Sign In | Subscribe

Enter your Sign on user name and password.

Forgot password?
  • Follow us on:
Start learning today, and be successful in your academic & professional career. Start Today!
Loading video...
This is a quick preview of the lesson. For full access, please Log In or Sign up.
For more information, please see full course syllabus of AP Chemistry
  • Discussion

  • Study Guides

  • Download Lecture Slides

  • Table of Contents

  • Transcription

  • Related Books & Services

Lecture Comments (38)

1 answer

Last reply by: Magic Fu
Mon Apr 17, 2017 9:49 AM

Post by Magic Fu on April 17, 2017

Since HCl , HBr and HI are all considered as strong acids then why HF is classified as a weak acid?

1 answer

Last reply by: Professor Hovasapian
Tue Feb 9, 2016 3:23 AM

Post by RHS STUDENT on February 8, 2016

Sir, is using the ICE chart legit in the AP exam. And also do I have to plug in the unit of some constant when asking short/long questions? Thanks

3 answers

Last reply by: Professor Hovasapian
Sat Jan 10, 2015 7:23 PM

Post by Stephen Donovan on January 9, 2015

In example 4, wouldn't the two non-dominant equations buffer the dominant one? Or is that not possible due to a lack of the cyanide and nitrite ions?

1 answer

Last reply by: Professor Hovasapian
Thu Jan 8, 2015 2:08 AM

Post by Delaney Kranz on January 7, 2015

How can you tell when it is a weak acid or a strong acid when you are only given the molarity?

1 answer

Last reply by: Professor Hovasapian
Thu Feb 27, 2014 7:25 PM

Post by Samiha Bushra on February 26, 2014

I just wanted to thank you. I think I aced my quiz because of you. This is coming from a girl that failed all three of her last quizes and had a not so great background on chemistry because I took it in summer school.

2 answers

Last reply by: William Dawson
Tue Dec 10, 2013 9:09 PM

Post by William Dawson on December 8, 2013

You do a wonderful job explaining this material. It helps tremendously to point out all the easy traps and pitfalls that typically derail students when attempting to organize these problems so that they can be comnuted. In general, however, it is better to make each problem a separate example , unless they are somehow dependent on each other. Making 1 example into 3 problems is no good. My chem prof does this and I get furious at him for it!

1 answer

Last reply by: Professor Hovasapian
Mon May 13, 2013 7:56 PM

Post by KyungYeop Kim on May 13, 2013

You said the Ka,3.5x10^8 , is too small and it means weak dissociation. How do you determine whether Ka is small or big? is there a certain number based on which to decide?

Thanks always.

15 answers

Last reply by: Professor Hovasapian
Fri May 3, 2013 9:24 PM

Post by Louise Barrea on April 22, 2013

Hi, I know this is probably not in the right topic, but there is a question I can't solve. I have to calculate the pH of a solution NH4CH3COO at 0,01 M with pK(NH4+)=9,25 and pK(CH3COOH)=4,75.

When I did the major species, I found: NH4+ CH3COO- H20

I tried to determine the major reaction looking at the pK, I found: CH3COO- + H20 = CH3COOH + OH
then I calculated pOH and did pH= 14-pOH

I still can't find the right answer. What did i do wrong?


1 answer

Last reply by: Ali Hashemi
Sun Dec 9, 2012 8:53 PM

Post by Ali Hashemi on December 9, 2012

What if the Ka values were closer in example 4, could you figure out the equilibrium concentrations for both reactions and add them together then figure out pH using the sum of the Hydrogen ion concentrations?

0 answers

Post by nazgol farrokhseyr on May 2, 2012

you are the best.You made chemistry much easier for me.Thank you so much

Related Articles:

pH of Weak Acid Solutions

  • Identify the major species in solution BEFORE any equilibrium is reached. Then use an ICE chart to find Eq. concentrations.
  • ICE charts must use molarities.

pH of Weak Acid Solutions

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.

  • Intro 0:00
  • pH of Weak Acid Solutions 1:12
    • pH of Weak Acid Solutions
    • Example 1
    • Example 2
    • Example 3
    • Example 4

Transcription: pH of Weak Acid Solutions

Hello, and welcome back to; welcome back to AP Chemistry.0000

In our last discussion, we introduced the basic notions of acids and bases; actually, we didn't talk about bases so much--we will actually talk about those mostly in the next lesson.0004

We introduced the notion of an acid, and we said an acid is something that has hydrogen ions that it can give up.0014

Often, what happens is: you will take a particular acid, like hydrogen chloride, drop it in water, and that will dissociate, and it will separate into hydrogen ions and chloride ions.0020

Or, it may not separate at all, which is part of the topic we are going to be talking about today: weak acids.0030

We talked a little bit about the pH scales, the relationship between the hydrogen ion concentration and the hydroxide ion concentration in water.0035

We said that each one of them is 10-7, to come up with a total of 10-14.0045

Right now, in a minute, I am actually going to write down some of the basic things we discussed, and then we are just going to launch right into a series of problems on how to handle the pHs of weak base solutions.0050

A weak base--I'm sorry, a weak acid--is something that doesn't dissociate completely; so let's just jump in and get started and see what we can work out.0064

OK, so, last time, we said that, remember, we have water, which acts as both an acid and a base.0073

H2O--it dissociates into H+ plus OH-.0080

We said that the equilibrium constant for this--because this is liquid water, this is aqueous, this is aqueous--so our Ka for this, which we ended up calling Kw for water, was just the hydrogen ion concentration, times the hydroxide ion concentration.0087

It always equals 1.0x10-14 (let me make that 14 a little bit more clear), at 25 degrees Celsius.0106

So, the Keq, the Ka...these things change at temperature, but normally, we are talking about 25 degrees Celsius.0119

This means that any aqueous solution, no matter what is in it, the hydrogen ion concentration times the hydroxide ion concentration is always going to equal 1.0x10-14.0125

That is very, very convenient, because, no matter what one of these is, we can always find the other--and that is the whole idea.0137

Today, we are going to talk about weak acids; and weak acids are ones that actually don't dissociate completely.0145

Let's talk about a strong acid first: hydrogen chloride--when we drop this is water, the dissociation reaction for this is H+ + Cl-; it's a strong acid--strong means it completely dissociates in water, just like sodium chloride (salt) or anything else that is just completely soluble.0153

Well, as it turns out, a weak acid doesn't do that; a weak acid behaves like this: like we said last time, HA--it will react a little bit with water, and it will actually establish a little bit of an equilibrium.0174

What you will end up getting is something like H3O+, plus the conjugate base of this weak acid.0187

Another way of writing this, which I personally prefer, is just the standard dissociation, without mentioning the water: H+ + A-.0195

Now, we said there is a Ka associated with this; and the Ka is equal to the H+ concentration.0205

And again, sometimes I will use the square brackets; sometimes I won't, simply to save myself a little bit of time.0210

But, these are all concentrations in moles per liter.0216

A- over HA: now, most of the time, for the weak acids, these Kas are going to be very, very small.0219

They are going to be things on the order of -4, -5, -8, -10, even -12 and -15.0227

What that means: because a Ka is small, a small equilibrium constant means that the reaction has not moved very far forward.0235

That means most of it is here; so, if I took, let's say, some hydrofluoric acid (which is not really hydrofluoric acid--it's hydrogen fluoride, and I dropped it in water to create a solution of hydrofluoric acid), it doesn't actually dissociate too much.0244

Yes, it does produce a little bit of this and this; it does come apart a little bit; but it is a weak acid, which means that most of the hydrofluoric acid will be floating around as HA, as the entire species, in aqueous solution.0260

You still can't see the Ha in there; it is still a homogeneous solution; but it is solvated, so it isn't a hydrogen...0274

This is why, when we talk about acids, acids are always that particular compound, found in water.0281

We just automatically presume that we are talking about an aqueous solution.0289

As it turns out, hydrogen fluoride, when you make it, is not an acid--it's just hydrogen fluoride.0293

Things become acids when you drop them in water; but, because we always find them in water, because we are always talking about aqueous chemistry, we just often refer to them as acids.0298

A weak acid is one that doesn't dissociate very much.0307

Now, we need to find a way to actually find the pH of these solutions; and that is pretty much what we are going to be doing today, in this lesson; we are just going to be doing several problems.0311

Some of them are exactly like the ones before--maybe slightly different; but the idea is to develop a sense of turning what we have discussed (as far as the basic definitions of pH acid)--now we want to quantify it--now we want to use some of the math behind it.0319

What you are going to discover is: these problems that we are going to do are exactly like the equilibrium problems that we did in the previous chapter.0336

This whole ICE chart--the Initial, the Change, the Equilibrium concentration--that is going to happen over and over and over and over again.0343

You are going to be doing a ton of these; and I am specifically concentrating on doing a number of these problems, because it is in the process of doing these problems that we actually get a sense of what is going on with these acid-base chemistries.0350

There is going to be a lot of commentary, a lot of talk.0362

This is where most of the deep chemistry is happening; and, if you understand this stuff, the AP exam will be an absolute piece of cake, I promise you.0365

This is where most of the problem takes place, and we want to make sure we really understand it, because it's beautiful chemistry; it's applicable chemistry; and...well, let's just do it.0375

OK, so we are going to start off, actually, with a strong acid, just to get a sense.0385

The first example is going to be: Calculate the pH of a 0.15 Molar nitric acid solution (nitric acid is HNO3), and the pH of a 1.5x10-10 Molar HCl solution.0389

OK, so let's do the first part: let's calculate the pH of a .15 Molar nitric acid solution.0420

All right, when we do these acid-base problems (strong acid, weak acid...later on when we work with weak base, strong base; and later on when we work with buffers), it is always the same thing.0426

We want to develop a nice, systematic approach, because, as the species in the solution get more complicated, it's all that much more important to keep track of what is going on.0437

We are going to use these problems to develop a sense of the chemistry; we are going to let the chemistry decide how the math works.0446

You need to understand the chemistry; if you don't understand the chemistry, the math is irrelevant.0451

It will mean nothing to you; but, if you understand the chemistry, the math is a piece of cake.0456

In fact, it is pure instinct; there is nothing here, mathematically, that you don't know or can't follow--there is nothing mysterious happening.0460

We just want to make sure you understand what is happening.0467

OK, so first of all, we notice that it is a strong acid; a strong acid means full dissociation--that means this is going to happen.0470

That means HNO3 is going to completely dissociate into H+ ion plus NO3-; that means H+ is going to be floating around freely, NO3 is going to be floating around freely, and there is going to be nothing that is left.0479

That is why there is an arrow pointing one way--this is not an equilibrium.0491

Strong acid--full dissociation.0494

So now, what we ask ourselves--always the first question we ask ourselves: What is the major species--list the major species in the solution.0497

Well, the major species in solution are...this will decide what the chemistry is, because there is always going to be something that is going to dominate the chemistry in the solution.0504

Well, major species: we have...of course, we have H2O--that is our solvent; we have H+ floating around; and we have NO3- floating around.0517

That is it; well, they are asking for the pH--what is pH?--remember, we said that pH is just a different way of dealing with the hydrogen ion concentration.0527

It is the negative log of the hydrogen ion concentration, in moles per liter.0537

Well, look at the major species we have in solution: we have H2O; we have H+ and NO3.0543

There are two sources of hydrogen ion: one of the sources is the dissociation of the nitric acid itself; well, the other source is the dissociation of water (remember, water also comes apart and produces a little bit of this).0548

But, because this is such a strong acid that all of these hydrogen ions...actually, what it is going to end up doing is: you know that this is (under normal conditions) just 10-7 molarity--the hydrogen ion concentration.0564

It is 7 orders of magnitude smaller than this; it is could completely neglect it; as a matter of fact, because this dissociates (remember Le Chatelier's Principle?), all of the hydrogen ion that is floating around from the dissociation of this strong acid actually pushes this equilibrium this way.0579

So, there is even less hydrogen ion coming from water itself.0598

For all practical purposes, we can ignore that; the dominant species here is this; our problem is already done.0602

It says we have .15 Molar HNO3; well, strong acid...0.15 moles per liter HNO3 produces .15 Molar H+ and .15 Molar NO3-, because it's 1:1, 1:1; that is the ratio.0611

This is easy; the pH is just minus the logarithm of 0.15; that is it, because the concentration of H+ in a .15 Molar HNO3 solution is .15 moles per liter of H+.0626

That is the free thing floating around; and we get 0.82.0642

That is it--nice and simple: strong acid--all you have to do is use the molarity of that acid.0648

Just take the negative log of it; you automatically have the pH, because you have full dissociation.0655

Now, let's do the second part, B: Now, we have a 1.5x10 to the negative (11 or 10?--what did we say here)...1.5x10-10 Molar HCl solution.0660

OK, again, let's see what the major species are in solution.0676

Well, again, we have HCl; it's a strong acid; it breaks up into H+ + Cl-; well, 1.5x10-10 Molar HCl is full dissociation.0680

That implies that it is 1.5x10-10 Molar H+.0696

Well, think about this for a minute (and again, this is why you don't just want to jump into the math--you are going to get the answer wrong if you just all of a sudden take the negative log of this, like we did the first one): look at this number.0701

1.5x10-10: that is a really tiny number; compared to the auto-ionization of water, water is 1.0x10-7; 1.0x10-7 is three orders of magnitude bigger than 1.5x10-10.0712

So, virtually, for all practical purposes, this hydrogen ion concentration that comes from this hydrochloric acid, which is a strong acid, is completely negligible.0730

The reason it is negligible is because the concentration of the H is so little.0741

Yes, it's a strong acid, so our instinct tells us, "Well, if it's a strong acid, then it must be the dominant force in the solution"; it's not.0746

There is so little of it that it doesn't have any effect; so for all practical purposes, the H+ concentration here is what it would be in water at 25 degrees Celsius--which is 10-7.0753

Well, the negative log of 10-7 is 7.0768

So, our pH is 7 in this particular case.0773

We took a look at the chemistry; we took a look at the numbers involved in the chemistry--the concentrations; that helps us decide what is going to happen next--what math to use.0777

Here, .15 is a pretty high concentration; 1.5x10-10: that is three orders of magnitude less than the normal concentration of just standard water--the hydrogen ion concentration in standard water.0787

We can ignore it, despite the fact that it is a strong acid.0801

Don't let names fool you; let the chemistry tell you what is happening--pull back from the problem and think about what is going on.0804

Science and chemistry--well, all of science, not just chemistry...there is a lot happening; you have to be able to keep track of multiple things.0811

It is not just about solving the problem; solving the problem is easy if you know what is happening.0820

If you don't know what is happening, solving the problem will always be a nightmare; you will always be a victim--you will always be sort of a slave to algorithmic procedures.0826

And once something--once a problem deviates from that algorithmic procedure, you are going to be lost.0835

You don't want to be lost; you want to be able to think--always think; that is the idea.0840

OK, so now, let's go on to weak bases; and this is where the mathematics and the situation actually gets interesting.0844

We are going to present--we are going to do the same thing: we are going to present a nice, systematic approach for all of these problems.0852

Weak acid, weak base; strong acid, strong base; buffer solution; the same steps are the habit we want to get into.0857

OK, so let's do Example 2: Calculate the pH of a 1.5 Molar hydrofluoric acid solution.0865

So again, we could say hydrogen fluoride solution; it's the same thing--it just means you have taken hydrogen fluoride, and you have dropped it into some water, and you have created a solution.0888

Hydrogen fluoride is your solute; water is your solvent; now, it becomes an acid.0897

Even though we speak about HF being hydrofluoric acid, it's only an acid when it has been dropped in water; it's very important to remember that.0902

OK, the Ka for hydrofluoric acid (because it is a weak acid, it has a Ka; remember, strong acids don't have Ka--the equilibrium lies far to the right, so the denominator is virtually 0) is 7.2x10-4.0911

That is a kind of a small number: 7.2x10-4; well, what does a small Ka mean?0927

A small Ka--it means that there is very little dissociation--that is the whole idea.0935

You need to know what these things mean; they are not just numbers and parentheses and concentrations; it means something, physically.0941

It means there is very little dissociation.0948

Very little dissociation--that means that HF...the dissociation of HF into H+ + F-...that means mostly, if we looked at this, it's going to be mostly hydrogen fluoride, not hydrogen ion and fluoride ion.0952

That is what the small Ka means.0966

OK, in other words, it is mostly here; the equilibrium is mostly here; in solution, it's still just HF floating around, not H+ and F- floating around freely; so, we will say mostly here.0969

Well, the next step is...let's list our major species to see what is going on.0981

Let me actually list the major species here: our major species in this solution are going to be...again, because there is very little dissociation, it's a weak acid, you are going to find mostly HF, and you are going to find H2O.0986

Both of these are weak acids: HF, as we said, has a Ka of 7.2x10-4; well, H2O has a Ka, which we call Kw; it is 1.0x10-14.1001

Well, between 10-4 and 10-14, this is 10 orders of magnitude bigger; it's huge--this is virtually nonexistent compared to this.1018

Because this is so much bigger, we can ignore any contribution that H2O does in its dissociation.1027

Its contribution of hydrogen ion is completely negligible--not even measurable, to be honest with you.1035

This is going to dominate the equilibrium in the solution.1040

So now, we want to see: at equilibrium, when everything has stopped and settled down, what is the pH of that solution?1043

Well, here is how we do it: We write down the equilibrium expression, HF goes to H+ + F-; we have I, we have C, and we have E (our initial concentration, change in concentration, and equilibrium concentration).1050

Our initial concentration was 1.5 Molar; there is, before the system has come to equilibrium, 0 and 0; so again, imagine, when you are doing these problems--you are taking something; you are dropping it in water; and it is before anything happens.1065

Before it starts to come apart, that is the initial conditions.1084

Now, HF is going to dissociate a little bit, so that means a little bit of it is going to disappear; and for every amount that disappears, that is how much H and F show up, right?1088

1:1, 1:1...when 1 molecule of this breaks up, it produces 1 ion of H+ and 1 ion of F-.1100

That is why we have -x, +x, +x, when they put +'s here.1108

And now, our equilibrium concentration is just adding these: 1.5-x; when everything has come to a stop, that is how much we have of the HF; that is how much we have of the H; and that is how much we have of the F-.1113

Now, we can write our expression.1126

Ka, which is equal to 7.2x10-4, is equal to...well, it's equal to the H+ concentration, times the F- concentration, over the HF concentration, at equilibrium.1128

That equals x, times x, divided by 1.5-x.1145

Now, let me rewrite this over here; so let's...part of one equation: 7.2x10-4; now, what we want to do...again, we want the pH.1155

We want to take hydrogen ion concentration and take the negative log of it; so we are looking for x.1166

We want to solve this equation for x.1172

Now notice, we are going to end up with a quadratic equation here; there are a couple of ways that we can handle this.1175

We can go ahead and (because you have graphing calculators, presumably--most of you) you can go ahead and use your graphing calculator to find the roots of this quadratic equation.1180

Or, you could solve the quadratic equation--turn this into a quadratic--multiply through and turn it into a quadratic--use the -b, plus or minus radical b2 minus 4ac over 2a to solve for x.1189

Or, we can actually simplify the procedure to make this a little better for us.1201

And again, the simplification is based on standing back and taking a look at the chemistry.1205

This is a weak acid; there is going to be very little dissociation here.1210

Even though there is an equilibrium, there is going to be very little, actually, of the H+ concentration--so little, in fact, that compared to 1.5 Molar--compared to this, this is going to be so tiny that the 1.5 probably won't even notice that it is gone.1214

So, we can simplify this expression by writing it as: 7.2x10-4 equals x2 over 1.5.1230

Now, when we find the value of x, we need to check to see if our assumption, if our approximation, is valid; and we will show you how to do that in just a minute.1245

Let's go ahead and just solve this first, and then we'll check to see if it's valid.1253

When we multiply through, we get x2 = 1.08x10-3, and we get x=0.033 moles per liter.1257

Well, this is the hydrogen ion concentration, because that was x in our ICE chart.1270

Now, we need to check if our approximation--our eliminating x from the denominator here--was valid.1275

Here is how you do it: most Kas that are listed in tables have an error of about plus or minus 5%.1280

Now, this value, .033, 0.033--if it is 5% or less than the original concentration that you subtracted from...1.5 times 100; you want to calculate a percentage; if this is 5% or less of this number, then our approximation is valid, meaning the difference is not really going to be noticeable--that your numbers are actually pretty good.1289

If it is above about 5 or 6 percent, that means you can't make this approximation--you have to solve the quadratic equation.1322

So again, we are taking a look at the chemistry, and we are letting the Ka tell us that we have a weak acid--very little dissociation.1329

So, this x, compared to 1.5, is probably going to be very, very tiny--so much so that we are probably not even going to notice a difference: 1.5, 1.49--it's going to be so small.1337

Well, if it falls into the 5% rule, then it's valid; if it doesn't, then it's not valid; so the first thing to do is go ahead and simplify, and check to see if this is the case.1349

In this particular case, we get 2.2%; well, 2.2% is definitely less than 5%, which means our approximation is valid, and we can use this value of x; it's perfectly good.1359

So, now that we have that, we have our H+ concentration of .033, our pH equals -log of 0.033, is equal to 1.48; there you go.1372

Now, I personally (just a little aside) don't care for pH; I think once you have found the hydrogen ion concentration, you are talking about concentration.1387

Taking a number and fiddling with it so that it has a more attractive number, like 1.48--I personally don't think 1.48 is more attractive or easier to deal with than .033, or 3.3x10-2.1396

It's a perfectly valid number; here, we know we're talking about concentration; pH--it's a little weird, because again, the lower the pH, the more acidic.1409

If you remember, you have a pH scale that runs from 0 to 14; 7 is neutral--that is the pH of plain water.1419

Below 7, we are talking about an acidic solution; above 7, you are talking about a basic solution.1427

The lower the number (6, 5, 4, 3, 2, 1)--that is more acidic; what that means is that there is a greater concentration of hydrogen ion.1434

It's a little strange: a lower number means more powerful; but again, this has just become part and parcel of the standard chemical practice in the industry, across the board.1444

But again, as long as you understand what is happening, that is what is important.1456

OK, so let's do another one; this time, let's calculate...let's see; this is going to be Example 3.1461

Let's calculate the pH of a 0.15 (this time, 0.15) Molar solution of HClO, hydrogen hypochloride--otherwise known as hypochlorous acid.1472

The Ka for this equals 3.5x10-8.1501

The first you want to notice is: look at the Ka; this is a very, very small number--that means we are talking about a very weak acid, very little dissociation.1507

When you take hydrogen hypochloride and drop it into water, it's going to stay hydrogen hypochloride; it's not going to come apart into hydrogen ion and hypochloride ion too much.1515

A little bit, it will; but not too much.1526

The fact that we have a number that we can measure means there is some dissociation; but again, this says that it is a weak dissociation.1529

OK, so now let's see which equilibrium is actually going to dominate in this.1536

We have two sources of hydrogen ion: we have the HClO, which breaks up into H+ + ClO-, potentially; and the Ka for this, as we just said, is 3.5x10-8.1542

Then, we have H2O; that is the major species in the water; that is H+ + OH-, but again, the Kw for this is 1.0x10 to the (oops, not 14) negative 14.1558

Well, 10 to the -8--even though it is a small number, it is a lot bigger than 10 to the -14; 6 orders of magnitude bigger.1577

The dominant reaction here, the dominant equilibrium, is this one.1584

This is virtually not even noticeable.1589

So, let's go ahead and do the problem.1596

Since the HClO equilibrium is going to dominate, we are going to write HClO, and I just am one of those people that likes to write everything--so, H+ + ClO- (I know that it's a little...I'm just writing it over here, but I just like to make sure that everything is clear).1598

This is initial, change, equilibrium (and let's erase some of these stray lines here); so, our initial concentration of HClO, before the system has come to equilibrium--when we just drop it in before it has come to equilibrium--this is 0.15 Molar.1619

There is no hydrogen ion, and there is no hypochloride ion.1635

Well, a certain amount of it dissociates; and it produces, for every mole that (we're not going to have that; we want this to be clean; +x) dissociates, it produces a mole of H+; it produces a mole of ClO-.1639

Equilibrium concentration is 0.15-x; this is +x; this is +x; and now, we can go ahead and write our equilibrium.1658

Our Ka is our hydrogen ion concentration, times our hypochloride ion concentration, over our hydrogen hypochloride.1671

When we do this, in terms of numbers, we get: 3.5x10-8, which is our Ka; it is equal to x times x, over 0.15-x.1681

And again, we want to simplify, so this is going to be approximately equal to x squared, over 0.15.1697

Then, we'll check to see if that approximation is actually valid.1705

When we solve this, we get x2 is equal to 5.25x10-9, and we get that the x, which is equal to the hydrogen ion concentration, is equal to 7.24x10-5.1709

Now, let's check the validity; so I'll write "check the validity of our approximation"--in other words, what we did here.1728

We'll take 7.24x10-5, and we'll divide it by the initial concentration, 0.15; so, when we do that and multiply by 100, it equals (let me drop it down a little bit here) 0.048%.1735

Wow, look at that; that means only 0.048% is actually dissociated; that is a very, very, very weak acid--virtually none of it has come apart.1753

This is clearly below 5%; our approximation is good; so we can use this number, and when we take the negative log of 7.24x10-5, we end up with (let's see what we ended up with here) 4.14.1765

4.14--and this our pH, and this is our hydrogen ion concentration.1788

Again, my preference is for the actual concentration; pH is just a number.1794

OK, that is good; so hopefully, you are getting a sense of what it is that we are doing; we are looking at Ka; we're choosing the major species; we're deciding which of those major species is going to dominate in the solution.1800

We know what chemistry is going to dominate; usually there is one species, one equilibrium, that dominates--everything else can be ignored.1812

And then, we just write out our ICE chart: Initial concentration, the Change equilibrium, and then we set it equal to the equilibrium constant, and we just deal with the math at that point; it's a simple algebra problem.1820

OK, now let's do a problem where we have a mixture of weak acids.1831

Same exact thing: it's going to be more species in water, but one of them is going to dominate; so let's do this.1834

Example 4: Calculate the pH of a solution that contains 1.2 moles per liter of hydrogen cyanide, or hydrocyanic acid, whose Ka is 6.2x10-10--very, very weak--and 4.0 Molar HNO2 (hydrogen nitrite or nitrous acid), whose Ka is equal to 4.0x10-4.1843

Also, calculate the concentration of the cyanide ion at equilibrium.1902

OK, so we want to find the pH; so we have a mixture--we have this certain amount of water; we drop in some 4-Molar hydrogen nitrite, or nitrous acid; we drop in 1.2-Molar hydrocyanic acid; we want to know what the pH of the solution is, and we want to know what the concentration of the free cyanide ion is in that solution.1925

OK, let's list our major species with their particular equilibriums.1949

Well, let me go back to blue here; so, major species in water.1955

That is what we are doing; we always want to do the major species to decide what chemistry is going to dominate.1963

Well, HCn is a weak acid; that means it is mostly HCn--it hasn't dissociated.1967

A strong acid dissociates--weak acids, not very much.1973

We also have HNO2 floating around in that, and we have H2O.1977

Well, the HCn equilibrium (H+ + Cn-)--we said that the Ka is equal to 6.2x10-10.1983

The (oops, wow, look at that; that is a crazy line; OK) HNO2 equilibrium (NO2-)--the Ka for this one is equal to 4.0x10-4.1996

And of course, we have the (oh, here we go again; wow)...last but not least, we have the H2O equilibrium, which is also another source of hydrogen ion, plus OH-; the Kw equals 1.0x10-14.2022

So now, let's compare: 10-10; 10-4; 10-14.2042

We can virtually ignore the 10-14--it's too tiny.2047

Between 10-4 and 10-10, this one--the HNO2; this is going to dominate the chemistry in this solution.2050

Because that is going to dominate, we can ignore any contribution of hydrogen ion from hydrogen cyanide, and we can ignore any contribution of H+ from water.2058

And again, we are doing this because there are three sources of hydrogen ion: some can come from the HCn; some can come from HNO2; some can come from H2O.2069

But, the Ka of the HCn and the H2O are so tiny that they are negligible, so this is what controls the chemistry of the solution; that is the chemistry we concentrate on.2078

Therefore, let's go with HNO2: H+ + NO2-; now, you are going to do the same thing that you did before.2089

You are going to make a little ICE chart: Initial, Change, Equilibrium; you are going to do your simplification; you are going to check it; everything is going to be fine; you are going to end up with a pH of...2103

I hope you actually run through and do this, based on the previous two examples.2113

So, for this one, you are going to end up with a pH of 1.40.2118

Now, the question is (the second part): How do we find the cyanide ion concentration?2126

OK, let's write down the cyanide equilibrium; we need to find the cyanide ion concentration, so we need to work with the cyanide equilibrium.2132

HCn goes to H+ + Cn-.2141

Ka equals 6.2x10-10.2147

Now, let's stop and think about what this means.2152

This is saying that, in a given solution, where you have cyanide ion, hydrogen ion, and HCn (hydrogen cyanide) floating around in solution--these three species floating around in some concentration each--the equilibrium concentration (in other words, the concentration of this, times the concentration of that, divided by the concentration of this) equals 6.2x10-10.2156

Remember, that is what an equilibrium expression is: equilibrium positions can change, but the constant stays the same--the relationship among these three--that is what equals this.2184

It doesn't matter where these H+ come from; this is talking about: at any given moment, if you have cyanide, hydrogen ion, and hydrocyanic acid in solution, the concentration of this times that, divided by the concentration of this, always equals that.2196

This gives us a way to find it; it doesn't matter where these H+ come from--they can come from HCn, or they can come from any other source, any other source.2215

In this case, the primary source of the hydrogen ion is the nitrous acid.2226

It is the dominant species in the water; it is the one that is going to give up its hydrogen ion and suppress the others.2232

So, at equilibrium, there is a certain concentration of hydrogen ion, and that concentration of hydrogen ion was the 1.4--well, the pH was 1.4; the 1.4 came from the negative log of the hydrogen ion concentration.2239

Now, let's go ahead and work this out.2257

This we know that the Ka is equal to H+ concentration, times Cn- concentration, over HCn concentration.2263

That equals 6.2x10-8; this is the equilibrium expression.2278

Well, let's see: let's go at equilibrium, we have...not 6.2x10-8; we have 6.2x10-10; yes, that is 10.2285

That means 6.2x10-10; now, we go back on our ICE chart, and we read off this, this, and this.2300

We ended up with: our equilibrium concentration is 0.04.2310

That was the hydrogen ion concentration that gave us the pH of 1.4.2321

We have the Cn- concentration, and, at equilibrium, the HCn concentration was 1.2-x.2325

OK, again, we are just dealing with a basic math problem, but as it turns out, we have a little bit of a problem; we have this x here...smaller than normal...suppresses the HCn dissociation...let's see here.2335

Let me see, how do I want to go ahead and explain this to make it most reasonable?2358

We have that; we have that; we have the cyanide concentration; OK.2364

Let me erase something here.2369

OK, so let me rewrite this here: HCn, H+ + Cn-.2378

Now, at equilibrium, we started off with 1.2 Molar of the HCn, right?2389

This was 1.2 Molar.2404

Now, we have the H+ concentration, which is 0.04; this is the equilibrium concentration.2407

This is what we want--this Cn- concentration.2412

Now, it's true that, of the HCn, some of it will dissociate; so the fact of the matter is, the equilibrium concentration is going to be 1.2-x, but again, remember what we said: the dominant equilibrium here was the nitrous acid.2417

It is so dominant that it is actually going to suppress this dissociation.2437

Anything that HCn would actually produce, any H+, is virtually nonexistent, because not only was it small to begin with, but because there is so much of the H+ from the nitrous acid, it's actually going to push this even more that way, so there is going to be even less.2442

So, for all practical purposes, we don't even have to worry about any HCn dissociating.2458

Therefore, our final equilibrium will end up actually being: 6.2x10-10 is equal to 0.04 (which was the hydrogen ion concentration we found from the HNO2 dissociation), times the Cn- concentration, over the HCn concentration (which was 1.2 Molar).2464

And then, when we solve for this, we end up getting a Cn- concentration (and this time, I will use the square brackets): 1.86x10-8 Molar.2490

As you can see, 1.86x10-8 Molar is so tiny--that means virtually...there is virtually no cyanide ion floating around.2504

That is based on this equilibrium.2515

So again, if you have a mixture of acids, one of those acids will dominate the equilibrium.2518

If you are asked for the concentration of a species from the other acid, the one that is not dominant, you can still just write out its equilibrium, and it just means that--in this particular case, we wanted cyanide; well, the equilibrium expression for HCn says that cyanide concentration, times H+, divided by this, equals this.2524

Well, we have had this from what we just calculated from the dominant species; we have this, which is the original concentration.2546

And then, all we have to do is solve for this, because we already have this.2555

1, 2, 3, 4; we have 1, 2, 3 of them; we solve for the fourth.2559

There you go; I hope that made sense.2564

OK, let's see: we have taken a look at some weak acid problems; we have listed the major species; we take a look at the major species to decide which one is dominant.2569

And usually, in this particular case (for weak acids), we just see which one has the higher Ka value.2582

Whichever one has the higher Ka value, that equilibrium, that acid, will dominate the chemistry of the solution.2588

We write its equation; we list its initial concentration, the change in the concentration, and the equilibrium concentration; and then, we put the equilibrium concentrations, in terms of x, into the equilibrium expression, and we solve for x.2595

More often than not, they are going to ask for calculating the hydrogen ion concentration or, actually, the pH.2610

More often than not, they will just straight-out ask for the pH.2616

So, thank you for joining us here at and AP Chemistry for weak acids.2619

Our next topic we are going to be discussing is going to be percent dissociation; we are going to spend a fair amount of time on weak bases.2624

Take care; goodbye.2631