Lecture 09. Acids and Bases. Pt. 6.
Arasasingham, Ramesh D.
 UCI Chem 1C is the third and final quarter of General Chemistry series and covers the following topics: equilibria, aqueous acid-base equilibria, solubility equilibria, oxidation reduction reactions, electrochemistry; kinetics; special topics. 
 Can I have everyone's attention please? Are there any questions before I begin? Great. So I'm just going to pick up where we stopped last time. And if you guys remember, last time we talked about the common ion effect. And we said that if you take a weak acid, for example, and if that weak acid is at equilibrium, and now if you disturb that equilibrium by adding a common ion that's common to that equilibrium, then that equilibrium would be disturbed. And if the equilibrium is disturbed, then we can predict how the system is going to react based on Le Chatelier's principle. Because we know what Le Chatelier's principle tells us is that if you disturb or stress the system at equilibrium, it's going to respond in a way in which it can counteract that stress. All right? Now another way of solving common ion effect problems is to have both of them to start with. So last time we took this problem where we had a weak acid. And so the weak acid here was HF, so we looked at HF equals in water. And because this is a weak acid, it's going to establish this equilibrium where KA is 7.2 times 10 to the negative 4. So rather than have a system where you have equilibrium being established and then you add a common ion, you can start with mixtures of both. And so this is an example of a problem where you have to begin with, you have one molar, so this is initial, you have one molar of this, and you have one molar of that. Do you see that? So the way you solve this problem is the same thing. You can have a system that's established and add a common ion, or you can start by having the weak acid and the conjugate base there. So now what you have is a mixture of both. In the past, we've looked at examples where you know how to calculate the pH of just the weak acid alone. If it was just the weak acid alone, you don't have any of the conjugate base to begin with. All right? In this example, what you have is you have both the weak acid and the common ion, which is the conjugate base, you've added a salt to that. And based on that, you can figure out what the pH of that solution is. So these are all slight distinctions of the approach that we take in solving different problems involving acids and bases and figuring out what the pH of the solution is. You can see that now this is a slight variation. Now today we're going to move on to looking at a phenomenon that is very similar to this common ion effect, and we call this buffer solutions or buffers. So today we're going to move on to looking at buffer solutions or buffers. And many of you have heard the term buffer solutions or buffers before. So we're going to look at buffered solutions or we call these buffers. And if you want to describe what that is, a buffered solution or buffer is any solution that maintains an approximately constant pH despite small additions of acid or base. So a buffered solution is any solution that maintains its pH despite small additions of acids or bases. Now we know typically if you take water and if you add an acid to water, what does that solution become? If you take pure water, which is neutral, and if you add an acid to pure water, what does that solution become? Acidic. And so what do we know about the pH? pH is going to go from 7 to a smaller value. If I take pure water, which is neutral, and add a strong base to it, what do we know about that solution? It is going to be basic. And so what does that tell me about pH? pH is going to go up. So that's if you just have pure water. The buffers are special solutions where if you add an acid, the pH does not change dramatically. There is just a slight change in pH. So it maintains the pH. Or if you take a buffer, which is a special solution, and add a strong base like hydroxide to it, usually if you take pure water and add a strong base, its pH is going to go up dramatically. Buffers are special solutions where if you add a strong base, or if you add a strong acid, its pH remains relatively unchanged. It will be altered slightly, but it maintains that pH. So it turns out these special solutions have some very distinct characteristics. So we need to define what that is. So we can say a buffer contains one, a weak acid, and a weak base that are conjugate to one another. So we said these are special solutions, and the characteristics of this special solution is that a buffer contains both a weak acid and its conjugate base together. So in other words, if we're mixing a weak acid and a weak base, what that means is you have a weak acid and the salt of its conjugate base. So if we want to mix a weak acid and its conjugate base, the conjugate base, if it's negatively charged, has to be a salt. So if I take HCN, if I take HCN, which is a weak acid, HCN, what is its conjugate base? HCN. HCN, conjugate base. CN minus. Got it? So if it's CN minus, you know that a charged piece is not going to be alone. So it needs a counter ion. So we add a group one metal to it. So it could be sodium cyanide. It could be potassium cyanide, and we call that the salt of its conjugate base. So a buffer is a weak acid. So pick HCN, HF, acidic acid, that's a weak acid, and its conjugate base. So if it's HCN, CN minus would be its conjugate base. If it's HF, what would its conjugate base be? F minus. If it's acidic acid, what is its conjugate base? Acetate. All right? So all of those are negatively charged ions, and therefore you have to have a salt. So it'll be a group one metal like sodium potassium. So a weak acid and the salt of its conjugate base, or a weak base like ammonia, and the salt of its conjugate acid. All right? So you can have two types. You can have a weak acid like HCN, HF, acidic acid, and the salt of its conjugate base. Or you can have a weak base that's neutral like ammonia. And if it's ammonia, what is the conjugate acid of ammonia in H4 plus? All right? So the salt of that would be in H4Cl. So the way to identify a buffer is to look for these characteristics. So if I have, for example, HF and NAF, all right? Where this is F minus, which is the conjugate base of HF. You can have HCN and over here NA-CN. If you have acidic acid, then over here you'll have CH3CO2NA. Each of these, this is a salt. And what do we know about salt? The moment you put a salt in water, what does a salt do? Dissociates. So it's going to release the anion, which is the conjugate base. So this would have CN minus. This would release CH3CO2 minus. And then if you take something like formic acid, which is that, then it would be the salt of formic acid. And therefore this would release that. Okay? Or if we're dealing with a weak base, then what we do is the acid form would be NH4Cl. All right? You would have NH3. So once again, this is an acid base, which are conjugate to one another. And this is a salt, so it will give you NH4+. And so what you end up with is NH4+, and NH3. These are a combination of weak acid and a weak base. All right? I want to look at that table that I gave you. If you take something like CH3NH3Cl, this is also a salt. So essentially what we're doing is we're replacing one of the hydrogens here with CH3. You see that? And this would be a combination of CH3NH2, all right? Where this would release the cation CH3NH3+, and so that would be. That would be the acid form, because this is the cation, and the chloride is there as the anion. So this would release the acid form, and that would be its conjugate base. All right? So don't get thrown off by the formulas, because you can kind of sit down and look at it and say if it's CH3NH3Cl, you know that that's a salt, and therefore it's got a cation, and the cation component would be that. And if you look at it, it's very similar to NH4+. The only difference is that one of those hydrogens have been replaced by CH3, all right? And therefore that would be the acid form, and that would be the basic form. So we said that that's the first criterion. If you want to have a buffer, you have to have both of those components, all right? And they have to be conjugate to one another. The second property that a buffer has to have is to the, or both components have to be in approximately equal amounts. In other words, their concentrations should be similar. For the best buffers, you want to make sure that they're both in approximately equal amounts. Okay? So if I'm taking a weak acid and my concentration is one molar, the best buffer would be my conjugate base is also one molar. So you have equal concentrations of both. You can, they can be slightly different from each other, but the ideal would be if their concentrations are equal to each other. Okay? The third is that both should be in substantial amounts, should be in substantial amounts. In other words, what that means is higher concentrations. All right? So when I say higher concentrations, one molar is considered to be quite a high concentration. All right? So if they're the order of one molar, two molar, then the more concentrated the both of them are, the better the buffer. Okay? So whenever you are asked to identify a buffer, you've got to have to look for this. You have to have a mixture of a weak acid and its conjugate base so that they're conjugate to one another. You have to have approximately equal concentrations and they should be in substantial amounts. All right? So these are the criteria for a good buffer, but now let's see how does a buffer work? So let's start looking at how does a buffer work? Now remember, if it's a good buffer, if you add acid, the pH should not change all that much. All right? So if we want to look at a buffer, we said a buffer is anything that has a weak acid and its conjugate base. So let's like the equilibrium. Remember, we're looking at acid-basic-coulibre, and the underlying theme for all of this is the equilibrium that we establish. So if I have a weak acid and its conjugate base, then let's start with a generic weak acid. I have HA, and if I place HA in water, I would establish the equilibrium where I have H3O plus aqueous plus A minus aqueous. And the form in which this equilibrium is written is in the form of an acid dissociation. So this equilibrium is described by KA. All right? So if I start with initial concentration, let's say I have one molar of this and one molar of that. And to begin with, these are the initial concentrations. I have initially to begin with one molar of the weak acid and one molar of the weak base. So they're approximately equal amounts. They're substantial amounts, and I have both of them there. And we know that when equilibrium is established, at equilibrium, now I can jump a step. This will be minus x plus x and so on. So at equilibrium, we know this would be one minus x. This would be x. This will be one plus x. And therefore, I can approximate this to one. This will be x. If KA for weak acid is always smaller, we can approximate and say x is small, and therefore that would be one. All right? So at equilibrium, we can say that this equals the initial concentration of HA, and this equals the initial concentration of A minus. Because the amount, the x is so small, because the weak acid KA is going to be really small, because x is going to be small, we're going to approximate it. So we're going to say the initial concentration hardly changes. Can you see that? And what is a sort of symbolic notation that we use to describe initial concentrations? We always put a subscript zero. And what does that subscript zero mean? It's at time zero. All right? And that's what put the zero at the bottom. Time zero means that no reaction has taken place. This is what you start with at the beginning, all right, before any reaction takes place. And so what we say is that because we're dealing with weak acids on their conjugate basis, KA is going to be really small, therefore we can approximate and say that the initial concentration doesn't change very much. All right? So now we're going to write the equilibrium expression, and we know KA equals the definition, which is the law of mass action, which is H3O minus A minus over HA. And we can say that this approximates to KA equals, this approximates to KA equals H3O plus the initial concentration of A minus over the initial concentration of HA, because we know that for weak acids, the initial concentration doesn't change by much, and we can always approximate that and say it's equal to the initial concentration. Okay? Now, remember we're looking at the effects of pH. All right? We want to look at how is the pH affected, and what does pH depend on? What is the definition of pH? Depends on hydronium ion concentration. So if we want to look at the effect of pH, we want to focus on hydronium ion, and so let's rearrange this equation so that we look at the hydronium ion concentration. So what we have is hydronium ion concentration, and if I rearrange this equation, you can say that equals KA. So if the pH is not altered, that means the hydronium ion concentration is not altered. All right? So if we add an acid and we say the pH does not change, what does that really mean? It means that the hydronium ion concentration is not altered by much. So we know that hydronium ion concentration depends on KA, which is a constant, and then the ratio of the acid to the base. All right? So if the ratio of these two is not altered too much, H3O plus will remain unchanged. Is that clear? So if we want to look at a buffer, and we have substantial amounts of both, and if the pH is not going to change, pH depends on a hydronium ion concentration, and therefore if that ratio is not altered too much, we know that the hydronium ion concentration will not change that much. And so that's how buffers work. It turns out that in buffers, because you have substantial amounts of this and this, when you add an acid or a base, this ratio is not altered all that much. All right? And as long as the ratio is not altered, then the hydronium ion concentration does not change, and therefore the pH remains unchanged. All right? So let's look at what happens if you add an acid or a base. So now let's take this scenario, and now let's say that if... So let's look at the first case. Let's say if H3O plus is added. All right? In other words, let's say something like 0.1 molar HCl is added. Okay? So all of you recognize that HCl is a strong acid, and we know that a strong acid dissociates completely. So if I put 0.1 molar HCl, what is the concentration of the hydronium ion that I'm adding? 0.1 molar. All right? So all I'm doing is I'm adding 0.1 molar hydronium ion. Now if I take 0.1 molar hydronium ion in pure water, what is the pH of 0.1 molar hydronium ion? 0.1 is what? 10 to the what? Negative 1. Negative 1. If it's 10 to the negative 1, what is that? 1. So that means 0.1 molar, if you take neutral water and add 0.1 molar to neutral water, it goes from pH 7 to? To pH 1. pH 1, that's a big change. Remember, it's a factor. Every change in pH unit is a change by a factor of 10, or by one order of magnitude. So now it changes by seven orders of magnitude. So that's a pretty strong acid. Now what happens is if I take this and add it in a buffer, now the pH of the buffer does not change. If you take pure water, it goes from seven to one. But if you put it in a buffer, the pH is not altered all that much. And the reason is that you have this equilibrium that we've established. Now, so we have this equilibrium, H2O plus A minus, and this equilibrium is established. But now I'm adding hydronium ion. So what am I doing to that? If I look at that equilibrium, I'm adding hydronium ion to that solution. So I'm adding a large amount of hydronium ion in there. So what happens to that equilibrium? Is there a common ion there? Is hydronium ion there already in the equilibrium? So now if I'm adding more hydronium ion, what happens to that equilibrium? I've disturbed it. So now Leicester's principles are going to jump in, and it's going to counteract that stress. So what happens is that this equilibrium, so to counteract the addition, so to counteract the added hydronium ion, what happens is that the hydronium ion plus A minus will react to give you HA plus H2O. So aqueous, aqueous, this will be an aqueous, and that'll be aqueous. So essentially what happens is that you've added too much hydronium ion, the system is going to respond by this reacting with this to give you. In this way, what it does is it absorbs the extra hydronium ion that you added, and the system will go back to equilibrium. When the system goes back to equilibrium, what happens is that this ratio, once it goes back to equilibrium, this ratio is altered only slightly. So the way this buffer acts is that it absorbs the added acid and reacts with it and forms the weak acid, and when the system goes back to equilibrium, because KA is a constant, do you see that? The ratio of products to reactants have to be maintained, because that's how KA is a constant. And so what happens is when it goes back to equilibrium, this ratio is only altered very slightly, and therefore because the ratio is altered only slightly, the pH remains relatively unchanged. As a result of this reaction, the ratio HA over A minus is relatively unchanged. And so if that ratio is unchanged, then the hydronium ion concentration is not changed all that much. Now the second scenario is that we added, if we added base. So now we understand how a buffer works, and we understand that when you add an added acid because it disturbs that equilibrium and the extra hydronium ion is absorbed and consumed in this reaction to give you HA, and the ratio is not altered that much, we know that the hydronium ion concentration does not change all that much. Now what happens if you add a base? So the second scenario is B, if OH minus is added. So we're going to add a strong base, and let's say we're adding 0.1 molar NaOH. So we know NaOH, KOH, LIOH, lithium hydroxide, these are all group 1 metal hydroxides, these are all strong bases because they're salts, and the moment you put them in water, they're going to dissociate completely. And so if you put 0.1 molar NaOH, the result is you're adding 0.1 molar hydroxide. If you take 0.1 molar hydroxide, what is the pH of 0.1 molar hydroxide? Remember hydroxide, 0.1 means 10 to the negative 1. So PoH is 1. If PoH is 1, what is pH? 13. So you can see that if I add this amount of sodium hydroxide into pure water, pure water is pH 7, and you add 0.1 molar sodium hydroxide, it goes from pH 7 to pH 13. That's a big change. It's 7 orders of magnitude, all right, and therefore 6 orders of magnitude, rather. And so there's a big change. If you add the same amount of hydroxide into a buffer, the pH is altered only very slightly. And let's look at why the pH is altered very slightly. Because if you take this equilibrium, once again, the buffer establishes an equilibrium. And this is the equilibrium that the buffer establishes. Now we're adding OH-. So added OH- reacts with the weak acid. So what happens is that it turns out an acid and a base in large amounts cannot coexist because they'll neutralize each other. So we have weak acids in large amounts, and we added a strong base in large amounts. And if in solution you have a weak, if you have any acid weak or strong, and if you have a base in large amounts, they will neutralize each other. They can exist in small amounts, like in water you have hydronium ion at 10 to the negative 7, and you have hydroxide 10 to the negative 7. Those are really small amounts, all right? But if you have really substantial amounts, like 1 molar and 0.1 molar, then what happens is large amounts of acid and base cannot coexist. They'll neutralize each other. And so when they neutralize each other, what you end up with is water and A-. So what happens is that when you add a strong base, now you have lots of the acid, which is the weak acid, and you have lots of bases. And we know that if you have substantial amounts of an acid and a base, they're going to neutralize each other. And when they neutralize each other, they form water. And so what happens is this reaction takes place, and so you can see the added OH- is consumed to produce A-. And when the system establishes equilibrium, so when equilibrium is established again, then the ratio of acid to base is relatively unchanged. All right? So can you see that if you have an acid added, the acid reacts with a conjugate base. So for added acid, you need substantial amounts of the conjugate base. If you add hydronium ion, remember we said hydronium ion reacts with A- to give you HF. So added acids are absorbed by the conjugate base. If you add a strong base, like hydroxide, it reacts with what? The acid part. So you need substantial amounts of the acid to absorb the base, and that's why we need substantial amounts of both. Exactly right, everybody. The buffer has substantial amounts of the weak acid and substantial amounts of the conjugate base, which is a salt. And you need both of those because one deals with any extra acid that's added, one will absorb that. So if you add extra acid, the conjugate base will absorb it, and the system will go back to equilibrium. If you add a strong base, the weak acid will react with it and absorb that. And so, and then it'll go back to equilibrium. And the net result is regardless whether you add a strong acid or a strong base, the ratio always goes back to what it was because K is always a constant. And we know that when equilibrium is established, the ratio of products to reactants have to be maintained if K is going to remain a constant. That ratio has to be a constant. Exactly right, everybody? And so that's what happens and that's how buffers behave. So now that what I need you to be able to do is whenever you're given a problem, you have to figure out is this a buffer or not. All right? And what you'll notice is a buffer is very similar to the common ion effect problem that we dealt with before because in a common ion effect problem, you had a weak acid and you had a weak base which is a common ion for that equilibrium and the buffer is the same thing. So let's take an example of a buffer problem and what we're going to do is we're going to calculate the pH of the buffer and then we're going to add an acid and we're going to calculate how the pH changes with added acid or base. Now once again, there's nothing new here, it kind of conforms to all of the different types of equilibrium problem that we've been working out. All right? So let's start by looking at part A. So we say, suppose one mole of formic acid and point five moles of sodium formate are added to water and diluted to one liter. All right? So we can start by saying that we have formic acid and the initial concentration of formic acid is one mole over one liter. Okay? That concentration is moles divided by the volume and that gives me one molar concentration. We're told that we have sodium formate. So you need to be able to recognize that now what we have is the salt of the conjugate base and this is point five moles in one liter which gives me point five zero zero molar. So to begin with, you can see that I have a mixture of a weak acid and the salt of its conjugate base. So what is that screen back at you? Buffer. All right? If you just had the weak acid, it would just calculate the pH of a weak acid. Now you have a mixture of both. Now ideally, if the concentrations of both were the same, that would be the best buffer. Do you see that? But the reason I'm using different numbers is because when we solve the math, if I just use one for both of those, then it's hard for you to see the changes that are taking place. So I'm using two different numbers so that you can track the changes that take place for each one of those. And an ideal buffer would be where you had both of them in approximately equal concentrations. So we're taking an example where you can see the salt is half as much as the acid. Now when we look at the salt, we know that the moment you put the salt in water, if I take this and the moment I dissolve it, you know that it's going to dissociate to give me the conjugate base plus the cation. And we know the concentration, if the initial concentration of this is 0.5, we know that the concentration of the conjugate base would be 0.5 and we know that the concentration of the sodium cation should be 0.5 because for every unit of the combined entity where it dissociates, it will give you one unit of the anion and one unit of the cation. So if we start with 0.5 molar of the combined unit, we know that the parts that come from that combined unit will have 0.5 molar concentration each. So now, like any equilibrium problem, we have to start with the equilibrium that we're looking at. We have the weak acid, formic acid, and they've given us Ka. So we can write down the equilibrium for that. And so we start by saying you have formic acid in water giving me. Can you tell me what the products of this would turn out to be? This is acting as an acid. This is the proton donor. This is the proton acceptor. So what would I end up with? H3O plus HCOO minus and my Ka is 1.77 times 10 to the negative 4. So that's a weak acid. Now my initial concentrations would be, we just calculated that. I have formic acid which is one molar. Not concerned about water. Hydrogen in mine is going to be 0 to begin with. But now I also have the conjugate base. So this is what I would start with. I have combined a weak acid with its salt. So before at t equals 0, my initial concentrations would be that. Now if I look at the change, this would be negative x plus x plus x. They're all one to one stoichiometry. Before at equilibrium, I would have 1 minus x, x and 0.5 plus x. So now you guys know how to solve the rest of this. Now I have to figure out what x is. And x is hydronium ion concentration. So if I want to calculate the pH, then once I figure out what x is, that gives me hydronium ion concentration. I can figure out what the pH is. So we can go pretty fast on this part because we've worked this so often. So we know that Ka equals 1.77 times 10 to the negative 4, which gives me the hydronium ion concentration times the formate ion concentration divided by formic acid, which is x times 0.500 plus x divided by 1 minus x. Now what do I do next? Approximate it because Ka is 10 to the negative 4. It's small and therefore I can make my approximation. And my approximation is going to be that 0.5 plus x is 0.5. It's just so small we can neglect it, and if x is small compared to 0.1, obviously it's got to be small compared to 1. So this is really small compared to this. Now 1 is a bigger number, so x has to be really small compared to that as well. So now I can say that 1.77 times 10 to the negative 4 equals x times 0.5 divided by 1. And so I can solve for x, and if I solve for x, the number comes out to be 3.54 times 10 to the negative 4 moles per liter. Now I have to check validity. Is my approximation valid? And can everybody see that I just need to pick one of these? Alright, I don't need to calculate validity for both because if I take the top one, if I say x is small compared to 0.5, it has to be obviously small compared to 1. So I'm going to take 3.54 times 10 to the negative 4 divided by 0.5 molar times 100. And if you work that out, it comes out to 0.0708%. In other words, it's 0.07. That's well below our threshold. So we know our approximation is valid. So if I know that the approximation is valid, now I know that my hydronium ion concentration is 3.54 times 10 to the negative 4 molar. And if that is the case, I can convert that to pH, and pH comes out to be 3.451. I start with a three significant figure number, so I'm going to end up with three decimals, and that would be the pH of this buffer. So that was our first part of the problem. How do you calculate the pH of a buffer? And you can see that by setting up the equations and solving in the manner in which we solved every equilibrium problem, you can figure out what the pH of that solution is. Now let's go to the second part. Suppose 0.1 moles of a strong acid such as HCl is added to the above mixture, calculate the pH of the resulting solution. And now for part B, now we have to figure out what happens if you add a strong acid. So once again, don't get thrown off by this problem because it just looks complicated. It's actually if you just follow the same principles that we've followed before, we should be able to come to a rational conclusion. So let's start with the hydronium ion concentration. So we have, we're adding, the initial concentration of HCl would be 0.1 moles in one liter of the, because we know that the buffer is one liter. So we start with 0.1 mole, and what we end up with is 0.10 molar concentration of HCl. So we've calculated the initial concentration of HCl that we added. We know that that is equal to what? Because it's a strong acid, that would equal the amount of hydronium ion that we added. Okay? Now, I want you to think about how buffers behave. If you add a strong acid to the buffer, how is the strong acid consumed? It reacts with what? The conjugate base. All right? In this case, HClO-. All right? So what's going to happen is that we have a limiting range of problem to begin with. So what we have is that the added acid, let me rewrite. The HClO- will react with the added acid to give you plus H2O. Does that make sense? That's how the added acid is consumed so that the pH is not altered very much. So what happens is that the instant you add the strong acid, the system is no longer at equilibrium. It's been disturbed. All right? So when it's disturbed, now it's going to respond by counteracting that disturbance by reacting. The hydronium ion would react with the conjugate base to give you formic acid and water. So if you want to look at this initial, at the beginning, what I have is initial before any reaction occurs. What do I have? To begin with, how much formic acid concentration did we have? The formic ion concentration is what? Is 0.5. All right? So this is 0.5. We added hydronium ion. And how much hydronium ion did we add? 0.1 molar. All right? And over here, we had one molar of that. Remember the buffer to begin with had 0.5 conjugate base and one molar weak acid. Now we added the strong acid in its 0.1 molar. So that means afterwards, or in the end, that is after any reaction occurs, you will have. So can you guys look at this? Now this becomes a limiting range of problem. All right? Between these two, which is limiting? You have 0.5. Lots of this. They react in a one-to-one ratio. Can everybody see that this is actually limiting? If this is limiting and it's all consumed, that means if it reacts with this, how much of this would be left over? Can you see that? So if this is all consumed, that means if one mole is consumed, if one mole of this is consumed and one mole of this is consumed, you're going to form one mole of that. Here what's happening is that because essentially what happens is you have 0.5 minus 0.1. All right? I end up with a two-significant number here. So this is two significant figures, two decimals, three decimals. We're subtracting. So you end up with a number that is two decimals here. Got it? Now since this is limiting, this is all consumed. So on the product side, what would you have? Can everybody see you end up with 1.10? Because on this side, you'll have 1 plus 0.10. So essentially what we're doing is we're looking at the reaction and now the added acid has been consumed. So because the added acid has been consumed, concentrations of formate and formic acid have changed. All right? So now we go back to the equilibrium. So now this, now once the equilibrium is established, what we will have is plus H2O giving you H3O plus aqueous plus HCOO minus aqueous. You know KA is 1.77 times 10 to the negative 4. But now once this reaction is complete, you can see my initial concentration of formic acid is 1.10. Zero water, we're not concerned about water. My drunamine is zero to begin with and what I have is 0.40 of formate. So the change here would be negative x plus x plus x. Therefore at equilibrium, I have 1 minus, 1.1 minus x. Not concerned about that, x and this would be 0.4 plus x. All right? And so now I can say KA equals 1.77 times 10 to the negative 4, which is the hydrionamine concentration times the formate ion concentration divided by the concentration of formic acid, which gives me x times 0.4 plus x divided by 1.10 minus x. Now I make the approximation, which is I'm going to say 0.4 plus x approximates to 0.4, 1.10 minus x approximates to 1.10. And once I do that, I can go back and say that 1.77 times 10 to the negative 4 equals x times 0.40 divided by 1.10. I can move fast here because you know this is the general form that we've solved every acid-base equilibrium problem and you solve for x. And if you solve for x, that number comes out to be 4.87. So it would be 4.9 because I'm keeping this to two significant figures because when we did that subtraction, all right, we ended up with a number that was 2 decimals and so that comes to that. 4.9 times 10 to the negative 4 molar. All right? And then if I check validity, validity would be x divided by 0.4 times 100, which is 4.9 times 10 to the negative 4 molar divided by 0.40 molar times 100, and that comes out to be 0.12%. So we know approximation is valid. And so we know what x is, x is 4.9 times 10 to the negative 4 molar, therefore pH. So x equals the hydronium ion concentration. And so pH is the negative log of that and it comes out to be 3.31. All right? So let's look at our take home lesson. If you remember when we had the buffer, it was 3.4, 5. All right? Without adding any acid, the buffer was 3.45. I added 0.1 molar of acid and the pH is now 3.31. So you can see the pH was not altered all that much. If I take pure water and add the same amount, you can see the pH changes by 7 orders of magnitude. If you take pure water and add 0.1 molar, it changes from pH 7 to pH 1, which is a big change. But in the buffer, you see that the pH is only altered very slightly. All right? And that's how a buffer works. Thanks for watching.
 Can I have everyone's attention please?  Are there any questions before I begin?  Great.  So I'm just going to pick up where we stopped last time.  And if you guys remember, last time we talked about the common ion effect.  And we said that if you take a weak acid, for example, and if that weak acid is at equilibrium,  and now if you disturb that equilibrium by adding a common ion that's common to that  equilibrium, then that equilibrium would be disturbed.  And if the equilibrium is disturbed, then we can predict how the system is going to  react based on Le Chatelier's principle.  Because we know what Le Chatelier's principle tells us is that if you disturb or stress  the system at equilibrium, it's going to respond in a way in which it can counteract that stress.  All right?  Now another way of solving common ion effect problems is to have both of them to start with.  So last time we took this problem where we had a weak acid.  And so the weak acid here was HF, so we looked at HF equals in water.  And because this is a weak acid, it's going to establish this equilibrium where KA is  7.2 times 10 to the negative 4.  So rather than have a system where you have equilibrium being established and then you  add a common ion, you can start with mixtures of both.  And so this is an example of a problem where you have to begin with, you have one molar,  so this is initial, you have one molar of this, and you have one molar of that.  Do you see that?  So the way you solve this problem is the same thing.  You can have a system that's established and add a common ion, or you can start by having  the weak acid and the conjugate base there.  So now what you have is a mixture of both.  In the past, we've looked at examples where you know how to calculate the pH of just the  weak acid alone.  If it was just the weak acid alone, you don't have any of the conjugate base to begin with.  All right?  In this example, what you have is you have both the weak acid and the common ion, which  is the conjugate base, you've added a salt to that.  And based on that, you can figure out what the pH of that solution is.  So these are all slight distinctions of the approach that we take in solving different  problems involving acids and bases and figuring out what the pH of the solution is.  You can see that now this is a slight variation.  Now today we're going to move on to looking at a phenomenon that is very similar to this  common ion effect, and we call this buffer solutions or buffers.  So today we're going to move on to looking at buffer solutions or buffers.  And many of you have heard the term buffer solutions or buffers before.  So we're going to look at buffered solutions or we call these buffers.  And if you want to describe what that is, a buffered solution or buffer is any solution  that maintains an approximately constant pH despite small additions of acid or base.  So a buffered solution is any solution that maintains its pH despite small additions of  acids or bases.  Now we know typically if you take water and if you add an acid to water, what does that  solution become?  If you take pure water, which is neutral, and if you add an acid to pure water, what  does that solution become?  Acidic.  And so what do we know about the pH?  pH is going to go from 7 to a smaller value.  If I take pure water, which is neutral, and add a strong base to it, what do we know about  that solution?  It is going to be basic.  And so what does that tell me about pH?  pH is going to go up.  So that's if you just have pure water.  The buffers are special solutions where if you add an acid, the pH does not change dramatically.  There is just a slight change in pH.  So it maintains the pH.  Or if you take a buffer, which is a special solution, and add a strong base like hydroxide  to it, usually if you take pure water and add a strong base, its pH is going to go up dramatically.  Buffers are special solutions where if you add a strong base, or if you add a strong acid,  its pH remains relatively unchanged.  It will be altered slightly, but it maintains that pH.  So it turns out these special solutions have some very distinct characteristics.  So we need to define what that is.  So we can say a buffer contains one, a weak acid, and a weak base that are conjugate to  one another.  So we said these are special solutions, and the characteristics of this special solution  is that a buffer contains both a weak acid and its conjugate base together.  So in other words, if we're mixing a weak acid and a weak base, what that means is you have  a weak acid and the salt of its conjugate base.  So if we want to mix a weak acid and its conjugate base, the conjugate base, if it's negatively  charged, has to be a salt.  So if I take HCN, if I take HCN, which is a weak acid, HCN, what is its conjugate base?  HCN.  HCN, conjugate base.  CN minus.  Got it?  So if it's CN minus, you know that a charged piece is not going to be alone.  So it needs a counter ion.  So we add a group one metal to it.  So it could be sodium cyanide.  It could be potassium cyanide, and we call that the salt of its conjugate base.  So a buffer is a weak acid.  So pick HCN, HF, acidic acid, that's a weak acid, and its conjugate base.  So if it's HCN, CN minus would be its conjugate base.  If it's HF, what would its conjugate base be?  F minus.  If it's acidic acid, what is its conjugate base?  Acetate.  All right?  So all of those are negatively charged ions, and therefore you have to have a salt.  So it'll be a group one metal like sodium potassium.  So a weak acid and the salt of its conjugate base, or a weak base like ammonia, and the  salt of its conjugate acid.  All right?  So you can have two types.  You can have a weak acid like HCN, HF, acidic acid, and the salt of its conjugate base.  Or you can have a weak base that's neutral like ammonia.  And if it's ammonia, what is the conjugate acid of ammonia in H4 plus?  All right?  So the salt of that would be in H4Cl.  So the way to identify a buffer is to look for these characteristics.  So if I have, for example, HF and NAF, all right?  Where this is F minus, which is the conjugate base of HF.  You can have HCN and over here NA-CN.  If you have acidic acid, then over here you'll have CH3CO2NA.  Each of these, this is a salt.  And what do we know about salt?  The moment you put a salt in water, what does a salt do?  Dissociates.  So it's going to release the anion, which is the conjugate base.  So this would have CN minus.  This would release CH3CO2 minus.  And then if you take something like formic acid, which is that, then it would be the  salt of formic acid.  And therefore this would release that.  Okay?  Or if we're dealing with a weak base, then what we do is the acid form would be NH4Cl.  All right?  You would have NH3.  So once again, this is an acid base, which are conjugate to one another.  And this is a salt, so it will give you NH4+.  And so what you end up with is NH4+, and NH3.  These are a combination of weak acid and a weak base.  All right?  I want to look at that table that I gave you.  If you take something like CH3NH3Cl, this is also a salt.  So essentially what we're doing is we're replacing one of the hydrogens here with CH3.  You see that?  And this would be a combination of CH3NH2, all right?  Where this would release the cation CH3NH3+, and so that would be.  That would be the acid form, because this is the cation, and the chloride is there as  the anion.  So this would release the acid form, and that would be its conjugate base.  All right?  So don't get thrown off by the formulas, because you can kind of sit down and look at it and  say if it's CH3NH3Cl, you know that that's a salt, and therefore it's got a cation,  and the cation component would be that.  And if you look at it, it's very similar to NH4+.  The only difference is that one of those hydrogens have been replaced by CH3, all right?  And therefore that would be the acid form, and that would be the basic form.  So we said that that's the first criterion.  If you want to have a buffer, you have to have both of those components, all right?  And they have to be conjugate to one another.  The second property that a buffer has to have is to the, or both components have to be in  approximately equal amounts.  In other words, their concentrations should be similar.  For the best buffers, you want to make sure that they're both in approximately equal amounts.  Okay?  So if I'm taking a weak acid and my concentration is one molar, the best buffer would be my  conjugate base is also one molar.  So you have equal concentrations of both.  You can, they can be slightly different from each other, but the ideal would be if their  concentrations are equal to each other.  Okay?  The third is that both should be in substantial amounts, should be in substantial amounts.  In other words, what that means is higher concentrations.  All right?  So when I say higher concentrations, one molar is considered to be quite a high concentration.  All right?  So if they're the order of one molar, two molar, then the more concentrated the both  of them are, the better the buffer.  Okay?  So whenever you are asked to identify a buffer, you've got to have to look for this.  You have to have a mixture of a weak acid and its conjugate base so that they're conjugate  to one another.  You have to have approximately equal concentrations and they should be in substantial amounts.  All right?  So these are the criteria for a good buffer, but now let's see how does a buffer work?  So let's start looking at how does a buffer work?  Now remember, if it's a good buffer, if you add acid, the pH should not change all that  much.  All right?  So if we want to look at a buffer, we said a buffer is anything that has a weak acid  and its conjugate base.  So let's like the equilibrium.  Remember, we're looking at acid-basic-coulibre, and the underlying theme for all of this is  the equilibrium that we establish.  So if I have a weak acid and its conjugate base, then let's start with a generic weak  acid.  I have HA, and if I place HA in water, I would establish the equilibrium where I have H3O  plus aqueous plus A minus aqueous.  And the form in which this equilibrium is written is in the form of an acid dissociation.  So this equilibrium is described by KA.  All right?  So if I start with initial concentration, let's say I have one molar of this and one  molar of that.  And to begin with, these are the initial concentrations.  I have initially to begin with one molar of the weak acid and one molar of the weak base.  So they're approximately equal amounts.  They're substantial amounts, and I have both of them there.  And we know that when equilibrium is established, at equilibrium, now I can jump a step.  This will be minus x plus x and so on.  So at equilibrium, we know this would be one minus x.  This would be x.  This will be one plus x.  And therefore, I can approximate this to one.  This will be x.  If KA for weak acid is always smaller, we can approximate and say x is small, and therefore  that would be one.  All right?  So at equilibrium, we can say that this equals the initial concentration of HA, and this  equals the initial concentration of A minus.  Because the amount, the x is so small, because the weak acid KA is going to be really small,  because x is going to be small, we're going to approximate it.  So we're going to say the initial concentration hardly changes.  Can you see that?  And what is a sort of symbolic notation that we use to describe initial concentrations?  We always put a subscript zero.  And what does that subscript zero mean?  It's at time zero.  All right?  And that's what put the zero at the bottom.  Time zero means that no reaction has taken place.  This is what you start with at the beginning, all right, before any reaction takes place.  And so what we say is that because we're dealing with weak acids on their conjugate basis,  KA is going to be really small, therefore we can approximate and say that the initial  concentration doesn't change very much.  All right?  So now we're going to write the equilibrium expression, and we know KA equals the definition,  which is the law of mass action, which is H3O minus A minus over HA.  And we can say that this approximates to KA equals, this approximates to KA equals H3O  plus the initial concentration of A minus over the initial concentration of HA, because  we know that for weak acids, the initial concentration doesn't change by much, and we can always  approximate that and say it's equal to the initial concentration.  Okay?  Now, remember we're looking at the effects of pH.  All right?  We want to look at how is the pH affected, and what does pH depend on?  What is the definition of pH?  Depends on hydronium ion concentration.  So if we want to look at the effect of pH, we want to focus on hydronium ion, and so  let's rearrange this equation so that we look at the hydronium ion concentration.  So what we have is hydronium ion concentration, and if I rearrange this equation, you can  say that equals KA.  So if the pH is not altered, that means the hydronium ion concentration is not altered.  All right?  So if we add an acid and we say the pH does not change, what does that really mean?  It means that the hydronium ion concentration is not altered by much.  So we know that hydronium ion concentration depends on KA, which is a constant, and then  the ratio of the acid to the base.  All right?  So if the ratio of these two is not altered too much, H3O plus will remain unchanged.  Is that clear?  So if we want to look at a buffer, and we have substantial amounts of both, and if the  pH is not going to change, pH depends on a hydronium ion concentration, and therefore  if that ratio is not altered too much, we know that the hydronium ion concentration  will not change that much.  And so that's how buffers work.  It turns out that in buffers, because you have substantial amounts of this and this,  when you add an acid or a base, this ratio is not altered all that much.  All right?  And as long as the ratio is not altered, then the hydronium ion concentration does not  change, and therefore the pH remains unchanged.  All right?  So let's look at what happens if you add an acid or a base.  So now let's take this scenario, and now let's say that if...  So let's look at the first case.  Let's say if H3O plus is added.  All right?  In other words, let's say something like 0.1 molar HCl is added.  Okay?  So all of you recognize that HCl is a strong acid, and we know that a strong acid dissociates  completely.  So if I put 0.1 molar HCl, what is the concentration of the hydronium ion that I'm adding?  0.1 molar.  All right?  So all I'm doing is I'm adding 0.1 molar hydronium ion.  Now if I take 0.1 molar hydronium ion in pure water, what is the pH of 0.1 molar hydronium  ion?  0.1 is what?  10 to the what?  Negative 1.  Negative 1.  If it's 10 to the negative 1, what is that?  1.  So that means 0.1 molar, if you take neutral water and add 0.1 molar to neutral water, it  goes from pH 7 to?  To pH 1.  pH 1, that's a big change.  Remember, it's a factor.  Every change in pH unit is a change by a factor of 10, or by one order of magnitude.  So now it changes by seven orders of magnitude.  So that's a pretty strong acid.  Now what happens is if I take this and add it in a buffer, now the pH of the buffer does  not change.  If you take pure water, it goes from seven to one.  But if you put it in a buffer, the pH is not altered all that much.  And the reason is that you have this equilibrium that we've established.  Now, so we have this equilibrium, H2O plus A minus, and this equilibrium is established.  But now I'm adding hydronium ion.  So what am I doing to that?  If I look at that equilibrium, I'm adding hydronium ion to that solution.  So I'm adding a large amount of hydronium ion in there.  So what happens to that equilibrium?  Is there a common ion there?  Is hydronium ion there already in the equilibrium?  So now if I'm adding more hydronium ion, what happens to that equilibrium?  I've disturbed it.  So now Leicester's principles are going to jump in, and it's going to counteract that  stress.  So what happens is that this equilibrium, so to counteract the addition, so to counteract  the added hydronium ion, what happens is that the hydronium ion plus A minus will react  to give you HA plus H2O.  So aqueous, aqueous, this will be an aqueous, and that'll be aqueous.  So essentially what happens is that you've added too much hydronium ion, the system is  going to respond by this reacting with this to give you.  In this way, what it does is it absorbs the extra hydronium ion that you added, and the  system will go back to equilibrium.  When the system goes back to equilibrium, what happens is that this ratio, once it goes  back to equilibrium, this ratio is altered only slightly.  So the way this buffer acts is that it absorbs the added acid and reacts with it and forms  the weak acid, and when the system goes back to equilibrium, because KA is a constant,  do you see that?  The ratio of products to reactants have to be maintained, because that's how KA is a  constant.  And so what happens is when it goes back to equilibrium, this ratio is only altered very  slightly, and therefore because the ratio is altered only slightly, the pH remains relatively  unchanged.  As a result of this reaction, the ratio HA over A minus is relatively unchanged.  And so if that ratio is unchanged, then the hydronium ion concentration is not changed  all that much.  Now the second scenario is that we added, if we added base.  So now we understand how a buffer works, and we understand that when you add an added acid  because it disturbs that equilibrium and the extra hydronium ion is absorbed and consumed  in this reaction to give you HA, and the ratio is not altered that much, we know that the  hydronium ion concentration does not change all that much.  Now what happens if you add a base?  So the second scenario is B, if OH minus is added.  So we're going to add a strong base, and let's say we're adding 0.1 molar NaOH.  So we know NaOH, KOH, LIOH, lithium hydroxide, these are all group 1 metal hydroxides, these  are all strong bases because they're salts, and the moment you put them in water, they're  going to dissociate completely.  And so if you put 0.1 molar NaOH, the result is you're adding 0.1 molar hydroxide.  If you take 0.1 molar hydroxide, what is the pH of 0.1 molar hydroxide?  Remember hydroxide, 0.1 means 10 to the negative 1.  So PoH is 1.  If PoH is 1, what is pH?  13.  So you can see that if I add this amount of sodium hydroxide into pure water, pure water  is pH 7, and you add 0.1 molar sodium hydroxide, it goes from pH 7 to pH 13.  That's a big change.  It's 7 orders of magnitude, all right, and therefore 6 orders of magnitude, rather.  And so there's a big change.  If you add the same amount of hydroxide into a buffer, the pH is altered only very slightly.  And let's look at why the pH is altered very slightly.  Because if you take this equilibrium, once again, the buffer establishes an equilibrium.  And this is the equilibrium that the buffer establishes.  Now we're adding OH-.  So added OH- reacts with the weak acid.  So what happens is that it turns out an acid and a base in large amounts cannot coexist  because they'll neutralize each other.  So we have weak acids in large amounts, and we added a strong base in large amounts.  And if in solution you have a weak, if you have any acid weak or strong, and if you have  a base in large amounts, they will neutralize each other.  They can exist in small amounts, like in water you have hydronium ion at 10 to the negative  7, and you have hydroxide 10 to the negative 7.  Those are really small amounts, all right?  But if you have really substantial amounts, like 1 molar and 0.1 molar, then what happens  is large amounts of acid and base cannot coexist.  They'll neutralize each other.  And so when they neutralize each other, what you end up with is water and A-.  So what happens is that when you add a strong base, now you have lots of the acid, which  is the weak acid, and you have lots of bases.  And we know that if you have substantial amounts of an acid and a base, they're going to neutralize  each other.  And when they neutralize each other, they form water.  And so what happens is this reaction takes place, and so you can see the added OH- is  consumed to produce A-.  And when the system establishes equilibrium, so when equilibrium is established again,  then the ratio of acid to base is relatively unchanged.  All right?  So can you see that if you have an acid added, the acid reacts with a conjugate base.  So for added acid, you need substantial amounts of the conjugate base.  If you add hydronium ion, remember we said hydronium ion reacts with A- to give you HF.  So added acids are absorbed by the conjugate base.  If you add a strong base, like hydroxide, it reacts with what?  The acid part.  So you need substantial amounts of the acid to absorb the base, and that's why we need  substantial amounts of both.  Exactly right, everybody.  The buffer has substantial amounts of the weak acid and substantial amounts of the conjugate  base, which is a salt.  And you need both of those because one deals with any extra acid that's added, one will  absorb that.  So if you add extra acid, the conjugate base will absorb it, and the system will go back  to equilibrium.  If you add a strong base, the weak acid will react with it and absorb that.  And so, and then it'll go back to equilibrium.  And the net result is regardless whether you add a strong acid or a strong base, the ratio  always goes back to what it was because K is always a constant.  And we know that when equilibrium is established, the ratio of products to reactants have to  be maintained if K is going to remain a constant.  That ratio has to be a constant.  Exactly right, everybody?  And so that's what happens and that's how buffers behave.  So now that what I need you to be able to do is whenever you're given a problem, you  have to figure out is this a buffer or not.  All right?  And what you'll notice is a buffer is very similar to the common ion effect problem that  we dealt with before because in a common ion effect problem, you had a weak acid and you  had a weak base which is a common ion for that equilibrium and the buffer is the same  thing.  So let's take an example of a buffer problem and what we're going to do is we're going  to calculate the pH of the buffer and then we're going to add an acid and we're going  to calculate how the pH changes with added acid or base.  Now once again, there's nothing new here, it kind of conforms to all of the different  types of equilibrium problem that we've been working out.  All right?  So let's start by looking at part A. So we say, suppose one mole of formic acid and  point five moles of sodium formate are added to water and diluted to one liter.  All right?  So we can start by saying that we have formic acid and the initial concentration of formic  acid is one mole over one liter.  Okay?  That concentration is moles divided by the volume and that gives me one molar concentration.  We're told that we have sodium formate.  So you need to be able to recognize that now what we have is the salt of the conjugate  base and this is point five moles in one liter which gives me point five zero zero molar.  So to begin with, you can see that I have a mixture of a weak acid and the salt of its  conjugate base.  So what is that screen back at you?  Buffer.  All right?  If you just had the weak acid, it would just calculate the pH of a weak acid.  Now you have a mixture of both.  Now ideally, if the concentrations of both were the same, that would be the best buffer.  Do you see that?  But the reason I'm using different numbers is because when we solve the math, if I just  use one for both of those, then it's hard for you to see the changes that are taking  place.  So I'm using two different numbers so that you can track the changes that take place  for each one of those.  And an ideal buffer would be where you had both of them in approximately equal concentrations.  So we're taking an example where you can see the salt is half as much as the acid.  Now when we look at the salt, we know that the moment you put the salt in water, if I  take this and the moment I dissolve it, you know that it's going to dissociate to give  me the conjugate base plus the cation.  And we know the concentration, if the initial concentration of this is 0.5, we know that  the concentration of the conjugate base would be 0.5 and we know that the concentration  of the sodium cation should be 0.5 because for every unit of the combined entity where  it dissociates, it will give you one unit of the anion and one unit of the cation.  So if we start with 0.5 molar of the combined unit, we know that the parts that come from  that combined unit will have 0.5 molar concentration each.  So now, like any equilibrium problem, we have to start with the equilibrium that we're looking  at.  We have the weak acid, formic acid, and they've given us Ka.  So we can write down the equilibrium for that.  And so we start by saying you have formic acid in water giving me.  Can you tell me what the products of this would turn out to be?  This is acting as an acid.  This is the proton donor.  This is the proton acceptor.  So what would I end up with?  H3O plus HCOO minus and my Ka is 1.77 times 10 to the negative 4.  So that's a weak acid.  Now my initial concentrations would be, we just calculated that.  I have formic acid which is one molar.  Not concerned about water.  Hydrogen in mine is going to be 0 to begin with.  But now I also have the conjugate base.  So this is what I would start with.  I have combined a weak acid with its salt.  So before at t equals 0, my initial concentrations would be that.  Now if I look at the change, this would be negative x plus x plus x.  They're all one to one stoichiometry.  Before at equilibrium, I would have 1 minus x, x and 0.5 plus x.  So now you guys know how to solve the rest of this.  Now I have to figure out what x is.  And x is hydronium ion concentration.  So if I want to calculate the pH, then once I figure out what x is, that gives me hydronium  ion concentration.  I can figure out what the pH is.  So we can go pretty fast on this part because we've worked this so often.  So we know that Ka equals 1.77 times 10 to the negative 4, which gives me the hydronium  ion concentration times the formate ion concentration divided by formic acid, which is x times 0.500  plus x divided by 1 minus x.  Now what do I do next?  Approximate it because Ka is 10 to the negative 4.  It's small and therefore I can make my approximation.  And my approximation is going to be that 0.5 plus x is 0.5.  It's just so small we can neglect it, and if x is small compared to 0.1, obviously it's  got to be small compared to 1.  So this is really small compared to this.  Now 1 is a bigger number, so x has to be really small compared to that as well.  So now I can say that 1.77 times 10 to the negative 4 equals x times 0.5 divided by 1.  And so I can solve for x, and if I solve for x, the number comes out to be 3.54 times 10  to the negative 4 moles per liter.  Now I have to check validity.  Is my approximation valid?  And can everybody see that I just need to pick one of these?  Alright, I don't need to calculate validity for both because if I take the top one, if  I say x is small compared to 0.5, it has to be obviously small compared to 1.  So I'm going to take 3.54 times 10 to the negative 4 divided by 0.5 molar times 100.  And if you work that out, it comes out to 0.0708%.  In other words, it's 0.07.  That's well below our threshold.  So we know our approximation is valid.  So if I know that the approximation is valid, now I know that my hydronium ion concentration  is 3.54 times 10 to the negative 4 molar.  And if that is the case, I can convert that to pH, and pH comes out to be 3.451.  I start with a three significant figure number, so I'm going to end up with three decimals,  and that would be the pH of this buffer.  So that was our first part of the problem.  How do you calculate the pH of a buffer?  And you can see that by setting up the equations and solving in the manner in which we solved  every equilibrium problem, you can figure out what the pH of that solution is.  Now let's go to the second part.  Suppose 0.1 moles of a strong acid such as HCl is added to the above mixture, calculate  the pH of the resulting solution.  And now for part B, now we have to figure out what happens if you add a strong acid.  So once again, don't get thrown off by this problem because it just looks complicated.  It's actually if you just follow the same principles that we've followed before, we  should be able to come to a rational conclusion.  So let's start with the hydronium ion concentration.  So we have, we're adding, the initial concentration of HCl would be 0.1 moles in one liter of  the, because we know that the buffer is one liter.  So we start with 0.1 mole, and what we end up with is 0.10 molar concentration of HCl.  So we've calculated the initial concentration of HCl that we added.  We know that that is equal to what?  Because it's a strong acid, that would equal the amount of hydronium ion that we added.  Okay?  Now, I want you to think about how buffers behave.  If you add a strong acid to the buffer, how is the strong acid consumed?  It reacts with what?  The conjugate base.  All right?  In this case, HClO-.  All right?  So what's going to happen is that we have a limiting range of problem to begin with.  So what we have is that the added acid, let me rewrite.  The HClO- will react with the added acid to give you plus H2O.  Does that make sense?  That's how the added acid is consumed so that the pH is not altered very much.  So what happens is that the instant you add the strong acid, the system is no longer at  equilibrium.  It's been disturbed.  All right?  So when it's disturbed, now it's going to respond by counteracting that disturbance by  reacting.  The hydronium ion would react with the conjugate base to give you formic acid and water.  So if you want to look at this initial, at the beginning, what I have is initial before  any reaction occurs.  What do I have?  To begin with, how much formic acid concentration did we have?  The formic ion concentration is what?  Is 0.5.  All right?  So this is 0.5.  We added hydronium ion.  And how much hydronium ion did we add?  0.1 molar.  All right?  And over here, we had one molar of that.  Remember the buffer to begin with had 0.5 conjugate base and one molar weak acid.  Now we added the strong acid in its 0.1 molar.  So that means afterwards, or in the end, that is after any reaction occurs, you will  have.  So can you guys look at this?  Now this becomes a limiting range of problem.  All right?  Between these two, which is limiting?  You have 0.5.  Lots of this.  They react in a one-to-one ratio.  Can everybody see that this is actually limiting?  If this is limiting and it's all consumed, that means if it reacts with this, how much  of this would be left over?  Can you see that?  So if this is all consumed, that means if one mole is consumed, if one mole of this  is consumed and one mole of this is consumed, you're going to form one mole of that.  Here what's happening is that because essentially what happens is you have 0.5 minus 0.1.  All right?  I end up with a two-significant number here.  So this is two significant figures, two decimals, three decimals.  We're subtracting.  So you end up with a number that is two decimals here.  Got it?  Now since this is limiting, this is all consumed.  So on the product side, what would you have?  Can everybody see you end up with 1.10?  Because on this side, you'll have 1 plus 0.10.  So essentially what we're doing is we're looking at the reaction and now the added  acid has been consumed.  So because the added acid has been consumed, concentrations of formate and formic acid  have changed.  All right?  So now we go back to the equilibrium.  So now this, now once the equilibrium is established, what we will have is plus H2O  giving you H3O plus aqueous plus HCOO minus aqueous.  You know KA is 1.77 times 10 to the negative 4.  But now once this reaction is complete, you can see my initial concentration of formic  acid is 1.10.  Zero water, we're not concerned about water.  My drunamine is zero to begin with and what I have is 0.40 of formate.  So the change here would be negative x plus x plus x.  Therefore at equilibrium, I have 1 minus, 1.1 minus x.  Not concerned about that, x and this would be 0.4 plus x.  All right?  And so now I can say KA equals 1.77 times 10 to the negative 4, which is the hydrionamine  concentration times the formate ion concentration divided by the concentration of formic acid,  which gives me x times 0.4 plus x divided by 1.10 minus x.  Now I make the approximation, which is I'm going to say 0.4 plus x approximates to 0.4,  1.10 minus x approximates to 1.10.  And once I do that, I can go back and say that 1.77 times 10 to the negative 4 equals  x times 0.40 divided by 1.10.  I can move fast here because you know this is the general form that we've solved every  acid-base equilibrium problem and you solve for x.  And if you solve for x, that number comes out to be 4.87.  So it would be 4.9 because I'm keeping this to two significant figures because when we  did that subtraction, all right, we ended up with a number that was 2 decimals and so  that comes to that.  4.9 times 10 to the negative 4 molar.  All right?  And then if I check validity, validity would be x divided by 0.4 times 100, which is 4.9  times 10 to the negative 4 molar divided by 0.40 molar times 100, and that comes out to  be 0.12%.  So we know approximation is valid.  And so we know what x is, x is 4.9 times 10 to the negative 4 molar, therefore pH.  So x equals the hydronium ion concentration.  And so pH is the negative log of that and it comes out to be 3.31.  All right?  So let's look at our take home lesson.  If you remember when we had the buffer, it was 3.4, 5.  All right?  Without adding any acid, the buffer was 3.45.  I added 0.1 molar of acid and the pH is now 3.31.  So you can see the pH was not altered all that much.  If I take pure water and add the same amount, you can see the pH changes by 7 orders of  magnitude.  If you take pure water and add 0.1 molar, it changes from pH 7 to pH 1, which is a big  change.  But in the buffer, you see that the pH is only altered very slightly.  All right?  And that's how a buffer works.  Thanks for watching.
General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__0_179.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__180_227.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__228_335.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__336_431.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__432_479.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__480_527.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__528_575.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__576_623.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__624_671.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__672_731.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__732_779.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__780_839.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__840_887.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__888_935.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__936_995.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__996_1055.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1056_1103.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1104_1151.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1152_1211.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1212_1259.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1260_1307.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1308_1415.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1416_1463.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1464_1511.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1512_1559.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1560_1607.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1608_1691.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1692_1835.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1836_1883.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1884_1931.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__1932_2663.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__2664_3395.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__3396_4091.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__4092_5075.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__5076_8987.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__8988_12395.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__12396_18179.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__18180_21323.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__21324_21467.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__21468_21671.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__21672_23435.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__23436_31379.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__31380_32279.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__32280_33623.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__33624_43271.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__43272_46535.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__46536_46727.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__46728_48323.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__48324_48623.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__48624_48779.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__48780_49727.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__49728_62087.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__62088_65351.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__65352_72443.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__72444_75455.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__75456_75515.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__75516_75575.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__75576_75623.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__75624_76091.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__76092_76139.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__76140_76331.png General_Chemistry_1C__Lecture_09__Acids_and_Bases__Pt__6__76332_76378.png
Can I have everyone's attention please? Are there any questions before I begin? Great. So I'm just going to pick up where we stopped last time. And if you guys remember, last time we talked about the common ion effect. And we said that if you take a weak acid, for example, and if that weak acid is at equilibrium, and now if you disturb that equilibrium by adding a common ion that's common to that equilibrium, then that equilibrium would be disturbed. And if the equilibrium is disturbed, then we can predict how the system is going to react based on Le Chatelier's principle. Because we know what Le Chatelier's principle tells us is that if you disturb or stress the system at equilibrium, it's going to respond in a way in which it can counteract that stress. All right? Now another way of solving common ion effect problems is to have both of them to start with. So last time we took this problem where we had a weak acid. And so the weak acid here was HF, so we looked at HF equals in water. And because this is a weak acid, it's going to establish this equilibrium where KA is 7.2 times 10 to the negative 4. So rather than have a system where you have equilibrium being established and then you add a common ion, you can start with mixtures of both. And so this is an example of a problem where you have to begin with, you have one molar, so this is initial, you have one molar of this, and you have one molar of that. Do you see that? So the way you solve this problem is the same thing. You can have a system that's established and add a common ion, or you can start by having the weak acid and the conjugate base there. So now what you have is a mixture of both. In the past, we've looked at examples where you know how to calculate the pH of just the weak acid alone. If it was just the weak acid alone, you don't have any of the conjugate base to begin with. All right? In this example, what you have is you have both the weak acid and the common ion, which is the conjugate base, you've added a salt to that. And based on that, you can figure out what the pH of that solution is. So these are all slight distinctions of the approach that we take in solving different problems involving acids and bases and figuring out what the pH of the solution is. You can see that now this is a slight variation. Now today we're going to move on to looking at a phenomenon that is very similar to this common ion effect, and we call this buffer solutions or buffers. So today we're going to move on to looking at buffer solutions or buffers. And many of you have heard the term buffer solutions or buffers before. So we're going to look at buffered solutions or we call these buffers. And if you want to describe what that is, a buffered solution or buffer is any solution that maintains an approximately constant pH despite small additions of acid or base. So a buffered solution is any solution that maintains its pH despite small additions of acids or bases. Now we know typically if you take water and if you add an acid to water, what does that solution become? If you take pure water, which is neutral, and if you add an acid to pure water, what does that solution become? Acidic. And so what do we know about the pH? pH is going to go from 7 to a smaller value. If I take pure water, which is neutral, and add a strong base to it, what do we know about that solution? It is going to be basic. And so what does that tell me about pH? pH is going to go up. So that's if you just have pure water. The buffers are special solutions where if you add an acid, the pH does not change dramatically. There is just a slight change in pH. So it maintains the pH. Or if you take a buffer, which is a special solution, and add a strong base like hydroxide to it, usually if you take pure water and add a strong base, its pH is going to go up dramatically. Buffers are special solutions where if you add a strong base, or if you add a strong acid, its pH remains relatively unchanged. It will be altered slightly, but it maintains that pH. So it turns out these special solutions have some very distinct characteristics. So we need to define what that is. So we can say a buffer contains one, a weak acid, and a weak base that are conjugate to one another. So we said these are special solutions, and the characteristics of this special solution is that a buffer contains both a weak acid and its conjugate base together. So in other words, if we're mixing a weak acid and a weak base, what that means is you have a weak acid and the salt of its conjugate base. So if we want to mix a weak acid and its conjugate base, the conjugate base, if it's negatively charged, has to be a salt. So if I take HCN, if I take HCN, which is a weak acid, HCN, what is its conjugate base? HCN. HCN, conjugate base. CN minus. Got it? So if it's CN minus, you know that a charged piece is not going to be alone. So it needs a counter ion. So we add a group one metal to it. So it could be sodium cyanide. It could be potassium cyanide, and we call that the salt of its conjugate base. So a buffer is a weak acid. So pick HCN, HF, acidic acid, that's a weak acid, and its conjugate base. So if it's HCN, CN minus would be its conjugate base. If it's HF, what would its conjugate base be? F minus. If it's acidic acid, what is its conjugate base? Acetate. All right? So all of those are negatively charged ions, and therefore you have to have a salt. So it'll be a group one metal like sodium potassium. So a weak acid and the salt of its conjugate base, or a weak base like ammonia, and the salt of its conjugate acid. All right? So you can have two types. You can have a weak acid like HCN, HF, acidic acid, and the salt of its conjugate base. Or you can have a weak base that's neutral like ammonia. And if it's ammonia, what is the conjugate acid of ammonia in H4 plus? All right? So the salt of that would be in H4Cl. So the way to identify a buffer is to look for these characteristics. So if I have, for example, HF and NAF, all right? Where this is F minus, which is the conjugate base of HF. You can have HCN and over here NA-CN. If you have acidic acid, then over here you'll have CH3CO2NA. Each of these, this is a salt. And what do we know about salt? The moment you put a salt in water, what does a salt do? Dissociates. So it's going to release the anion, which is the conjugate base. So this would have CN minus. This would release CH3CO2 minus. And then if you take something like formic acid, which is that, then it would be the salt of formic acid. And therefore this would release that. Okay? Or if we're dealing with a weak base, then what we do is the acid form would be NH4Cl. All right? You would have NH3. So once again, this is an acid base, which are conjugate to one another. And this is a salt, so it will give you NH4+. And so what you end up with is NH4+, and NH3. These are a combination of weak acid and a weak base. All right? I want to look at that table that I gave you. If you take something like CH3NH3Cl, this is also a salt. So essentially what we're doing is we're replacing one of the hydrogens here with CH3. You see that? And this would be a combination of CH3NH2, all right? Where this would release the cation CH3NH3+, and so that would be. That would be the acid form, because this is the cation, and the chloride is there as the anion. So this would release the acid form, and that would be its conjugate base. All right? So don't get thrown off by the formulas, because you can kind of sit down and look at it and say if it's CH3NH3Cl, you know that that's a salt, and therefore it's got a cation, and the cation component would be that. And if you look at it, it's very similar to NH4+. The only difference is that one of those hydrogens have been replaced by CH3, all right? And therefore that would be the acid form, and that would be the basic form. So we said that that's the first criterion. If you want to have a buffer, you have to have both of those components, all right? And they have to be conjugate to one another. The second property that a buffer has to have is to the, or both components have to be in approximately equal amounts. In other words, their concentrations should be similar. For the best buffers, you want to make sure that they're both in approximately equal amounts. Okay? So if I'm taking a weak acid and my concentration is one molar, the best buffer would be my conjugate base is also one molar. So you have equal concentrations of both. You can, they can be slightly different from each other, but the ideal would be if their concentrations are equal to each other. Okay? The third is that both should be in substantial amounts, should be in substantial amounts. In other words, what that means is higher concentrations. All right? So when I say higher concentrations, one molar is considered to be quite a high concentration. All right? So if they're the order of one molar, two molar, then the more concentrated the both of them are, the better the buffer. Okay? So whenever you are asked to identify a buffer, you've got to have to look for this. You have to have a mixture of a weak acid and its conjugate base so that they're conjugate to one another. You have to have approximately equal concentrations and they should be in substantial amounts. All right? So these are the criteria for a good buffer, but now let's see how does a buffer work? So let's start looking at how does a buffer work? Now remember, if it's a good buffer, if you add acid, the pH should not change all that much. All right? So if we want to look at a buffer, we said a buffer is anything that has a weak acid and its conjugate base. So let's like the equilibrium. Remember, we're looking at acid-basic-coulibre, and the underlying theme for all of this is the equilibrium that we establish. So if I have a weak acid and its conjugate base, then let's start with a generic weak acid. I have HA, and if I place HA in water, I would establish the equilibrium where I have H3O plus aqueous plus A minus aqueous. And the form in which this equilibrium is written is in the form of an acid dissociation. So this equilibrium is described by KA. All right? So if I start with initial concentration, let's say I have one molar of this and one molar of that. And to begin with, these are the initial concentrations. I have initially to begin with one molar of the weak acid and one molar of the weak base. So they're approximately equal amounts. They're substantial amounts, and I have both of them there. And we know that when equilibrium is established, at equilibrium, now I can jump a step. This will be minus x plus x and so on. So at equilibrium, we know this would be one minus x. This would be x. This will be one plus x. And therefore, I can approximate this to one. This will be x. If KA for weak acid is always smaller, we can approximate and say x is small, and therefore that would be one. All right? So at equilibrium, we can say that this equals the initial concentration of HA, and this equals the initial concentration of A minus. Because the amount, the x is so small, because the weak acid KA is going to be really small, because x is going to be small, we're going to approximate it. So we're going to say the initial concentration hardly changes. Can you see that? And what is a sort of symbolic notation that we use to describe initial concentrations? We always put a subscript zero. And what does that subscript zero mean? It's at time zero. All right? And that's what put the zero at the bottom. Time zero means that no reaction has taken place. This is what you start with at the beginning, all right, before any reaction takes place. And so what we say is that because we're dealing with weak acids on their conjugate basis, KA is going to be really small, therefore we can approximate and say that the initial concentration doesn't change very much. All right? So now we're going to write the equilibrium expression, and we know KA equals the definition, which is the law of mass action, which is H3O minus A minus over HA. And we can say that this approximates to KA equals, this approximates to KA equals H3O plus the initial concentration of A minus over the initial concentration of HA, because we know that for weak acids, the initial concentration doesn't change by much, and we can always approximate that and say it's equal to the initial concentration. Okay? Now, remember we're looking at the effects of pH. All right? We want to look at how is the pH affected, and what does pH depend on? What is the definition of pH? Depends on hydronium ion concentration. So if we want to look at the effect of pH, we want to focus on hydronium ion, and so let's rearrange this equation so that we look at the hydronium ion concentration. So what we have is hydronium ion concentration, and if I rearrange this equation, you can say that equals KA. So if the pH is not altered, that means the hydronium ion concentration is not altered. All right? So if we add an acid and we say the pH does not change, what does that really mean? It means that the hydronium ion concentration is not altered by much. So we know that hydronium ion concentration depends on KA, which is a constant, and then the ratio of the acid to the base. All right? So if the ratio of these two is not altered too much, H3O plus will remain unchanged. Is that clear? So if we want to look at a buffer, and we have substantial amounts of both, and if the pH is not going to change, pH depends on a hydronium ion concentration, and therefore if that ratio is not altered too much, we know that the hydronium ion concentration will not change that much. And so that's how buffers work. It turns out that in buffers, because you have substantial amounts of this and this, when you add an acid or a base, this ratio is not altered all that much. All right? And as long as the ratio is not altered, then the hydronium ion concentration does not change, and therefore the pH remains unchanged. All right? So let's look at what happens if you add an acid or a base. So now let's take this scenario, and now let's say that if... So let's look at the first case. Let's say if H3O plus is added. All right? In other words, let's say something like 0.1 molar HCl is added. Okay? So all of you recognize that HCl is a strong acid, and we know that a strong acid dissociates completely. So if I put 0.1 molar HCl, what is the concentration of the hydronium ion that I'm adding? 0.1 molar. All right? So all I'm doing is I'm adding 0.1 molar hydronium ion. Now if I take 0.1 molar hydronium ion in pure water, what is the pH of 0.1 molar hydronium ion? 0.1 is what? 10 to the what? Negative 1. Negative 1. If it's 10 to the negative 1, what is that? 1. So that means 0.1 molar, if you take neutral water and add 0.1 molar to neutral water, it goes from pH 7 to? To pH 1. pH 1, that's a big change. Remember, it's a factor. Every change in pH unit is a change by a factor of 10, or by one order of magnitude. So now it changes by seven orders of magnitude. So that's a pretty strong acid. Now what happens is if I take this and add it in a buffer, now the pH of the buffer does not change. If you take pure water, it goes from seven to one. But if you put it in a buffer, the pH is not altered all that much. And the reason is that you have this equilibrium that we've established. Now, so we have this equilibrium, H2O plus A minus, and this equilibrium is established. But now I'm adding hydronium ion. So what am I doing to that? If I look at that equilibrium, I'm adding hydronium ion to that solution. So I'm adding a large amount of hydronium ion in there. So what happens to that equilibrium? Is there a common ion there? Is hydronium ion there already in the equilibrium? So now if I'm adding more hydronium ion, what happens to that equilibrium? I've disturbed it. So now Leicester's principles are going to jump in, and it's going to counteract that stress. So what happens is that this equilibrium, so to counteract the addition, so to counteract the added hydronium ion, what happens is that the hydronium ion plus A minus will react to give you HA plus H2O. So aqueous, aqueous, this will be an aqueous, and that'll be aqueous. So essentially what happens is that you've added too much hydronium ion, the system is going to respond by this reacting with this to give you. In this way, what it does is it absorbs the extra hydronium ion that you added, and the system will go back to equilibrium. When the system goes back to equilibrium, what happens is that this ratio, once it goes back to equilibrium, this ratio is altered only slightly. So the way this buffer acts is that it absorbs the added acid and reacts with it and forms the weak acid, and when the system goes back to equilibrium, because KA is a constant, do you see that? The ratio of products to reactants have to be maintained, because that's how KA is a constant. And so what happens is when it goes back to equilibrium, this ratio is only altered very slightly, and therefore because the ratio is altered only slightly, the pH remains relatively unchanged. As a result of this reaction, the ratio HA over A minus is relatively unchanged. And so if that ratio is unchanged, then the hydronium ion concentration is not changed all that much. Now the second scenario is that we added, if we added base. So now we understand how a buffer works, and we understand that when you add an added acid because it disturbs that equilibrium and the extra hydronium ion is absorbed and consumed in this reaction to give you HA, and the ratio is not altered that much, we know that the hydronium ion concentration does not change all that much. Now what happens if you add a base? So the second scenario is B, if OH minus is added. So we're going to add a strong base, and let's say we're adding 0.1 molar NaOH. So we know NaOH, KOH, LIOH, lithium hydroxide, these are all group 1 metal hydroxides, these are all strong bases because they're salts, and the moment you put them in water, they're going to dissociate completely. And so if you put 0.1 molar NaOH, the result is you're adding 0.1 molar hydroxide. If you take 0.1 molar hydroxide, what is the pH of 0.1 molar hydroxide? Remember hydroxide, 0.1 means 10 to the negative 1. So PoH is 1. If PoH is 1, what is pH? 13. So you can see that if I add this amount of sodium hydroxide into pure water, pure water is pH 7, and you add 0.1 molar sodium hydroxide, it goes from pH 7 to pH 13. That's a big change. It's 7 orders of magnitude, all right, and therefore 6 orders of magnitude, rather. And so there's a big change. If you add the same amount of hydroxide into a buffer, the pH is altered only very slightly. And let's look at why the pH is altered very slightly. Because if you take this equilibrium, once again, the buffer establishes an equilibrium. And this is the equilibrium that the buffer establishes. Now we're adding OH-. So added OH- reacts with the weak acid. So what happens is that it turns out an acid and a base in large amounts cannot coexist because they'll neutralize each other. So we have weak acids in large amounts, and we added a strong base in large amounts. And if in solution you have a weak, if you have any acid weak or strong, and if you have a base in large amounts, they will neutralize each other. They can exist in small amounts, like in water you have hydronium ion at 10 to the negative 7, and you have hydroxide 10 to the negative 7. Those are really small amounts, all right? But if you have really substantial amounts, like 1 molar and 0.1 molar, then what happens is large amounts of acid and base cannot coexist. They'll neutralize each other. And so when they neutralize each other, what you end up with is water and A-. So what happens is that when you add a strong base, now you have lots of the acid, which is the weak acid, and you have lots of bases. And we know that if you have substantial amounts of an acid and a base, they're going to neutralize each other. And when they neutralize each other, they form water. And so what happens is this reaction takes place, and so you can see the added OH- is consumed to produce A-. And when the system establishes equilibrium, so when equilibrium is established again, then the ratio of acid to base is relatively unchanged. All right? So can you see that if you have an acid added, the acid reacts with a conjugate base. So for added acid, you need substantial amounts of the conjugate base. If you add hydronium ion, remember we said hydronium ion reacts with A- to give you HF. So added acids are absorbed by the conjugate base. If you add a strong base, like hydroxide, it reacts with what? The acid part. So you need substantial amounts of the acid to absorb the base, and that's why we need substantial amounts of both. Exactly right, everybody. The buffer has substantial amounts of the weak acid and substantial amounts of the conjugate base, which is a salt. And you need both of those because one deals with any extra acid that's added, one will absorb that. So if you add extra acid, the conjugate base will absorb it, and the system will go back to equilibrium. If you add a strong base, the weak acid will react with it and absorb that. And so, and then it'll go back to equilibrium. And the net result is regardless whether you add a strong acid or a strong base, the ratio always goes back to what it was because K is always a constant. And we know that when equilibrium is established, the ratio of products to reactants have to be maintained if K is going to remain a constant. That ratio has to be a constant. Exactly right, everybody? And so that's what happens and that's how buffers behave. So now that what I need you to be able to do is whenever you're given a problem, you have to figure out is this a buffer or not. All right? And what you'll notice is a buffer is very similar to the common ion effect problem that we dealt with before because in a common ion effect problem, you had a weak acid and you had a weak base which is a common ion for that equilibrium and the buffer is the same thing. So let's take an example of a buffer problem and what we're going to do is we're going to calculate the pH of the buffer and then we're going to add an acid and we're going to calculate how the pH changes with added acid or base. Now once again, there's nothing new here, it kind of conforms to all of the different types of equilibrium problem that we've been working out. All right? So let's start by looking at part A. So we say, suppose one mole of formic acid and point five moles of sodium formate are added to water and diluted to one liter. All right? So we can start by saying that we have formic acid and the initial concentration of formic acid is one mole over one liter. Okay? That concentration is moles divided by the volume and that gives me one molar concentration. We're told that we have sodium formate. So you need to be able to recognize that now what we have is the salt of the conjugate base and this is point five moles in one liter which gives me point five zero zero molar. So to begin with, you can see that I have a mixture of a weak acid and the salt of its conjugate base. So what is that screen back at you? Buffer. All right? If you just had the weak acid, it would just calculate the pH of a weak acid. Now you have a mixture of both. Now ideally, if the concentrations of both were the same, that would be the best buffer. Do you see that? But the reason I'm using different numbers is because when we solve the math, if I just use one for both of those, then it's hard for you to see the changes that are taking place. So I'm using two different numbers so that you can track the changes that take place for each one of those. And an ideal buffer would be where you had both of them in approximately equal concentrations. So we're taking an example where you can see the salt is half as much as the acid. Now when we look at the salt, we know that the moment you put the salt in water, if I take this and the moment I dissolve it, you know that it's going to dissociate to give me the conjugate base plus the cation. And we know the concentration, if the initial concentration of this is 0.5, we know that the concentration of the conjugate base would be 0.5 and we know that the concentration of the sodium cation should be 0.5 because for every unit of the combined entity where it dissociates, it will give you one unit of the anion and one unit of the cation. So if we start with 0.5 molar of the combined unit, we know that the parts that come from that combined unit will have 0.5 molar concentration each. So now, like any equilibrium problem, we have to start with the equilibrium that we're looking at. We have the weak acid, formic acid, and they've given us Ka. So we can write down the equilibrium for that. And so we start by saying you have formic acid in water giving me. Can you tell me what the products of this would turn out to be? This is acting as an acid. This is the proton donor. This is the proton acceptor. So what would I end up with? H3O plus HCOO minus and my Ka is 1.77 times 10 to the negative 4. So that's a weak acid. Now my initial concentrations would be, we just calculated that. I have formic acid which is one molar. Not concerned about water. Hydrogen in mine is going to be 0 to begin with. But now I also have the conjugate base. So this is what I would start with. I have combined a weak acid with its salt. So before at t equals 0, my initial concentrations would be that. Now if I look at the change, this would be negative x plus x plus x. They're all one to one stoichiometry. Before at equilibrium, I would have 1 minus x, x and 0.5 plus x. So now you guys know how to solve the rest of this. Now I have to figure out what x is. And x is hydronium ion concentration. So if I want to calculate the pH, then once I figure out what x is, that gives me hydronium ion concentration. I can figure out what the pH is. So we can go pretty fast on this part because we've worked this so often. So we know that Ka equals 1.77 times 10 to the negative 4, which gives me the hydronium ion concentration times the formate ion concentration divided by formic acid, which is x times 0.500 plus x divided by 1 minus x. Now what do I do next? Approximate it because Ka is 10 to the negative 4. It's small and therefore I can make my approximation. And my approximation is going to be that 0.5 plus x is 0.5. It's just so small we can neglect it, and if x is small compared to 0.1, obviously it's got to be small compared to 1. So this is really small compared to this. Now 1 is a bigger number, so x has to be really small compared to that as well. So now I can say that 1.77 times 10 to the negative 4 equals x times 0.5 divided by 1. And so I can solve for x, and if I solve for x, the number comes out to be 3.54 times 10 to the negative 4 moles per liter. Now I have to check validity. Is my approximation valid? And can everybody see that I just need to pick one of these? Alright, I don't need to calculate validity for both because if I take the top one, if I say x is small compared to 0.5, it has to be obviously small compared to 1. So I'm going to take 3.54 times 10 to the negative 4 divided by 0.5 molar times 100. And if you work that out, it comes out to 0.0708%. In other words, it's 0.07. That's well below our threshold. So we know our approximation is valid. So if I know that the approximation is valid, now I know that my hydronium ion concentration is 3.54 times 10 to the negative 4 molar. And if that is the case, I can convert that to pH, and pH comes out to be 3.451. I start with a three significant figure number, so I'm going to end up with three decimals, and that would be the pH of this buffer. So that was our first part of the problem. How do you calculate the pH of a buffer? And you can see that by setting up the equations and solving in the manner in which we solved every equilibrium problem, you can figure out what the pH of that solution is. Now let's go to the second part. Suppose 0.1 moles of a strong acid such as HCl is added to the above mixture, calculate the pH of the resulting solution. And now for part B, now we have to figure out what happens if you add a strong acid. So once again, don't get thrown off by this problem because it just looks complicated. It's actually if you just follow the same principles that we've followed before, we should be able to come to a rational conclusion. So let's start with the hydronium ion concentration. So we have, we're adding, the initial concentration of HCl would be 0.1 moles in one liter of the, because we know that the buffer is one liter. So we start with 0.1 mole, and what we end up with is 0.10 molar concentration of HCl. So we've calculated the initial concentration of HCl that we added. We know that that is equal to what? Because it's a strong acid, that would equal the amount of hydronium ion that we added. Okay? Now, I want you to think about how buffers behave. If you add a strong acid to the buffer, how is the strong acid consumed? It reacts with what? The conjugate base. All right? In this case, HClO-. All right? So what's going to happen is that we have a limiting range of problem to begin with. So what we have is that the added acid, let me rewrite. The HClO- will react with the added acid to give you plus H2O. Does that make sense? That's how the added acid is consumed so that the pH is not altered very much. So what happens is that the instant you add the strong acid, the system is no longer at equilibrium. It's been disturbed. All right? So when it's disturbed, now it's going to respond by counteracting that disturbance by reacting. The hydronium ion would react with the conjugate base to give you formic acid and water. So if you want to look at this initial, at the beginning, what I have is initial before any reaction occurs. What do I have? To begin with, how much formic acid concentration did we have? The formic ion concentration is what? Is 0.5. All right? So this is 0.5. We added hydronium ion. And how much hydronium ion did we add? 0.1 molar. All right? And over here, we had one molar of that. Remember the buffer to begin with had 0.5 conjugate base and one molar weak acid. Now we added the strong acid in its 0.1 molar. So that means afterwards, or in the end, that is after any reaction occurs, you will have. So can you guys look at this? Now this becomes a limiting range of problem. All right? Between these two, which is limiting? You have 0.5. Lots of this. They react in a one-to-one ratio. Can everybody see that this is actually limiting? If this is limiting and it's all consumed, that means if it reacts with this, how much of this would be left over? Can you see that? So if this is all consumed, that means if one mole is consumed, if one mole of this is consumed and one mole of this is consumed, you're going to form one mole of that. Here what's happening is that because essentially what happens is you have 0.5 minus 0.1. All right? I end up with a two-significant number here. So this is two significant figures, two decimals, three decimals. We're subtracting. So you end up with a number that is two decimals here. Got it? Now since this is limiting, this is all consumed. So on the product side, what would you have? Can everybody see you end up with 1.10? Because on this side, you'll have 1 plus 0.10. So essentially what we're doing is we're looking at the reaction and now the added acid has been consumed. So because the added acid has been consumed, concentrations of formate and formic acid have changed. All right? So now we go back to the equilibrium. So now this, now once the equilibrium is established, what we will have is plus H2O giving you H3O plus aqueous plus HCOO minus aqueous. You know KA is 1.77 times 10 to the negative 4. But now once this reaction is complete, you can see my initial concentration of formic acid is 1.10. Zero water, we're not concerned about water. My drunamine is zero to begin with and what I have is 0.40 of formate. So the change here would be negative x plus x plus x. Therefore at equilibrium, I have 1 minus, 1.1 minus x. Not concerned about that, x and this would be 0.4 plus x. All right? And so now I can say KA equals 1.77 times 10 to the negative 4, which is the hydrionamine concentration times the formate ion concentration divided by the concentration of formic acid, which gives me x times 0.4 plus x divided by 1.10 minus x. Now I make the approximation, which is I'm going to say 0.4 plus x approximates to 0.4, 1.10 minus x approximates to 1.10. And once I do that, I can go back and say that 1.77 times 10 to the negative 4 equals x times 0.40 divided by 1.10. I can move fast here because you know this is the general form that we've solved every acid-base equilibrium problem and you solve for x. And if you solve for x, that number comes out to be 4.87. So it would be 4.9 because I'm keeping this to two significant figures because when we did that subtraction, all right, we ended up with a number that was 2 decimals and so that comes to that. 4.9 times 10 to the negative 4 molar. All right? And then if I check validity, validity would be x divided by 0.4 times 100, which is 4.9 times 10 to the negative 4 molar divided by 0.40 molar times 100, and that comes out to be 0.12%. So we know approximation is valid. And so we know what x is, x is 4.9 times 10 to the negative 4 molar, therefore pH. So x equals the hydronium ion concentration. And so pH is the negative log of that and it comes out to be 3.31. All right? So let's look at our take home lesson. If you remember when we had the buffer, it was 3.4, 5. All right? Without adding any acid, the buffer was 3.45. I added 0.1 molar of acid and the pH is now 3.31. So you can see the pH was not altered all that much. If I take pure water and add the same amount, you can see the pH changes by 7 orders of magnitude. If you take pure water and add 0.1 molar, it changes from pH 7 to pH 1, which is a big change. But in the buffer, you see that the pH is only altered very slightly. All right? And that's how a buffer works. Thanks for watching.
special solutions, and the characteristics of third is that both should be in substantial minus aqueous. And the form in which this equilibrium. And this is the equilibrium that the general form that we've solved every acid-base