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 II

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 1:27
- Example 1: Question
- Example 1: ICE Chart
- Example 1: Major Species Upon Addition of OH⁻, But Before Rxn
- Example 1: Equilibrium, ICE Chart, and Final Calculation
- Summary
- Another Look at Buffering & the Henderson-Hasselbalch equation
- Example 2
- Example 3

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### Transcription: Buffers, Part II

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

*Today, we are going to continue our discussion of buffer solutions; it is a very, very important application of acid-base chemistry.*0004

*We are actually just going to start off with the second example that was going to be part of the last lesson, but I decided to hold it forward until this lesson to sort of split them up a little bit.*0011

*If you remember, last lesson, the exercise that we did--the last example--we had a buffer solution that we created at a given pH, and then we added some sodium hydroxide to that buffer solution, and we wanted to see what the change in pH is.*0021

*This next example is going to be something like that; however, we are not going to be concerned with what the pH of the buffer solution is before we add the sodium hydroxide.*0036

*It's just going to be: I have this buffer solution made of this and this; a certain amount of sodium hydroxide is added; what is the final pH?*0047

*So, there again, it is just a question of realizing what the question is asking before you actually jump into solving the problem.*0054

*Again, there is no algorithmic process, as far as higher-level science is concerned; you have to let each situation decide how you are going to go about it.*0063

*This is why it's very, very important to understand what is going on, to use your intuition to pull back from the problem and let what you know about what is happening physically decide what the math is going to look like.*0074

*Let's just start our example and see what we can do and gain a little bit better sense of this whole thing called a buffer solution.*0085

*This example is going to be Example 1; we will say: 0.020 moles of NaOH is added to 1.5 liters of a buffer solution (sol'n) made of 0.10 Molar propanoic acid and 0.10 Molar sodium propanoate.*0093

*The prop- prefix means three carbons, which actually doesn't really matter in this problem; that is the nice thing about these acid-base problems--you can sort of give it any symbol you want.*0150

*Ultimately, what you are concerned with is just that one hydrogen that comes off or jumps on; that is it.*0157

*Everything else doesn't matter; it's just some species--some conjugate base, if you will.*0163

*OK, so what is the final pH?--that is what we want to know.*0168

*What is the final pH of the solution?--and the K _{a} of (I'm going to abbreviate it HPr for) propanoic acid is 1.3x10^{-5}.*0172

*That is the K _{a}--again, K_{a}; this is an acid.*0192

*All right, so again, notice that, unlike the previous problem, we are not asking for the change in the pH; we are asking for just "What is the final pH of the solution?"*0195

*We have a buffer solution that we prepared; it's going to be at a given pH; we are going to add some sodium hydroxide to it, and then we are going to find out what the final pH is.*0205

*When you are adding acid or base to a buffer solution, this is where you are going to end up doing two problems: you are going to do the stoichiometry first, and then you are going to do the equilibrium part.*0214

*We will see what we mean in just a minute.*0224

*So, let's see: let's think about this for a second: in preparing the buffer solution, before we actually add the sodium hydroxide, here is what actually happened.*0226

*You ended up with...the problem that we would do, but we don't necessarily have to do, is: you are dealing with: HPr dissociates into H ^{+} + Pr^{-}.*0237

*OK, and here is where you have these two initial concentrations: let's say you have an Initial, a Change, and an Equilibrium.*0252

*The initial concentration of the propanoic acid, the HPr, is 0.10 Molar; there is no hydrogen ion; and this is 0.10 Molar also, because it is right there: .10 Molar sodium propanoate; the propanoate ion, because sodium propanoate is fully soluble, so it's going to be floating around freely.*0259

*This is going to dissociate a bit; this is going to show up; this is going to show up; and we are going to end up with equilibrium concentrations of 0.10+x.*0282

*Now, these are the equilibrium concentrations (and we can go ahead and write the K _{a} and do the whole bit).*0296

*However, because we know that, see, this is a buffer solution, and the buffer solution is nothing more than a common ion situation where you are taking a weak acid (in this case), and you are adding some salt of its conjugate base; well, it's a common ion effect--the propanoate is the common ion.*0302

*And we know that the common ion--because this shows up; because this isn't 0 to start off with--it is going to push the equilibrium that way.*0321

*In other words, it's going to suppress this HPr dissociation; so x is going to be, not only small under normal circumstances, but now that you have a common ion (Pr), it's going to be even smaller.*0329

*This x is going to be virtually nonexistent; so, for all practical purposes, we don't have to go through this part to find out what the initial concentrations of propanoic acid and propanoate are for the next part of the problem, the actual buffering that we do when we actually add the sodium hydroxide.*0342

*We can just take the initial concentrations of .1 Molar of the propanoic acid, and the .1 Molar of the sodium propanoate; again, this is all before we add the sodium hydroxide; these would be the concentrations that we start with in the next part.*0361

*But, we don't have to do that; these are pretty much approximately equal to 0.010, because x is so very tiny--there is very little dissociation: approximately equal to 010.*0380

*So again, we can skip this part when we are presented with a buffering problem like this.*0393

*Again, .1 Molar, compared to the amount that is going to dissociate, is really, really tiny.*0397

*Now, you wouldn't be able to do this if you were talking about, let's say, .001 Molar, because then you are talking about really dilute concentrations, and the amount that dissociates is going to be a fair percentage of the amount that you started off with, so you can't do this.*0403

*But, for the most part--for most buffering problems--in fact, almost all buffering problems, the amount of acid and conjugate base that you add is going to be very high, compared to what dissociates.*0420

*OK, so now, let's go ahead and start the problem as if we are adding the sodium hydroxide.*0434

*And now, our major species: so we always look at the major species in solution upon addition of the sodium hydroxide, but before any reaction takes place, because we need to decide what reaction is going to dominate the equilibrium here.*0443

*The major species upon addition (and I'm sorry I keep repeating myself, but I find that repetition is actually the best way to solidify something in the mind), but before reaction (I'm sorry, "upon addition of sodium hydroxide" or "hydroxide"--"upon addition of OH ^{-}, but before reaction"): we have propanoic acid floating around, .10 Molar; we have propanoate floating around, .10 Molar.*0458

*Well, we certainly have water floating around--that is the solvent that the thing is floating around in; we have sodium ion, because it's a strong base, fully dissociated; so the sodium ion is fully free; and, of course, we have hydroxide ion.*0496

*OK, hydroxide ion is a very, very strong base; now again, this is always going to happen like this; any time you add hydroxide ion, the first thing that is going to happen is: you have to account for the stoichiometry of this strong base reacting with any proton source, with any source of hydrogen ion, before the system comes to equilibrium.*0510

*The first thing that is going to happen is: this hydroxide is going to react with this acid in this particular reaction.*0528

*It's going to be OH ^{-} + HPr; this is the buffering part; it's going to form HOH, water, + Pr^{-}; that is what the buffering does.*0536

*This propanoic acid is one of the buffering components; this and these are buffering--when you add a base, this reacts with this in order to sequester it, in order to bind it.*0551

*It basically eliminates it from solution, so it's not floating around freely; so what is floating around freely in solution is Pr ^{-}, instead of OH^{-}.*0561

*That is the whole idea behind a buffer; it wants to swallow up any hydroxide that you add in.*0569

*OK, so let's do a Before, After, and a Change; so this is the stoichiometry part.*0574

*Again (let's do this...let's write this in red), this is the stoichiometry part; we do the stoichiometry before we do the equilibrium to find the pH.*0579

*Let me go back to blue; so, Before: the concentration of OH ^{-}...well, it is .020 Molar NaOH; strong base, fully dissociated; so, this is 0.020.*0594

*Well, I guess we are going to do red, after all; OK.*0607

*This is in moles; when we do stoichiometry, we do it in moles; when we do equilibrium, we do it in concentration (moles per liter).*0610

*Let me say that again: when you are doing the stoichiometry, you work in moles; when you are working in equilibrium, when you do the equilibrium part, you work in concentrations, which is moles per liter--which is why, notice, I have this 1.5 liter--I'm going to use that when I do the equilibrium part.*0617

*So, OK, the hydroxide reacts with this source of protons; it is completely used up--before anything happens, I have that much floating around; I have that much (actually, you know what, no; this is not true).*0634

*We said that we have .10 Molar of the HPr, and .10 Molar of the propanoate; but we have 1.5 liters of it--it's sitting around in 1.5 liters.*0648

*Therefore, 1.5 liters, times 0.10 moles per liter, gives me 0.15--0.15 moles of HPr.*0660

*Water doesn't matter; and the same thing with propanoate--it's sitting in 1.5 liters, and I have 0.10 moles per liter of it, right here.*0674

*OK, I work in moles--this is the stoichiometry part; so I put 0.15 moles of the propanoate, before anything happens.*0686

*Well, the change: all of the hydroxide reacts with this--it's a very strong base, so every bit of the hydroxide is going to pull a hydrogen ion off of this, producing that.*0694

*This is going to be -0.020, -0.020; because, for every one of these that pulls one of these off, one of these is used up; it's 1:1.*0707

*Water doesn't matter; this is going to be +0.020 (let me get rid of these random marks here).*0720

*So now, our After: after the stoichiometry has taken place, we have no hydroxide left--it has fully reacted; we have 0.15-0.02, for a total of 0.13 moles; and here we have the water, which doesn't matter; and here, we have 0.15+0.02, which is going to be 0.17 mol of propanoate.*0729

*Moles: this is how many moles of HPr is now in solution; this is how many moles of propanoate is in solution.*0760

*Now, the system is going to come to equilibrium.*0769

*So now, we do the Equilibrium part: now, the Equilibrium part is: What are the major species after everything has happened?*0773

*Major species: always do the major species, so you know what is going on--what has just happened.*0789

*Well, you still have HPr in solution; you still have Pr ^{-} in solution; it's sitting in water--that is your solvent; you still have the Na^{+} in solution; but now, notice: there is no OH^{-}.*0795

*It has been used up; we have accounted for it in the stoichiometry; it has reacted with that to produce new concentrations of HPr and Pr ^{-}.*0807

*Now, these two will dominate the equilibrium; and here is the equilibrium that will take place.*0815

*The normal weak-acid equilibrium: H ^{+} + Pr^{-} (this is the initial; this is the change; this is the equilibrium).*0820

*Now, notice: we had .13 moles and .17 moles of the propanoic acid and the sodium propanoate, respectively.*0831

*In equilibrium, we work in concentrations; so, the concentration of propanoic acid is 0.13 moles, divided by 1.5 liters; it's equal to .0867 moles per liter.*0840

*OK, we had moles; we did the stoichiometry; but it's floating around in 1.5 liters of solution.*0863

*So again, equilibrium part: we have to deal in concentrations.*0869

*We don't have to deal in concentrations when we do the ICE chart, but eventually, we have to divide by it, so it's better to just do it now, with the ICE chart; it is a good way to separate.*0877

*When you are doing stoichiometry, use moles; when you are doing equilibrium, use concentrations--it's a good way to separate the two so you don't lose your way.*0885

*OK, there is no hydrogen ion floating around just yet; here, we have 0.17 moles that we found from the stoichiometry, floating around now in 1.5 liters.*0892

*This is 0.1133, if I'm not mistaken; yes, there we go; OK.*0902

*So now, this is going to dissociate a little; this is going to show up; this is going to show up; and here we have 0.0867-x; this is going to be x; this is going to be 0.1133+x; and then, of course, we have...now, we do our K _{a}.*0909

*K _{a} is equal to the hydrogen ion concentration, times the propanoate concentration, divided by the concentration of propanoic acid.*0934

*We have: 1.3x10 ^{-5} is equal to x times 0.1133+x, all divided by 0.0867-x; well, again, x is going to be pretty small compared with these numbers; so we are probably OK if we go ahead and drop the x from the numbers in parentheses.*0943

*We have 0.1133 on top, and we have 0.0867 on the bottom; when we solve for x (oops, no; we definitely don't want those stray lines; OK), which is the hydrogen ion concentration (right?--that is our x, right there), we get x is equal to 9.9x10 ^{-6}; that is in molarity, moles per liter.*0973

*When we take the negative log of that, we get a pH of 5.0.*1006

*So, there you have it: under these circumstances, in this buffer solution, once we add our sodium hydroxide, and allow the hydroxide to be completely eaten up by the propanoic acid, once the amounts of propanoic acid and propanoate adjust and come to equilibrium again to reach the equilibrium constant of 1.3x10 ^{-5}, based on this ICE chart, our final pH is going to be 5.0.*1011

*That is it; so, a quick review: one thing that you are always going to be doing with these buffer solutions--again, don't think of what I'm about to write down as "let me do this, let me do this, let me do this!"--you don't want to do it mindlessly.*1040

*You want to take a look at the situation and let the situation tell you what to do.*1059

*But, in general, it is always good to follow some standard practices, in order to be consistent.*1063

*And, the first thing you want to do is: you want to check your major species upon addition of acid or base.*1070

*And then, you want to do your stoichiometry problem; in other words, you want to show the hydroxide that you have added, if you have added base reacting with the propanoic acid.*1089

*Or, if you have added acid, you want to show the hydrogen ion reacting with the propanoate.*1098

*You want to do that stoichiometry first; and again, this is in moles--you work in moles with stoichiometry--it's the whole idea; it's what stoichiometry is all about.*1104

*3: (1, 2, 4--that's nice; I can count!) And then, you want to do the equilibrium problem.*1112

*This is going to be in molarity--very, very important.*1124

*That is it: hopefully, these two problems have given you a sense of what is going on with buffers; but again, if you are still a little hazy about it, don't worry about it, because we have some more discussion to do about buffers.*1130

*Now, let's take another look (let me go back to blue ink here) at buffering.*1141

*All right, so a buffer solution--before we actually add any acid or base--we said, again, is based on the following equilibrium.*1159

*It's based on the acid in equilibrium with H ^{+}, plus A^{-}.*1171

*And again, we have some initial concentration; a change; and an equilibrium concentration; so let's just say that our initial concentration of the HA is a.*1179

*And let's say that the initial concentration of the A ^{-} is b.*1192

*Again, right?--weak acid, plus some salt of its conjugate base; so there is some of each, and this is 0.*1195

*Well, there is going to be a change this way; this is going to be -x; this is going to be +x; this is going to be +x.*1201

*We are going to a-x here; x here; and we are going to have b+x here.*1207

*This is always going to happen like this.*1211

*Again, because the common ion effect is so small, for the most part, this is going to be equal to a, and this is going to be equal to b, because this x is going to be really tiny.*1215

*So, for all practical purposes, we don't need to go through this; we can just take the initial concentrations as is.*1227

*We don't have to do this first part to find what these concentrations are for the stoichiometry part of the problem.*1233

*Well, check this out; this is actually kind of interesting; this is sort of a different way of finding the pH of a buffer solution and/or a buffer solution to which something is added, and let's just sort of run through it.*1240

*You know that the K _{a} is equal to the hydrogen ion concentration, times the conjugate base concentration, over the acid concentration; now, let me rearrange this equation.*1256

*When I do rearrange it (so let me just write "rearrange")--and let me solve for the hydrogen ion concentration, because that is really what we are finding most of the time--we are concerned with what the pH is.*1269

*If I multiply this and divide by that, I get the following: I get that the hydrogen ion concentration is equal to the K _{a}, times the concentration of the acid itself, divided by the concentration of the conjugate base.*1282

*Notice: this is a constant; this is some number that is dependent on a constant.*1300

*This right here--the concentration of acid, divided by the concentration of conjugate base--this is what controls the pH: it's the ratio of the amount of acid to the amount of conjugate base.*1305

*This is just a number; it's some number divided by another number; it is a number in and of itself, a number times a constant.*1321

*There is a direct relationship between this and this; so the pH depends on the relative concentrations of the acid and its conjugate base.*1327

*That is what is going on here.*1338

*So now, let's do a little bit of math to this.*1340

*Let's go ahead and take the logarithm of both sides, the negative logarithm.*1343

*Well, the negative logarithm of the left-hand side is the pH; if I take the negative logarithm of the right-hand side and expand it according to the properties of logarithms (the log of something times something is the log of something plus the log of something, and the log of something divided by something is the log of the top minus the log of the bottom--remember those properties from precalculus and calculus?), I end up with the following equation.*1347

*I end up with: the pK _{a}, plus the log of A^{-}, divided by...this is called the Henderson-Hasselbalch equation, and it is sort of a shortened version of the problems that we have been doing.*1372

*We'll use it in a minute; so let me just write, actually, what it is.*1396

*Henderson-Hasselbalch equation; it's not a new, different thing that we have been doing; all that we have done is take what we normally do, when we set up this equality, and we solve for x, which is H ^{+}.*1400

*All I have done is actually rearrange it, and then take the logarithm of both sides, and then I have this thing that says that the pH of a buffer solution is equal to the pK _{a} of the acid used in that buffer, plus the logarithm of the conjugate base concentration, divided by the actual acid concentration.*1415

*Here is what is important to remember: base on top; acid on the bottom; base on top; acid on the bottom, in the Henderson-Hasselbalch equation.*1438

*Now, I have to be perfectly honest with you: oh, before I actually discuss the equation itself, here we have assumed...notice the assumption that we have made: we have assumed that this concentration here and this concentration here--those are these concentrations, that they are essentially the same as the concentrations that we started the problem with.*1445

*When we start the problem, these are the concentrations that we start with; this comes to equilibrium, and, of course, some of the acid dissociates; some of the conjugate base shows up; so, these concentrations are a-x and b+x.*1468

*But, because x is so small because of the common ion effect, this equilibrium concentration of a and b is essentially the same as the initial concentration of a and b--because x is so small, and because a and b will usually be quite large.*1485

*So, for that reason, that is why this equation is valid; this is the assumption that we have made.*1499

*But again, notice: it is no different than anything that we have done--we have just rearranged things, and we are taking pHs and pK _{a}s instead of actually doing the math.*1505

*Now, I personally, personally--I don't like the Henderson-Hasselbalch equation; I don't like it for many, many reasons.*1514

*I think that, if you understand the chemistry, that is fine; you are welcome to use the equation; but make sure you understand the chemistry and what is going on.*1522

*If you don't, this is just one more equation that is going to end up getting in the way and giving you grief.*1532

*If you go through the process that we have been going through, making the ICE chart, making the before and after chart, you are always going to get the right answer.*1537

*Will it take a little bit longer?--sure, but so what--it's not going to take 20 minutes longer; it's just going to take an extra couple of minutes.*1545

*The best part about doing it that way is that you see everything that is happening.*1551

*I'm a big fan of seeing everything that is going on; I don't like shortcuts; I don't like symbols that look pretty, because things that look pretty tend to hide things, and in math, it's no different than science.*1557

*Just because something looks elegant, that means something is hidden; so this is a very nice equation and expresses a nice relationship between pH, pK _{a}, and concentration of base, and concentration of acid.*1568

*However, you need to understand where it comes from; it comes from the problems that we have been doing the entire time--it comes from this right here.*1581

*It comes from the equilibrium constant expression; there is nothing new.*1588

*So, we are going to go ahead and use the Henderson-Hasselbalch equation in some problems that we do, because it is something that does come up, especially for those of you in the biological sciences.*1592

*They tend to use this equation more than they use the actual process.*1602

*It is important to know, from a practical standpoint; but really, please make sure you understand the chemistry.*1606

*That is what we want you to know, because the rest will happen naturally.*1613

*And, if you don't remember the equation, you don't have to; you know the chemistry--the chemistry will guide you.*1617

*OK, enough of my little lecturing here; let me go ahead and just start an example.*1622

*Let's see, let's do...this is going to be Example 2, if I'm not mistaken: OK, calculate the pH of a buffer made from 0.70 Molar lactic acid, and 0.30 Molar sodium lactate.*1629

*The K _{a} for lactic acid, which I will abbreviate as HLac, is equal to 1.4x10^{-4} (I know my notation can be a bit sloppy; sorry about that).*1672

*OK, lactic acid, by the way, happens to be the stuff that builds up in your muscles when your muscles start burning sugar, not in the presence of oxygen.*1689

*Anaerobic respiration, basically, is what is happening; when you start to do some rapid exercise very, very quickly, it takes a couple of minutes for your body to sort of get the oxygen needed to the cells to actually get it to burn the sugar aerobically (meaning in the presence of oxygen), but during these few minutes, those 2 minutes to 5 minutes, during that time, if there is a lot of rigorous exercise, your body will actually do it anaerobically.*1705

*One of the byproducts of anaerobic respiration, or anaerobic oxidation, is lactic acid; so when your muscles start to get sore--you know that sore feeling, and all of a sudden you can't move anymore--that is lactic acid building up in your muscles.*1733

*So, just to know where it comes from--it is an important acid, but again, it's just an acid; so you can use it to form a buffer solution.*1748

*Now, major species: we always want to do our major species, because we want to know the chemistry.*1757

*Major species: we have the lactic acid (the HLac); we have the sodium lactate; we have the Lac; so we also have the sodium; and, of course, we have the water.*1764

*Well, 1.4x10 ^{-4}; water is 1.0x10^{-14}; so yes, this is going to dominate the equilibrium.*1774

*Let's just use the Henderson-Hasselbalch equation; OK.*1785

*We said that the HLac concentration at equilibrium is approximately equal to the HLac initial concentration.*1790

*That was the whole idea behind the equation; it equals 0.70 Molar--that is right here.*1800

*And the Lac, the lactate, equilibrium concentration is approximately equal to the lactate concentration initially, because again, there is going to be so little dissociation, so little lactate extra formed, it's going to be virtually the same as you started.*1809

*There we go; so we take those numbers; we write our Henderson-Hasselbalch equation; it's going to be pK _{a}, plus the log of the base over the acid.*1826

*Lac ^{-} over HLac; these are our concentrations; we end up with a pK_{a}--so we take negative log of 1.4x10^{-4}, plus the log of 0.30, divided by 0.70 (right?--the base over the acid), and we end up with 3.85 plus a negative 0.37, which gives us a final pH of 3.48.*1836

*So, I know what you are thinking: "Wow, that was really, really quick; why did we have to go through all that headache before?--why couldn't we just have introduced the Henderson-Hasselbalch equation and used this always?"*1877

*Well, a couple of reasons: 1) We need to understand the chemistry in what is happening, which is why those other problems seemed a lot slower than they were.*1887

*Now that we have sort of gotten a sense of what is going on, now that we can use the Henderson-Hasselbalch, as long as we understand where it is coming from, you are welcome to use it.*1895

*I personally--to this day, I don't use this.*1905

*I actually go through the rigorous process; that is just a personal preference of mine.*1907

*But, you are welcome to use it; there it is; it's quick; it works beautifully; so there you go.*1913

*OK, let's do another one, just to get a little bit more practice with it.*1921

*Example 3: OK, a buffer solution contains (oops--yes, a buffer; so when we say "buffer," we mean a buffer solution)--a buffer contains 0.30 Molar ammonia (NH _{3}; wow) and 0.45 Molar NH_{4}Cl (ammonium chloride).*1930

*What is the pH?--and the K _{b} for ammonia is 1.8x10^{-5}.*1960

*Now, notice: in this case, we have a buffer solution that is not made with a weak acid and the salt of its conjugate base; it is made with a weak base, ammonia, and the salt of its conjugate acid, ammonium chloride--the conjugate acid being ammonium, NH _{4}^{+}.*1976

*So now, it's slightly reversed; and notice, this is K _{b}, not K_{a}; when we use the Henderson-Hasselbalch equation, here is another reason why I don't like using it, because in this situation, you can't use it directly.*1994

*You actually have to calculate the K _{a} and then use it.*2006

*The Henderson-Hasselbalch equation is pH=pK _{a}, plus the log of the base over the acid.*2010

*It's not a problem; it's just an extra step; but the idea is, the way we did it initially, with the ICE chart and everything--you can always just do it.*2017

*That never changes; that is the nice thing about the sort of--what I call the more rigorous method.*2024

*But, that is not a problem; we can go ahead; and so, we will use the Henderson-Hasselbalch equation; but again, it has to be K _{a}; so, we know that there is a relationship between K_{a}--that is (K_{a})(K_{b})=K_{w}, which means that K_{a} is equal to K_{w} over K_{b}.*2030

*In this case, it's going to be 1.0x10 ^{-14} (that is the equilibrium dissociation constant for water), divided by the K_{b} here, which is 1.8x10^{-5}; and we end up with a number, which is 5.6x10^{-10}.*2051

*OK, now, we write down the Henderson-Hasselbalch equation: pH=pK _{a} plus the log of base over acid.*2071

*In this case, the base is NH _{3}; that is your weak base.*2084

*Its conjugate acid is ammonium; so remember: base over acid (let me just write that here; this is the base on top; this is the acid on the bottom).*2088

*That is the Henderson-Hasselbalch equation.*2101

*OK, so that means it equals--the pK _{a} is--the negative log of the K_{a} which we just found, 5.6x10^{-10}, plus the logarithm of the ammonia concentration, which is 0.30 Molar, over the acid concentration (NH_{4} is the conjugate acid); this is fully soluble, which means that .45 moles per liter produces .45 moles per liter of ammonium; equals 9.25 plus a negative 0.18, for a final answer of 9.07.*2103

*There you go; that is it; using the Henderson-Hasselbalch equation, you ended up...the only thing you have to be careful of is, of course, you have to work with K _{a}, because the Henderson-Hasselbalch equation requires that you work with K_{a}.*2151

*So, if you are dealing with a buffer solution that has a weak base and the salt of its conjugate acid, you are going to be given K _{b} for that weak base; you just need to calculate the K_{a} and put it in there, and you will get your final answer.*2169

*Now, of course, you could have done this with an ICE chart and K _{b}; in fact, I'm going to do it really quickly, just so you see how you actually could have done this.*2183

*Let me...so could have done the following: we could have said: OK, what are our major species?*2191

*Well, our major species are ammonia, water, ammonium, and chloride; well, what is going to dominate the equilibrium?*2200

*What is going to dominate the equilibrium is that, because again, the K _{b} for H_{2}O, which--well, for H_{2}O, we are talking about 1.0x10^{-14}; this is 1.8x10^{-5}; it's so much bigger than that, so these are going to dominate.*2217

*Here is what happens: you get: NH _{3} + HOH; it is going to be in equilibrium with NH_{4}^{+} + OH^{-}.*2234

*We are going to end up with a basic solution, because we are creating hydroxide.*2247

*Well, our Initial, our Change, and our Equilibrium: our initial ammonia concentration was 0.30; water doesn't matter; 0 for hydroxide--nothing is there; our initial ammonium concentration was 0.45 Molar, right?--that is exactly it.*2251

*And now, again, the change: this is going to diminish; water doesn't matter; this is going to show up; this is going to show up; we are going to have 0.30-x; this doesn't matter; this is going to be 0.45+x; and this is +x.*2271

*Well, we have: K _{b} is equal to the OH^{-} concentration, times the NH_{4}^{+} concentration, over the NH_{3} concentration.*2289

*You have: 1.8x10 ^{-5} is equal to x, times 0.45+x, over 0.30-x; and again, it's actually going to be so small, because of this common ion effect, that you can pretty much ignore it.*2304

*It's going to be x, times 0.45, divided by 0.3; and here, we get x equal to an OH (now, this is where you have to be careful; we are looking for the H ^{+}; we have the OH^{-}--we have to do that first)--equals 1.2x10^{-5}, which gives me a pOH of 4.92, which gives me a pH is equal to 14.00-4.92 (let me make sure that minus sign is legible), and we get a pH of 9.08, which matches our pH that we got doing the Henderson-Hasselbalch equation.*2324

*For that one, we got 9.07; but again, 9.07, 9.08--you ended up in the same place.*2371

*This is sort of the rigorous procedure; you have the Henderson-Hasselbalch equation, which is just a rearrangement of this for acids--because Henderson-Hasselbalch requires K _{a}.*2376

*pH=pK _{a} + log(base/acid).*2388

*That is it.*2395

*Thank you for joining us here at Educator.com for AP Chemistry.*2397

*We have one more lesson of buffering solutions to discuss.*2401

*We'll see you next time; take care--goodbye.*2404

1 answer

Last reply by: Professor Hovasapian

Thu Jan 25, 2018 4:03 AM

Post by peter alabi on January 24 at 12:01:58 AM

Hi, Dr. Hovasapian.

Why does buffering capacity change with temperature? Is it because Ka is like any other equilibrium constant, and thus change with temperature. If so can I use the equation (lnk2/ln1)= R/H((1/T1)-(1/T2)) to find the changes in ka as a function of temp?

Thank a lot for such great lectures.

sincerely, Peter A.

1 answer

Last reply by: Professor Hovasapian

Fri Feb 26, 2016 11:10 PM

Post by Nadan Cha on February 26, 2016

Hello Professor,

For the Henderson Hasselbalch equation, which is pH = pKa + log([A-]/[HA]), are [A-] and [HA] equilibrium concentrations? Or are they initial concentrations?

Also, can we not use the Henderson Hasselbalch equation for problems that have acid/base added to the buffer solution? Is it just for finding out the pH of a buffer solution with no acid/base added (ex: 0.020 mol NaOH added to 1.5L buffer solution)?

Thank you so much for offering wonderful lessons =)

3 answers

Last reply by: Professor Hovasapian

Tue Apr 7, 2015 10:56 PM

Post by Lyngage Tan on April 1, 2015

on the last problem is true that there is also a reaction NH4 -> NH3 + H+ happening at equilibrium. i used this reaction and solved for Ka then plugging in the equilibrium expression i got [H+]= 8.4 x 10^-10 and a pH of 9.075.

3 answers

Last reply by: bob singh

Tue Mar 24, 2015 9:13 PM

Post by bob singh on March 23, 2015

Around 37:00, why don't we use a before,change, and after chart? Why are we using an ice chart?

3 answers

Last reply by: Professor Hovasapian

Mon Feb 16, 2015 3:06 PM

Post by Hemant Srivastava on February 8, 2015

On the first example,at first equilibrium of HPr <=> H+ + Pr-,

you had said we could ignore the x because it was very small, but I didn't ignore it, and on my calculations I got the following results:

[HPr] at equilibrium (before NaOH added) is 0.10 M and [Pr-] at equilibrium (before any NaOH added) is 0.10 M, but there is a little bit of hydrogen (since x = 1.3 x 10^-5).

In the stoichiometry, there is no H+, so that doesn't matter, but after the NaOH is added, on the last equilibrium reaction of HPr, there is initially H+ in the beaker (from dissociation of HPr before the NaOH was added), and you can't write 0 M of H+ as initial amount of hydrogen ion for the last equilibrium since 1.3 x 10^-5 M H+ was present before any reaction happened. You had chosen to ignore that, but when find x, the x value ([H+]) is smaller than the initial amount of hydrogen. X is on the scale of 10^-6, but the original H+ concentration was on the scale of 10^-5! And so that small amount that you assumed negligible at beginning shouldn't be negligible because it changes your pH to 4.8 vs. the pH you got which was 5.0. Is that right?

1 answer

Last reply by: morgan franke

Thu May 2, 2013 9:31 PM

Post by morgan franke on May 2, 2013

why did you use -log before your ka constant in example three?