:13 AM

In example 2, how did you know that HPO4 was the conjugate of weak acid H3PO4? In my head, I think I might have chosen that it was the conjugate of -PO4, or conjugate of H2PO4...

1 answer

Last reply by: Professor Franklin Ow
Sat Apr 26, 2014 5:16 PM

Post by Anthony Mendoza on April 25 at 01:15:02 AM

Hi Professor Ow
,
Could I get some clarification on how increased electronegativity makes it a stronger acid? HF has a strong electronegativity difference, as in the F- pulls strongly on the H+, so I would imagine it would't want to let it go (and thus dissociate and thus not be that strong).

I'm trying to compare HF to the others on the list, but it's difficult because aren't the rest bases. They already have a negative charge, so obviously aren't going to want to dissociate. But how about comparing HF to HCl. HCl is a strong acid, yet has lower ElectroNegativity (since F- is supposed to be the most EN?). This part still confuses me.

Thanks for your help! Your explanations are making class a lot more manageable!

1 answer

Last reply by: Professor Franklin Ow
Sat Apr 26, 2014 5:15 PM

Post by Melyssa James on April 2 at 12:36:29 PM

For your first example, after you determined the pH, i don't see where you calculated the percent ionization??

2 answers

Last reply by: Professor Franklin Ow
Fri Feb 28, 2014 11:40 AM

Post by Jack Miars on February 26 at 05:19:09 PM

Can you always make the assumption that x is negligible? If not, what situations cause it to be accounted for?

1 answer

Last reply by: Professor Franklin Ow
Mon Feb 10, 2014 10:58 AM

Post by Miley Xiao on February 9 at 01:43:07 PM

what does x at 13:38 stands for? and why are CH3CO2 and H3O the same since he wrote two xs?

0 answers

Post by Cheng Jiang on May 16, 2013

very gucci

0 answers

Post by Max Mayo on April 2, 2013

great job

Acid-Base Chemistry

  • Bronsted-Lowry acid-base chemistry involves a loss/gain of a proton to/from water.
  • Conjugate pairs only differ by one proton and are inversely related in terms of acidity/basicity.
  • Acidity and basicity can be quantified primarily via pH, Ka ,pKa and percent ionization.
  • The structure of a molecule can have a significant impact on how acidic it can be.
  • ICE tables are easily applied to an acid-base equilibrium situation.

Acid-Base Chemistry

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

Transcription: Acid-Base Chemistry

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