For more information, please see full course syllabus of AP Chemistry

For more information, please see full course syllabus of AP Chemistry

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### 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

### AP Chemistry Online Prep Course

### 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^{-} + H_{2}O.*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 pK _{a} + the log of A^{-} over HA, the concentration of the base part over the acid.*0388

*And now, we want to say the K _{a} 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 pK _{a} (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 pK _{a}, which is 4.74; interesting.*0453

*How about solution B?*0459

*Well, pH equals pK _{a} (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 H _{2}O 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=pK _{a}, 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=pK _{a}, 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=pK _{a} 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 pK _{a} 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 pK _{a}, 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=pK _{a}.*1472

*Therefore, if you can find an acid that has a pK _{a} whose pH is close to that pK_{a}, 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 pK _{a} 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 pK _{a} 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 K _{a} of 1.35x10^{-3}.*1561

*Our second choice is propanoic acid, and our K _{a} 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 K _{a} of 6.4x10^{-5}.*1626

*And finally, we have something called hypochlorous acid, and it has a K _{a} 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 pK _{a} 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: pK _{a} + 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 pK _{a} (oh, wait; we have to...what am I doing?--we haven't even chosen an acid yet!--we don't know what a pK_{a} is!--OK).*1741

*So, let's go back; so, for acid A, which is monochloroacetic acid, we have a pK _{a}...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 pK _{a} is going to be 2.87; B: our pK_{a} is going to be 4.89--that looks pretty good; C--benzoic acid: the pK_{a} is going to be 4.19; and D: our pK_{a} 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 pK _{a}, 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 pK _{a} 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 10^{x}.*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 pK _{a} 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 pK _{a} 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

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? :))