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Lecture Comments (8)

0 answers

Post by Bhupen Khanolkar on May 16, 2015

Dear Prof. Hovasapian,

Thank you very much for your help in this concept. I was a little confused on how you did Example 3 of this lecture. When calculating the molarity of NaPr, wouldn't you have to find the moles of Pr- that come from the dissociation of HPr, and subtract that from the total Pr- concentration to give the moles of Pr- that come from the dissociation of NaPr, and then find the molarity? When I solved it this way, I found the concentration of Pr- (from HPr)= 6.43*10^(-4). The concentration of Pr- (from NaPr) was 0.09691 mol, and dividing that by 0.250L gave me a molarity of 0.387. I was wondering if this method was correct.

Please let me know soon.

Thank you very much

1 answer

Last reply by: Professor Hovasapian
Wed Feb 4, 2015 10:08 PM

Post by Ravi Bala on February 4, 2015

Hello Prof. Raffi,

I really feel like I understand the concept behind buffers very well. I think it's for this reason why I don't really like using the henderson equation. However, I just want to make sure that I can always avoid using that equation when calculating for pH, or are there certain situations in where the problem can only be solved using that equation?

Also I am just wondering, considering we won't get a calculator for the AP chem multiple choice, what will the acid base questions look like on there?

1 answer

Last reply by: Professor Hovasapian
Thu Oct 30, 2014 1:26 AM

Post by Long Tran on October 28, 2014

Hi Prof
on the example 1 calculating STOICH part a, the changing of H+ is 0. could you explain ?

0 answers

Post by Chemutai Shiow on March 23, 2014

In Example 3, you use 150 L instead of 150 mL(what was in the question) which threw of the math but the concept was the same. No biggie

1 answer

Last reply by: Professor Hovasapian
Wed Nov 7, 2012 1:38 PM

Post by noha nasser on November 7, 2012

Hello prof Raffi,
im a little confused here about the ka of the HAc in example 1 at the begining you gave us = 1.8x10(-5) and then in the solution you used 4.74 where did the 4.74 come from? :))

Buffers, Part III

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
  • Buffers 0:25
    • Buffer Capacity Part 1
    • Example 1
    • Buffer Capacity Part 2
    • Example 2
    • Example 3

Transcription: Buffers, Part III

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

Today, we are going to close out our discussion of buffering.0003

I know that we spent a fair amount of time on this, but it is important; it is an important application of acid-base chemistry, and so, today, we are going to round it out.0007

We are going to discuss something called buffer capacity and do some applications of buffer capacity.0016

Let's just start with a definition, and go forward from there.0023

Buffer capacity: buffer capacity is the amount of H+ or OH- a buffer solution can absorb without significant change in pH.0029

So again, you can create a buffer solution in any way you like, but the idea is: the more of the components in that buffer that you have that can actually swallow up any excess hydroxide or hydrogen ion, the greater your buffer capacity.0074

That means it has a greater capacity to actually buffer; it gives you a broader range of lack of change; that is the whole idea--that is the whole idea behind what a buffer is.0090

A buffer gives you a range where things can happen without anything changing in the overall system.0101

Recall: a buffer solution is HA, A-.0110

So, added base (which is OH-; anytime you add a base, you are essentially adding OH-) reacts with...these are the two things that are in the buffer.0129

Added base reacts with the HA; so HA essentially absorbs the OH- so it is not floating around freely in solution, changing the pH.0145

Here is how it does it: the HA reacts with the OH- to create A- + H2O.0179

It takes the OH-, and it pulls it out of solution; it pulls it out of solution as water, so that this floats around instead of the OH---that is what a buffer does.0190

Now, added H+ reacts with A-; so A- absorbs the H+, and it does so as follows.0199

All of these minuses and plusses...A- + H+ produces HA; any added acid is taken up by the A-, and it is locked away as the weak acid--it is locked away as the acid itself, the undissociated form, so you don't have any of this free hydrogen ion floating around, changing the pH.0224

Let's go ahead and do an example; I think it will be the best.0251

So, Example 1: Calculate the change in pH when 0.010 mol of gaseous hydrogen chloride is added to 1.0 liter of solution A, which is 4.0 Molar acetic acid, and a 4.0 Molar sodium acetate, and solution B is 0.040 Molar acetic acid and 0.040 Molar sodium acetate.0256

OK, so we have two buffer solutions: the first buffer solution is made with 4 Molar of the acetic acid and 4 Molar sodium acetate; another buffer solution is made with .04 Molar acetic acid and .04 Molar sodium acetate.0321

To both of these solutions (separately), we bubble in .010 moles of gaseous hydrogen chloride.0338

Now, notice: I said hydrogen chloride, not hydrochloric acid; as a gas, hydrogen chloride is a molecule; and when you bubble it into water, it dissolves because it's a strong acid in aqueous solution, and then you free H+ and Cl.0348

So, gaseous hydrogen chloride is added to 1 liter of each of these solutions; what is the change in pH?0362

OK, so what we are investigating here is this thing called buffer capacity, in case you wanted to know what it is that we are actually doing.0368

Well, so let's go ahead and use the Henderson-Hasselbalch equation to actually calculate the pH of both of these solutions before the hydrogen chloride is added.0376

So, let's write the Henderson-Hasselbalch equation up here: pH is equal to pKa + the log of A- over HA, the concentration of the base part over the acid.0388

And now, we want to say the Ka of acetic acid, of HAc, is equal to 1.8x10-5 (I didn't write that in the previous page, but there it is).0407

OK, so, well, if we take solution A...so, solution A: the pH is equal to the pKa (so the negative log of 1.8x10-5, which is 4.74), plus the logarithm of...well, the concentration of the acetate is 4 Molar; the concentration of the acetic acid is 4 Molar; well, this is...4/4 is 1; what is the logarithm of 1?--it's 0.0419

It goes to 0; so the pH equals the pKa, which is 4.74; interesting.0453

How about solution B?0459

Well, pH equals pKa (the same thing: 4.74), plus the log (this time, it is the log of 0.040, over .040; .040 Molar sodium acetate, .040 Molar acetic acid).0465

Well, this is still 1; the logarithm of 1 is still 0; it hasn't changed; so again, notice, we have different molarities of these solutions, but the ratio is 1; the logarithm of 1 is 0; so both of these have a pH of 4.74.0482

It's kind of interesting, isn't it?0497

Now, we want to see which one of these has a greater capacity to absorb anything that we add, whether it's acid or base (in this case, we are adding acid, not base) and remain close to 4.74, or as close as possible.0499

OK, so now, we do our major species upon addition of H+, but before any reaction.0513

We are doing accounting here; we are accounting for what is inside the solution, so we can decide what is happening.0534

Well, we have the acetic acid; it's a weak acid, so it's mostly in this form; we have dropped in the acetate, so we also have this floating around.0539

We have H2O floating around, but that is the solvent; we have sodium--that is the sodium acetate; and we have the H+, and we have Cl-, because that is the hydrochloric acid--the hydrogen chloride--that we have bubbled into it; now it's dissociated, because it is a strong acid in aqueous solution, so now we can call it hydrochloric acid.0547

OK, now, let's go ahead and do the stoichiometry.0569

We always have to do the stoichiometry, right?--when we add an acid or a base to a buffer solution, we do the stoichiometry before we do the equilibrium part.0575

The Henderson-Hasselbalch equation is for equilibrium part; it's not for the stoichiometry--there is no shortcut for the stoichiometry; we simply have to do the stoichiometry.0592

OK, the stoichiometry is the following: you have this floating around in solution, and this is going to react with the best source of base here (the best source of base is either this or this).0600

It is not going to react with this--it is not going to combine with this; it just separated from that.0614

It is going to combine with this; these two are going to dominate the stoichiometry.0619

H is going to react completely with this according to the following: it is going to be: Ac- + H+ is going to form HAc.0623

We have a Before; we have the Change; and we have the After--stoichiometry, not equilibrium.0637

Well, before anything happens, the concentration of acetate is 4.0 molarity; the H+ concentration is 0.010, because that is how many moles were bubbled in, and we have 4.0 moles of the acetic acid.0647

Well, what change is going to take place?--well, all of this H is going to react with this completely; this reaction goes to completion--that is what the stoichiometry part of these buffer problems is.0677

You are assuming that this reaction of H + Ac- (or, if you add base, OH- plus the HAc) goes to completion, which means that all of this is going to be 0, which means that all of the H+ is going to be used up, and since this is 1:1, that means this is going to diminish by .010.0689

This is going to show up; so this is going to be +0.010.0714

Our final concentration, before the system comes to equilibrium, is going to be 3.99 moles; 0 moles; 4.01 moles.0719

These are moles, OK?--this is one liter of solution; that is why we had 4 Molar; but we are actually...because it's in one liter of solution, it's in moles.0737

OK, now let's see if we can...all right, now we are ready to do the equilibrium part, but we don't have to do the equilibrium part; we can use the Henderson-Hasselbalch equation--that is the equilibrium part.0747

Let us do that: we get: pH=pKa, plus the log of acetate concentration, over acetic acid concentration; pH = 4.74 plus the log of...now, we said it's moles, 3.99 moles, but we are in one liter, so the molarity is 3.99/1, so it's the same.0761

That is the nice thing about working with one liter of solution.0795

3.99: that is the acetate concentration; 4.01: that is the HAc concentration; it equals 4.74, plus log of (oh, you know what, I didn't even do the numbers; but actually, it doesn't matter; if you were to do this number, you would find that it's very, very small, so I'll just put a little box here for whichever number you get when you take the logarithm of) 3.99/4.01: you are going to find, to two significant figures, the answer is 4.74: no change in pH.0799

Very interesting: the reason there is no change in pH is because we had so much of the acetate and this, and we had it in a 1:1...the ratio here is 1, so that this changed very little down; this changed very little up; this ratio remained .996, .997, virtually 1.0842

That is why, as far as to two significant figures, there is virtually no change.0864

The pH didn't change here when we added .010 moles of acid to this solution.0868

.010 moles of acid is not...it sounds like a little, but it's actually a fair amount of acid; and yet, the pH didn't change at all--this has a high buffer capacity.0875

Now, let's do the same for the solution B.0885

Solution B: OK, well, we want to do the stoichiometry part.0891

Stoichiometry: again, we have acetate, plus any of the acid that we want to absorb, forming HAc; we have a Before; we have a Change; and we have an After.0900

This time, the Before concentration...number of moles; I'm sorry...the concentration is .040 moles per liter; we have one liter of the solution, so we have .040 moles.0920

We have 0.010 moles of the H+ floating around; and we have 0.040 moles, minus 0.010 (because this reaction is going to go to completion, so all of the H+ is going to be used up, which means that .010 moles is going to disappear from here); this is going to be -0.010, and this is going to be +0.010.0933

Here, we get 0.03; here, we get 0; here, we get 0.05; here we go.0965

These are how many moles of the acetate and the acetic acid that we have; well, again, since we are dealing with one liter of solution, it is just these things, divided by 1 liter, so the moles and the molarity is the same.0976

Now, let's go ahead and do pH=pKa, plus the log...actually, you know what, I'm just not going to rewrite the equation; you already know it by now, so I'm just going to write out pH is equal to 4.74, plus the logarithm of 0.03, divided by 0.05, and we end up with a pH of 4.52.0989

OK, now, in solution A, we went from a pH of 4.74 to a pH of 4.74.1022

In the second solution we had, we went from a pH of 4.74 to a pH of 4.52.1032

That is a reasonable change in pH under the circumstances, and the reason that this particular buffer solution did not regulate the change in pH was because it did not have as much buffer capacity as the other solution.1040

Buffer capacity means the amount of OH or H+ that can be absorbed without a significant change in pH.1058

This is significant: .52 to .74--that is .22 units.1065

Well, that means the buffering components, which are your base, your acetate, and your acetic acid--your acetate and your acetic acid--the amounts of these are so high, compared to what you add, that any change in these concentrations leaves this value close to 1.1069

But, the change here, even though it can act as a buffer--it is buffering here; it's not a problem--I mean, you are not getting a huge change--but now, the ratio has gone from 1 to .03 to .05; so let's take a look at what these ratios are.1088

This ratio here, for solution A...the ratio of A to HA is equal to 3.99 over 4.01, is 0.996; the ratio hasn't changed all that much from the 1.1105

Here, you have a ratio of 0.03 over 0.05, and the ratio is 0.6; it has gone from 1 (.04 over .04) to .03/.05; it has gone down to .6.1127

That is a 40% drop in the ratio; therefore, even though it is buffering, its capacity to buffer is not as good as this one, because you don't have as much of the acetate and the acetic acid--you have less of it.1143

Because you have less of it, the more of it that is used, it changes the ratio that much more of an amount.1159

That is what is going on here.1167

So, let me write this down (wow, it would be really, really nice if I could actually spell; OK): the greater concentration of buffering components A- and HA in solution gives a greater buffering capacity.1169

So, pH=pKa plus the logarithm of concentration of A-, over the concentration of HA; these are the components of the buffer, the weak acid and its conjugate base.1226

When there are enough of those--when there are enough of those to where they can actually absorb fair amounts of added acid or base, this value, this ratio, changes little.1244

When this ratio changes little, the pH changes little.1257

That is the whole idea behind buffer capacity.1261

OK, so let's close this out with a nice practical application of buffers.1264

In the lab, when preparing a buffer solution, you are choosing the pH you want beforehand; that is the whole idea.1270

So again, a buffer solution is a solution of a specific pH, prescribed, predetermined pH, that you want before you run your experiment--before you add acid or base to it.1306

Let's say you are running an experiment for a particular cell culture, and you need a pH of, I don't know, 4.6.1317

Who knows what it is?--4.6, 7.2, 3 point...you can choose any pH you want.1326

Well, the question is: OK, so let's say you have chosen a particular pH; how do you actually choose an acid and its conjugate base, or a base and a conjugate acid, that actually gives you the pH that you want?1332

Well, we are going to answer that question right now.1344

OK, so the question is: how do you choose which weak acid or base to use?1347

And the answer is: You choose the acid with a pKa as close as possible to your predetermined pH, to your desired pH.1375

And here is why--it's actually very, very obvious: pH equals pKa, plus the log of the base over the acid component; these are our concentrations.1406

We said that we want this number to change as little as possible--this ratio; so, when you add acid or base, these are going to...one is going to go up; one is going to go down.1418

Well, we want it to change as little as possible at our given pH; well, if we want this to change as little as possible, we want the ratio of this to be as close to 1 as possible, because we want as much of this and as much of this as possible.1433

We don't want something like 10 of this and 1 of this...you know, for every one of these, ten of these; or for every 1 of these, 10 of these, because then, any change that is made is going to get even bigger in those directions.1450

So, when this value--when this ratio--is equal to 1; well, when this ratio is equal to 1, it has the greatest buffer capacity and the greatest resistance to changes in pH, but a logarithm of 1 is 0.1461

Because the logarithm of 1 is 0, pH=pKa.1472

Therefore, if you can find an acid that has a pKa whose pH is close to that pKa, you can pretty much arrange it in such a way that these will always be close to 1.1477

That is your answer: if you want...the pH of the buffer solution that you want...you want to choose an acid that has a pKa as close to that pH as possible.1490

If you want a buffer solution of a pH of 3.6, you need to find an acid that has a pKa as close to 3.6 as possible.1500

So, let's go ahead and do an example, and I think it will all make sense.1512

All right, let's see: Example...I think this is going to be Example 2: OK, a chemist needs a solution buffered at 4.80, and can choose from the following.1520

OK, our first choice is monochloroacetic acid; it has a Ka of 1.35x10-3.1561

Our second choice is propanoic acid, and our Ka for that is 1.3x10-5; we saw it in a previous example.1582

Our third choice is benzoic acid...very common, actually, benzoic acid...most of you probably use something called...when you put sunscreen on, that is actually paraamina benzoic acid; that is benzoic acid that has had some amino group attached to it.1595

That is what you are actually putting on; that is what sunscreen is--it absorbs ultraviolet light.1621

But, the standard acid, benzoic acid, has a Ka of 6.4x10-5.1626

And finally, we have something called hypochlorous acid, and it has a Ka of 3.5x10-8.1633

OK, well, which acid works best?1648

Which acid works best?...that is only part; and: What is the ratio that puts the pH at 4.80?1653

What is the ratio of concentration of base to acid that puts the pH at 4.80?1669

Now again, chances are very, very slim that you are going to find an acid that has exactly a pKa of the pH that you are looking for.1688

There is going to be some difference in the ratio to adjust, based on the Henderson-Hasselbalch equation.1695

So here, we want to know which acid works best; and once you have chosen the acid, what is the ratio of base to acid (the A-, the HA) that actually puts the pH at the 4.80 that we want?1701

OK, well, let's start with our Henderson-Hasselbalch equation: pKa + log(A-/HA).1717

I hope you will forgive me, but I am so tired of putting these parentheses on there; I'm just going to leave them like that--the parentheses and those brackets--they drive me crazy; we have done these enough to know that we are dealing with concentrations, so it shouldn't be a problem.1730

OK, we want the pH to be 4.80, and the pKa (oh, wait; we have to...what am I doing?--we haven't even chosen an acid yet!--we don't know what a pKa is!--OK).1741

So, let's go back; so, for acid A, which is monochloroacetic acid, we have a pKa...we just take the negative log of these values that we have right here (OK, let me circle them in red); we just take the negative log of these--that is the p.1754

That pKa is going to be 2.87; B: our pKa is going to be 4.89--that looks pretty good; C--benzoic acid: the pKa is going to be 4.19; and D: our pKa is equal to 7.46.1776

We want 4.8, is our target, right?--target pH; so it looks like we are going to go with propanoic acid.1798

That is the one we want to use: propanoic acid.1805

OK, so now that we have chosen HPr, now we can go ahead and find out what our ratio is.1809

So, let's move on: let's do pH is equal to pKa, plus the log of base over acid, Pr- over HPr concentrations.1817

You know what, I had better just "bite the bullet" and do it.1831

OK, we want 4.80 equals...the pKa is 4.89; plus the log of Pr- over HPr; go ahead and do the math here, and we end up with...well, you know what, let me just go ahead and do it.1836

-0.09=log(Pr- concentration/HPr concentration), and anytime you have a log, the inverse function of a logarithm function is the exponentiation with a base 10, so 10x.1854

So we raise both sides to the power of 10; that gets rid of the log function; we get a ratio of 10-0.09=0.813.1872

So, when we pick propanoic acid, we want to have a solution that has a pH of 4.8.1893

We pick propanoic acid because the pKa is 4.89; it is close to it--that means that the adjustment we have to make, to bring it down from the 4.89 to the 4.8--the ratio is going to be .813; that is nice--that is still pretty close to 1.1900

It's close enough to where we can actually get some good buffering out of it; that is the whole idea.1917

So now, we will continue this example; now we will do Example 3.1923

I'm just going to basically continue our discussion of this; this is a typical problem that you might see in a free response question that has multiple parts to it.1931

Example 3: Now, I want to prepare a 250-milliliter buffer solution like the previous problem.1942

In other words, I want to prepare the buffer solution at 4.8, using my propanoic acid.1969

OK, if I use 150 milliliters of a 0.20 Molar HPr (propanoic acid), what is the molarity of the sodium propanoate solution I must use to achieve the desired 4.80 pH?1975

OK, so we said we have a propanoic acid solution, and we know that the ratio has to be .813; that was our previous example.2025

Now, I want to actually prepare the buffer solution; I want to prepare a buffer solution, 250 milliliters of that buffer solution.2032

Well, I use 150 milliliters of a .20 propanoic acid; I know that I have to add another 100 milliliters of the sodium propanoate; but what is the molarity of the sodium propanoate solution that I must use in order to achieve the desired pH?2039

See...please, it is very, very important that you understand what this question is asking: I want the pH of 4.8; I have already calculated the fact that I have a ratio of .813; I am preparing 250 milliliters of that buffer solution.2060

I am using 150 milliliters of a .2 Molar HPr; well, 250 minus 150 leaves me 100 milliliters of solution of sodium propanoate.2075

However, what is the molarity of that sodium propanoate I need in order to get to a pH of 4.80?2083

This is actually just a stoichiometry problem, believe it or not.2090

Let's go ahead and do this: well, we know that our (actually, let me use the actual identities of species) ratio of propanoate to HPr is equal to 0.813.2093

OK, well, I have 0.150 liters (that is 150 milliliters), times 0.20 moles per liter of the propanoic acid.2112

That means there are .03 moles in that 150 milliliter: .030 moles of (wait a minute, I'm using HPr...yes, that is right) the HPr.2128

OK, now this 0.03 mol of HPr is in 250 milliliters, OK?--because again, I am adding another 100 liters of solution, so the Henderson-Hasselbalch equation talks about...I am talking about the molarity in the whole solution.2151

I know that I have used .030 moles, because I have used 150 milliliters of a .2 Molar; but now, it's swimming around in 250 milliliters.2179

0.03 moles, divided by 0.250 liters, gives me 0.12 Molar HPr.2187

Well, I know that my Pr- concentration, divided by my HPr concentration (which I just found is 0.12 Molar) is equal to 0.813.2203

Therefore, my propanoate concentration has to be 0.813 times 0.12, equals 0.09756 (I'm not worried about significant figures here) molarity.2217

I also know that I am going to have 100 milliliters of this, because I had 150 milliliters of the propanoic acid solution; I want a total volume of 250; so I need 100 milliliters.2237

Now, I can go, and I can actually make...I can take some solid sodium propanoate; I can drop it into solution, depending on...I weigh it out in terms of grams, because now I know what concentration I need.2249

I hope that this made sense.2263

I needed a buffer solution at a specific pH; I knew that, if I can find an acid that has a pKa close to that pH, and I knew that the ratio of the base to the acid is going to be .813, and based on the rest of the problem (given the volumes that I had and the molarity that I was dealing with), I was able to calculate the molarity of the sodium propanoate solution that I would need in order to achieve that 4.80 pH.2265

I would definitely recommend going through this problem a couple of times, just to make sure you understand what it is that we did.2297

This is sort of a whole bunch of problems tied together.2303

OK, well, this pretty much covers our discussion of buffer solutions.2307

Next time that we see each other, we are going to start our discussion of acid-base titrations and titration curves.2312

Thank you for joining us here at Educator.com.2319

We will see you next time; goodbye.2321