Lecture 12. Aqueous Equilibria Pt. 1.
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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. 
 Let's go ahead and start. Did anybody have any questions before I begin? All right, so if you guys remember, last time we started looking at acid-base titrations and we said the reason we're looking at acid-base titrations in class is because it allows us to apply all of the principles of acid-base chemistry that we've looked at previously. And so if we can do these calculations, it illustrates that we have a good grasp of all the underlying principles that we've been studying up to this point. And we said when you look at acid-base titrations, there are three types of acid-base titrations. You have strong acid-strong base or strong acid, sorry, weak acid-strong base or weak base-strong acid. And last time we took the first of this that was looking at strong acid versus strong base. And if you guys remember, the example that we took was that we put the strong base in the burette. We had 0.1 molar of the strong base in the burette and we put 100 milliliters of the strong acid in the flask. And as I said last time, usually if you're carrying this out in the laboratory, your goal is to figure out the concentration of the acid in the flask because remember you always put the solution with the unknown concentration in the flask. But because this is more of a mental challenge, what we want to do is we know the concentrations in the flask as well. Now we're going to calculate what the pH would be. Usually you measure the pH using the pH meter. Here what we're going to do is we're going to give you the concentration and volume in the flask as well and let's see whether we could calculate the pH that you would measure if you put a pH meter in there. And we took three scenarios. We calculated the pH of the strong acid before you added any base. Then we looked at a case where now we added 30 milliliters of the base. And so the second scenario was calculating the pH after 30 milliliters of the base had been added. And we said that once 30 milliliters of base had been added to the flask, now we had 130 milliliters. So the total volume in the flask after adding the 30 milliliters would be 130 milliliters. And then the last scenario that we looked at was what happens after you've passed the equivalence point. And so now we looked at a case where we had added 100 and 0.5 milliliters of base. And so the total volume would be 200 and 0.5 milliliters. Now last time when I wrote this down, the third scenario, I wrote down the total volume rather than give you the amount of base that had been added. So if you can turn back to your notes and make sure that you write down that the amount of base added was 100.05 milliliters. Because what I wrote down last time when we worked the problem was I gave you the total volume but I did not give you the amount of base that was added. So once you add 100.05 milliliters of base to the 100 milliliters of acid in the flask, your total volume is 200.05. So can you make that change in your notes and make sure you note down that we added 100.05 milliliters of the base and the number that I gave you in that problem is the total volume. And some of you were confused by that so I want to clarify that. Now today we're going to move on to looking at the second class and these are the ones where now we're going to look at where we have a weak acid versus strong base. So we're going to look at a titration where you have a strong base and let's say we have the same base which is 0.1 molar sodium hydroxide. But now in the flask instead of having HCl we're going to have a weak acid and the weak acid we're going to look at is 100 milliliters of 0.1 molar. Let's say we're looking at acidic acid and this is a weak acid because we know Ka equals 1.76 times 10 to the negative 5. So now we're going to do the same calculations that we did before but we're going to look at a weak acid. So before we start doing anything we need to know what kind of reactions are taking place. Remember we said an acid and a base in large amounts cannot coexist. They're going to give you water. So now what we have is instead of having a strong acid we know that dissociates completely we have a weak acid. What do we know about the weak acid? It does not dissociate completely. In fact only about 0.1% dissociates. So the major part of the weak acid almost 99.9% remains undissociated. So the major species in the weak acid will be the undissociated acid. So if we were to write down the complete ionic equation for the reaction that takes place we know that in the flask we have a weak acid and now the weak acid remains undissociated. 99.9% of the weak acid remains in this form. Only about 0.01% actually dissociates to give you the conjugate base and the hydronium ion. So the major species would be the weak acid. But now we're adding a strong base to this. So if I add a strong base that dissociates completely. So the instant I add sodium hydroxide I know sodium hydroxide is a strong base. It dissociates completely to give me that. So this is what I would have at the beginning. Before any reaction takes place at time 0 the instant you add the base to the acid at time 0 these are the species that will be there in solution. You're going to have the undissociated weak acid and the dissociated strong base. Now we know that an acid and a base cannot coexist. So we have a weak acid and we have a strong base and we have substantial amounts of both. So the weak acid and the strong base cannot coexist. They're going to give you water. What's going to happen? This hydroxide strong base will pull off this hydrogen atom here so that you end up with the conjugate base, water and Na+. So remember if you have a weak acid we said acids and bases in substantial amounts cannot coexist. They're going to react and neutralize each other to give you water. So you have a large amount of weak acid and you have a large amount of hydroxide so the hydroxide will react with the weak acid to give you water and its conjugate base. Now in this case if you look at the spectator ions you can see that there's only one spectator ion. This is the spectator ion and that's the spectator ion. So you can see that Na+, is the spectator ion. In the previous example we had two spectator ions. We had chloride ions and sodium ions as spectators but here we have only one spectator ion which is the sodium. And we said if you take the spectator ions out because remember they're there at the beginning, they're there at the end, they're not really involved in the reaction. So if you take the spectator ions out what you end up with is the net ionic equation. And the net ionic equation summarizes exactly what's going on. So you have a weak acid, it's going to react with a strong base and we know both of these can't coexist. They're going to neutralize each other to give you water. And so what we end up with is the conjugate base plus water. So simply put if you want to summarize the process or the reaction that actually takes place when you add a strong base to a weak acid, the net ionic equation summarizes it. The weak acid reacts to a strong base to give you conjugate base and water. Now this reaction, remember in titrations the reactions have to be clean, quantitative and rapid. And so this is a reaction that takes place completely, it takes place cleanly and it takes place quantitative. That means if you're looking at a limiting reactant and excess reactant and so on, this reaction would take place completely and what you end up with is the excess reactant left over. Now usually up to this point you would have seen the general reaction. And the general reaction is quite misleading. So if you look at the general reaction, you have to kind of read between the lines. If you look at the general reaction, reactant is the weak acid, base is the strong base sodium hydroxide giving you the salt, which is this, plus water. All right? But what the general reaction doesn't show you is that the strong base actually dissociates completely. The strong base doesn't stay as an AOH. We know that once the products are formed you end up with water and salt. So usually acid-base reactions result in the formation of water and salt. The salt that you form is the sodium salt of the acetate anion, but because it's a soluble salt you know that it dissociates completely. So even though you write it like that, we know that it has dissociated completely. All right? So this is what the reactions look like and overall the net ionic equation is what summarizes exactly what's going on in that solution. So now we want to do the same thing. We're going to calculate the pH when you add certain amounts of sodium hydroxide to it. But before that I want to show you what the titration curve looks like. Now to compare the two, we'll start by looking at the strong acid. So let me just make this smaller. So this is the strong acid versus strong base. This is what we've seen before for hydrochloric acid and NaOH. And we said that has an S-shaped curve and that's shown here. Now if instead of having the acid in the flask, if I have the base in the flask and I'm adding the strong acid, you can see now that the curve is the mirror image of this. So here to begin with, before you add any of the titrant, the pH is 1. If instead in the flask if you have a strong base, you can see now before any titrant is added all you have is 0.1 molar NaOH. 0.1 molar NaOH will have a pH of 13 because hydroxide ion is 10 to the negative 1. Therefore, a hydronium ion would be 10 to the negative 13 and if that would give you a pH of 13. You see that? So if you just look at the pH before you add any of the titrant, you know whether you have a strong acid or a strong base in the flask and you can see that this curve is actually a mirror image of that. And so the whole process is now reversed. Now we're going to move on to looking at a weak acid. Now that titration curve looks like this. So what you have here is this is the pH and over here the volume of base added and remember we said the shape of this curve, if it's a strong acid strong base, is an S-shaped curve of sigmoidal. So you see that for a strong acid you see that it is S-shaped. But if you have a weak acid you actually have two S's. See there's one S here and the second S here. So if you have a curve that has two S's in there, that means that now you're looking at a weak acid versus a strong base. So that's the difference between the plot for a strong acid versus a weak acid. In fact, it depends on, this curvature depends on the K of the weak acid. So here I have a picture that kind of shows you what this looks like. So way down here this represents the strong acid, that's the typical S-shaped curve. Now this is another strong acid over here but now you can see as the acids get weaker and weaker and weaker you can see that that curve here becomes more pronounced. And so the end result is that what I want you to remember is that if you take a weak acid like acidic acid and sodium hydroxide this is what that curve looks like. Okay, not very good. And if you want to see a picture of what this plot looks like if you were to collect the data in the laboratory this is what it looks like. So here what we have is once again a curve and these are actual data points that were collected. So you can see that as you add alkyl-tosodium hydroxide the dots represent the actual titration and you can see that there is a first curve here and then one like that. If it were like this that would be a strong acid, strong base. So this is a titration curve that corresponds to a weak acid versus a strong base. Now if you look at the equivalence point so this is the equivalence point and remember one of the things that we said was if you take a strong acid, strong base at the equivalence point the solution will be neutral because all the strong acid has been neutralized by all the strong base and all you have in that solution is salt and water. The salt is sodium chloride, the pH is neutral, water the pH is neutral therefore all you have in there is neutral solutions and therefore the way to tell where the equivalence point is you look for pH 7 and that will tell you the equivalence point. Now in this instance the equivalence point will always be greater than 7. The solution is going to be basic and the reason is that if you look at the net ionic equation so here we are if you look at this net ionic equation you know that the weak acid reacts with hydroxide to give you acetate ion. So can you see that here what we have is the conjugate base of a weak acid. At the equivalence point you have water which is neutral and you have the conjugate base of a weak acid. The conjugate base of a weak acid is acidic, basic or neutral. It's going to be basic and therefore the pH has to be greater than 7. Do you see that? So for weak acid strong base titrations at the equivalence point the pH will be basic. It has to be above 7 and you can see in this plot it's around 8. So this would be the equivalence point right here which is above 8. Now in this region you can see that it's almost flat. Can everybody see that that plot the pH hardly changes in this region. The reason that the pH hardly changes in this region is because as you add strong base to acidic acid what you're forming is acetate anion. So in this region what we have is a weak acid and it's conjugate base. Whenever you have a mixture of a weak acid and it's conjugate base in substantial amounts in approximately equal amounts what does it act like? A buffer. So can everybody see that in that region it's acting like a buffer because you have both the weak acid and you have the weak base and because both of them are coexisting that region is going to act like a buffer and if it acts like a buffer you can see that the pH hardly changes and so that's why we see this flat region here. This is a flat region where the pH hardly changes. Here is another example of that flat region where the pH hardly changes. So what we're going to be doing is now we're going to do these same calculations that we did with strong acids and strong bases. Now we're going to look at it in terms of weak acid strong base. And so we're going to go on the presumption that we have.1 molar sodium hydroxide in there. We're going to have 100 milliliters of.1 molar acidic acid in the flask. So let's start calculating this. So once again we have to go through the whole process. So we start by looking at the first scenario where we calculate the pH before any base is added. Before any base is added what we have is.1 molar acidic acid and we know that the KA for the acid is 1.76 times 10 to the negative 5. Now all of you know how to calculate the pH of a weak acid. How do we do that? Remember that's the first kind of problems that we work. We have acidic acid. You can set up the equilibrium. So what we have is we have acidic acid aqueous in water giving me the... so this is the acid, this is the base, this is the proton donor, this is the proton acceptor. So what you end up with is the conjugate base and hydronium ion. All of you remember that. Okay and we know KA is 1.76 times 10 to the negative 5. We say at the beginning the concentration of acidic acid is.1 molar. We're not interested in water. Zero, zero. The change would be minus x, not interested in water, plus x, plus x. At equilibrium we know this would be.1 minus x, we know this would be x, we know that KA equals 1.76 times 10 to the negative 5 which equals the concentration of the acetate ions times the concentration of the hydronium ion divided by the concentration of acidic acid which is x squared over.1 minus x. We can approximate that to x squared divided by.1. You guys know the process. So if you solve for this and validate your approximation and so on, I'm not going to go through the whole process because we've worked problems like this so many times and you know how to calculate the pH of a weak acid. But we know if you calculate x, x gives us the hydronium ion concentration which turns out to be 1.33 times 10 to the negative 3 molar and therefore if you calculate the pH of that solution it comes out to be 2.877. So I want you to try this at home and make sure you know the process. So all we have is a weak acid in the flask and we know how to calculate the pH of a weak acid once you know the concentration. So that's the pH before we add any base. Now we're going to go on to the second scenario where we look at what happens when we add 30 milliliters. So the second scenario is we're going to calculate pH after 30 milliliters of NaOH is added. Now we said in these titrations because the volume is constantly changing because we're calculating the pH in the flask and because we're incrementally adding sodium hydroxide and because the volume is changing. In other words, you're diluting that solution we need to, we can't think of everything in concentration because concentration is moles divided by the volume and if the volume keeps changing that means the concentration is changing. And therefore in these calculations we always calculate the number of moles because the number of moles of the acid does not change. As the reaction takes place you can figure out the number of moles of the acid that's consumed and then we can figure out the number of moles that's left over and if we can figure out the number of moles then we can calculate the new concentration. So we're going to use the same strategy that we used last time. So we're going to start by calculating the initial number of moles of acidic acid. So we want to know how many moles of the acid are actually in the flask before any reaction takes place. And that is 0.100 moles per liter times we had 100 milliliters which is 0.100 liters. We're taking everything in three significant figures. So that comes out to 1 times 10 to the negative 2 moles of acidic acid. So before any reaction takes place this is the number of moles of acidic acid that you had. Okay, likewise we're going to calculate the initial number of moles of hydroxide ion that we added. So this is the number of moles of hydroxide that's in the 30 milliliters that you determined. Now how do I convert concentration to moles? Remember we added 30 milliliters of NaOH to the flask. So if I want to calculate the number of moles of hydroxide in the 30 milliliters what do I need to know? How do I calculate moles if I have the concentration? Times the volume. Okay, so I'm going to take concentration times volume. So I have 0.1 moles per liter of sodium hydroxide times I added 30 milliliters so that becomes it goes to three significant figures. So it becomes 0.03 liters and that gives me 3 times 10 to the negative 3 moles of hydroxide. So what I've calculated is the number of moles of each of the reactants before any reaction took place. Now I need to figure out after this reaction is completed how much each component do I have. So in order to do this I need to write the net ionic equation. What was the net ionic equation for this? We know that it's a weak acid so we start with a weak acid. So we have acidic acid which is the weak acid. Okay, the other reactant was hydroxide ion. It reacts cleanly, rapidly and quantitatively to give me the conjugate base which is the acetate anion plus water. So we're going to start by saying before any reaction takes place what I have is I just calculated that I have 1 times 10 to the negative 2 moles of acidic acid and that I have 3 times 10 to the negative 3 moles of hydroxide and I have none of that. Okay, I'm not concerned about water because water is the solvent. So now we're going to let the reaction takes place. So what we have is after reaction occurs. Now this is a limiting reactant problem. You can see that this is excess, this is limiting. So that means that if I subtract this from this what I end up with is 7 times 10 to the negative 3 moles. All of this is consumed and the reason it's consumed is because this is the limiting reactant and this is the excess reactant. So if all of this reacts you can see on the product side I end up with 3 times 10 to the negative 3 moles. So once this reaction has taken place we know that we have some unreacted acidic acid left over and we formed the conjugate base. So now what do I have there? If I have a weak acid in this conjugate base how is that system going to act? If you have a weak acid and it's conjugate base together in approximately equal amounts what do we know? It's going to act as a buffer. If it's a buffer how do you calculate the pH of a buffer? You can use the Henderson-Hasselbalch equation. So now I know the moles of the acidic acid. I know the moles of its conjugate base. If I'm going to apply the Henderson-Hasselbalch equation I need to know concentration. So how do I convert moles to concentration? Divide by the volume. What is my volume? 100 milliliters I added. 30. So now my total volume is what? 130. So now I'm going to go calculate the new concentration. So I'm going to start by calculating the concentration of the excess reactant that's left over which is the moles divided by the new volume which is 130 liters of 0.13 liters or 130 milliliters and that comes out to be 5.38 times 10 to the negative 3, 10 to the negative 2 molar and then I'm going to calculate its conjugate base which is 3 times 10 to the negative 3 moles divided by 0.13 liters which gives me 2.31 times 10 to the negative 2 molar. So now I know the concentration of the weak acid and its conjugate base. Now I can say pH equals pKa minus log the concentration of the acid form divided by the concentration of the acetate anion after the titration is completed. So we're going to assume that that's our new initial and so this is acidic acid so this is 4.754 minus log concentration of the acid is 5.38 times 10 to the negative 2 divided by the concentration of the acetate which is 2.31 times 10 to the negative 2 and so that gives me a pH of 4.387. So now we've calculated what the pH of that solution is in that flask. Can everybody see that what we're doing is the same stuff that we've done before except that now we have a lot more things going on and I want you to be able to understand this at the same level. There's a lot of detail there and you need to kind of understand that in this instance what's happening is there's a reaction taking place after that reaction is complete we want to calculate what are the species in that solution in the flask and then we need to calculate the concentration of those and from that we can figure out what the pH is. Now as homework can I ask you guys to calculate what the pH of that solution would be if you've added let's say 45 milliliters. If you added 30 milliliters now I want you to try to calculate what would the pH of that solution turn out to be if you have 45 milliliters. Now we're going to look at the next scenario which is at the equivalence point. The next one would be so we looked at after you add 30 milliliters now let's look at pH at the equivalence point. So we need to calculate what the equivalence point is remember what is the equivalence point we call this equivalence point or the stoichiometric point and what does that represent? Somebody tell me what happens at the equivalence point? Equal amount in other words the number of moles of the acid should equal the number of moles of the base in other words all of the acid has been completely neutralized by the base and usually when you calculate the equivalence point the underlying principle is that the number of moles in this case the number of moles of acidic acid should equal the number of moles of hydroxide that you have in that solution. They have both reacted in a one to one stoichiometry because the balanced equation shows us that they react in a one to one ratio. So if all of the weak acid has been neutralized by the strong base then you have to have the same number of moles of both and we know that moles is what? Concentration times volume. So on this side we have concentration of acidic acid times the volume of acidic acid and over here you have concentration of hydroxide which is sodium hydroxide times the volume of hydroxide which is the volume of sodium hydroxide that you added. So at the equivalence point if I want to calculate the volume of hydroxide that's there that would be the concentration of acidic acid times the volume of acidic acid divided by the concentration of hydroxide which is we know concentration of acidic acid is 0.1 moles per liter times the volume was that we had 100 milliliters and we know that we had 0.1 moles per liter of base so can you see that your volume of the base should be 0.1 milliliters. Now sometimes in the laboratory you guys remember sometimes in the lab they give you this equation M1 V1 equals M2 V2 this is the same thing. M is the molarity, volume is the volume of acid that you added and then M2 V2 would be the molarity of the base times the volume of the base that you added. You have to be careful because you can only apply that if the acid and the base react in a one to one ratio alright so that you have one acidic proton alright and therefore the hydroxide will react with that one acidic proton and therefore it's a one to one ratio alright so what do we know? How much base do I need to add at the equivalent point? You can see that I have to add 100 milliliters alright so if I add when I've added 100 milliliters base I would be at the equivalent point alright so if I want to calculate pH now I've got to look at this so pH at the equivalent point is remember the reaction we're looking at is the weak acid reacting with the base to give you the conjugate base plus liquid water okay and we said before reaction we had one times 10 to the negative 2 moles remember in the flask when you have 100 milliliters of.1 molar that's the number of moles of acidic acid that you had to start with so that was 100 milliliters times.1 molar. Now we had at the equivalent point we have to have 100 milliliters so can somebody tell me how many moles of hydroxide did I add? We had 100 milliliters.1 molar in AOH so what would be the number of moles of NaOH? What do you think? Louder it's the same thing do you understand that because at the equivalent point the number of moles have to be the same you see the moles of acid have to be the same as the moles of base so good so you should see that that you're at the equivalent point so you're going to have the same number of moles so can everybody see that this would be 10 to the negative 2 moles and we're going to have 0 of this alright now after reaction so now the reaction is going to take place this is the titration you've added the base to the acid they cannot coexist they're going to react cleanly rapidly and quantitatively so the reaction is going to go to completion so now can you see these are in the same amount so if all of this reacts with all of that you're going to end up with 1 times 10 to the negative 2 moles of the product so remember we said at the equivalent point all the acid has been used up all the base has been used up so what are we left in solution at the end of the titration the conjugate base so now we need to calculate the pH of the conjugate base how do you calculate the pH of a weak base alright so I need the concentration now alright so I'm going to take the number of moles and calculate the concentration and therefore that would be the concentration of the conjugate base which is 1 times 10 to the negative 2 moles divided by the volume what is my volume I had 100 millilitres in the flask what's the volume of base that I added 100 so what is my total volume 200 okay so I'm going to divide by the total volume which is 0.200 liters which gives me 0.5 00 moles per liter so now we go back to working remember this is a conjugate base how do I calculate the pH of this is actually a salt can you see that it's the it's the salt of the conjugate base of the weak acid acidic acid I know the initial concentration so can everybody write down how you would calculate the pH of that solution this is applying what we did like that the first few weeks of class what so if you're going to calculate the pH of a weak base what is the starting point what do you need to write down what do you start with what do you need to write down how do you calculate the pH of a weak base all of you did really well on the test so you need to know this you have to draw the equilibrium and use the ice chart I see that so so we're going to write down the equilibrium first so this is acidic acetate anion which is the salt of the conjugate base in water giving me the weak acid what would I put here OH minus great and this represents KB did they give me KB no they give me all the KA so I need to convert KA to KB therefore KB equals one times 10 to negative 14 divided by 1.76 times 10 to negative 5 which gives me the KB which is 5.68 times 10 to negative 10 you guys understand that I can move really fast because we've done many problems of this type once again we're just applying what we've learned before okay so if I want to write the ice chart I start with my initial concentration would be point zero five zero zero I'm not concerned about water I have zero zero to begin with but I want to keep figure out what the changes would be negative x plus x plus x not concerned about water at equilibrium I know this will be point zero five zero minus x x x okay so we're looking at KB because this is the formation of hydroxide from the weak conjugate base therefore KB equals 5.68 times 10 to negative 10 which is concentration of the acidic acid times the concentration of hydroxide divided by the concentration of the weak base alright so that gives me x squared over point zero five zero zero minus x alright and as all of you know we can approximate this to x squared over point zero five zero now since you guys know how to solve for this I would like you to try this at home you have to check the validity of your approximation but if you calculate x after you check for a validity of approximation and so on x will come out to be 5.33 times 10 to the negative 6 molar so we know that the hydroxide ion concentration equals x which is 5.33 times 10 to the negative 6 molar remember we need to calculate pH so I need to convert that to hydronium ion concentration which is Kw divided by the hydroxide ion concentration which is 1 times 10 to the negative 14 divided by 5.33 times 10 to the negative 6 molar this will be molar squared and therefore if you work that out it comes out to be 1.88 times 10 to the negative 9 molar therefore pH equals 8.727 remember we said at the equivalence point the pH has to be greater than 7 because that solution is basic alright and so we figured out what the pH of that solution is at the equivalence point okay now I need you to be able to calculate so if you look at the fourth scenario you need to be able to calculate pH after 100.05 milliliters of NaOH is added okay now the solution to this is very similar to the strong acid because once we pass the equivalence point now all of the acid has been consumed completely so now all you have is a strong base alright and so the solution for this part will be exactly the same as the solution for the previous part because what you have is now at the equivalence point you have added 100 milliliters of the base now you have an excess of 0.05 alright so in that solution now you have a strong base and a weak base and the strong base is much stronger than the weak base you see that so the major contribution is going to come from the strong base so if you have 0.05 milliliters of excess hydroxide all you have to do is calculate the number of moles in 0.05 milliliters of hydroxide divide by the new total volume which is 200.05 and that gives you the concentration of the strong base and because the strong base dissociates completely you know what the hydroxide concentration is and so you know the hydroxide concentration that comes from that extra 0.05 that was not neutralized alright and you can calculate what the pH of that solution is okay so you should be able to solve these kinds of problems and in discussion for next week actually it's this week isn't it the discussions actually get you to solve some of these problems I need you to be able to look at these different types of problems and to be able to solve them okay now a couple of things I want to point out so we're going to go back to this graph and I want to point out some things on this graph so if you look at this titration curve this is the point at which there is no base added so you should be able to calculate the pH of that weak acid okay we know this region where it flattens out is called the buffer region okay so in this region where it's flat we call that the buffer region because that's the region where it's going to act as a buffer so whenever you have mixtures of the weak acid and it's conjugate base it's going to act as a buffer okay now this is the equivalence point alright and we know that the equivalence point the pH will be greater than 7 okay now if you take we already calculated that you have to add 100 milliliters to reach the equivalence point so exactly half of that is called the half equivalence point alright and at the half equivalence point what you realize is so at the half equivalence point what you realize is that the concentration of the acidic acid that's left over equals the concentration sorry CH3CO2 minus can I just rewrite that okay because it's not very clear and I don't want to rush that so at the half equivalence point we know that the concentration of the acidic acid that's left over in the flask the excess amount should equal the concentration of the acetate ion alright so you know that pH equals pKa minus log of the acid over the base and these two are equal to each other what do we know pH equals pKa minus log 1 which is pKa minus 0 so we know at the half equivalence point pH should equal pKa so if you're at the half equivalence point so you figured out that the equivalence point is 100 milliliters if you're at the half equivalence point that's the perfect buffer you have equal amounts of both alright and therefore the pH should equal pKa so if that's the case you don't need to do any calculations if you get a question saying you know the point in the equivalence point they ask you what is the pH of the solution at this point you don't have to do any math because you know pH equals pKa and this is acidic acid so pKa acidic acid is what 4.75 so you know that the pH at that point will be 4.75 okay now I want to finish up by just quickly going through this question so this is an old question from an older exam so I just pulled it this morning and I want you guys to take a look at this and I'll put a copy of this on the class website okay so why is that here we are so the question is consider the titration of a generic weak acid with a strong base that gives the following titration curve so you have already seen this curve you know that it's a weak acid strong base curve okay on the curve indicate the points that correspond to the following the first is the equivalence point so all of you know that the equivalence point of the stoichiometric point has to be this got it so that's the equivalence point the region with maximum buffering so you want to show an arrow that kind of shows where this is plateaus out and flattens out this is the region of maximum buffering somewhere between this and this that region got it pH equals pKa where would that be you look at the equivalence point it's a little above 25 all right and it's 30 here so it's somewhere between so it's it's actually somewhere here right in the middle so you have to look at this value and divide that exactly by half and that will give you where pH equals pKa and that will be that point there okay pH depends only on the initial concentration so this should be a subscript okay and so if the pH depends on only the initial concentration of the weak acid where would that be zero before you add any base that will be the point at which okay pH depends only on the conjugate base at the equivalence point because the equivalence point is the only point at which you have exclusively just the conjugate base lastly pH depends only on the amount of excess strong base added where would that be where does the pH depend on the amount of strong base anything past the equivalence point so once you pass this all of these points once you pass this anything above that will depend on the concentration of the hydroxide okay so we'll stop there for today and next time we'll move on to the next topic.
 Let's go ahead and start.  Did anybody have any questions before I begin?  All right, so if you guys remember, last time we started looking at acid-base titrations  and we said the reason we're looking at acid-base titrations in class is because it allows us  to apply all of the principles of acid-base chemistry that we've looked at previously.  And so if we can do these calculations, it illustrates that we have a good grasp of all  the underlying principles that we've been studying up to this point.  And we said when you look at acid-base titrations, there are three types of acid-base titrations.  You have strong acid-strong base or strong acid, sorry, weak acid-strong base or weak  base-strong acid.  And last time we took the first of this that was looking at strong acid versus strong base.  And if you guys remember, the example that we took was that we put the strong base in  the burette.  We had 0.1 molar of the strong base in the burette and we put 100 milliliters of the  strong acid in the flask.  And as I said last time, usually if you're carrying this out in the laboratory, your  goal is to figure out the concentration of the acid in the flask because remember you  always put the solution with the unknown concentration in the flask.  But because this is more of a mental challenge, what we want to do is we know the concentrations  in the flask as well.  Now we're going to calculate what the pH would be.  Usually you measure the pH using the pH meter.  Here what we're going to do is we're going to give you the concentration and volume in  the flask as well and let's see whether we could calculate the pH that you would measure  if you put a pH meter in there.  And we took three scenarios.  We calculated the pH of the strong acid before you added any base.  Then we looked at a case where now we added 30 milliliters of the base.  And so the second scenario was calculating the pH after 30 milliliters of the base had  been added.  And we said that once 30 milliliters of base had been added to the flask, now we had 130  milliliters.  So the total volume in the flask after adding the 30 milliliters would be 130 milliliters.  And then the last scenario that we looked at was what happens after you've passed the  equivalence point.  And so now we looked at a case where we had added 100 and 0.5 milliliters of base.  And so the total volume would be 200 and 0.5 milliliters.  Now last time when I wrote this down, the third scenario, I wrote down the total volume  rather than give you the amount of base that had been added.  So if you can turn back to your notes and make sure that you write down that the amount  of base added was 100.05 milliliters.  Because what I wrote down last time when we worked the problem was I gave you the total  volume but I did not give you the amount of base that was added.  So once you add 100.05 milliliters of base to the 100 milliliters of acid in the flask,  your total volume is 200.05.  So can you make that change in your notes and make sure you note down that we added  100.05 milliliters of the base and the number that I gave you in that problem is the total  volume.  And some of you were confused by that so I want to clarify that.  Now today we're going to move on to looking at the second class and these are the ones  where now we're going to look at where we have a weak acid versus strong base.  So we're going to look at a titration where you have a strong base and let's say we have  the same base which is 0.1 molar sodium hydroxide.  But now in the flask instead of having HCl we're going to have a weak acid and the weak  acid we're going to look at is 100 milliliters of 0.1 molar.  Let's say we're looking at acidic acid and this is a weak acid because we know Ka equals  1.76 times 10 to the negative 5.  So now we're going to do the same calculations that we did before but we're going to look  at a weak acid.  So before we start doing anything we need to know what kind of reactions are taking place.  Remember we said an acid and a base in large amounts cannot coexist.  They're going to give you water.  So now what we have is instead of having a strong acid we know that dissociates completely  we have a weak acid.  What do we know about the weak acid?  It does not dissociate completely.  In fact only about 0.1% dissociates.  So the major part of the weak acid almost 99.9% remains undissociated.  So the major species in the weak acid will be the undissociated acid.  So if we were to write down the complete ionic equation for the reaction that takes place  we know that in the flask we have a weak acid and now the weak acid remains undissociated.  99.9% of the weak acid remains in this form.  Only about 0.01% actually dissociates to give you the conjugate base and the hydronium ion.  So the major species would be the weak acid.  But now we're adding a strong base to this.  So if I add a strong base that dissociates completely.  So the instant I add sodium hydroxide I know sodium hydroxide is a strong base.  It dissociates completely to give me that.  So this is what I would have at the beginning.  Before any reaction takes place at time 0 the instant you add the base to the acid at  time 0 these are the species that will be there in solution.  You're going to have the undissociated weak acid and the dissociated strong base.  Now we know that an acid and a base cannot coexist.  So we have a weak acid and we have a strong base and we have substantial amounts of both.  So the weak acid and the strong base cannot coexist.  They're going to give you water.  What's going to happen?  This hydroxide strong base will pull off this hydrogen atom here so that you end up with  the conjugate base, water and Na+.  So remember if you have a weak acid we said acids and bases in substantial amounts cannot  coexist.  They're going to react and neutralize each other to give you water.  So you have a large amount of weak acid and you have a large amount of hydroxide so the  hydroxide will react with the weak acid to give you water and its conjugate base.  Now in this case if you look at the spectator ions you can see that there's only one spectator  ion.  This is the spectator ion and that's the spectator ion.  So you can see that Na+, is the spectator ion.  In the previous example we had two spectator ions.  We had chloride ions and sodium ions as spectators but here we have only one spectator ion which  is the sodium.  And we said if you take the spectator ions out because remember they're there at the  beginning, they're there at the end, they're not really involved in the reaction.  So if you take the spectator ions out what you end up with is the net ionic equation.  And the net ionic equation summarizes exactly what's going on.  So you have a weak acid, it's going to react with a strong base and we know both of these  can't coexist.  They're going to neutralize each other to give you water.  And so what we end up with is the conjugate base plus water.  So simply put if you want to summarize the process or the reaction that actually takes  place when you add a strong base to a weak acid, the net ionic equation summarizes it.  The weak acid reacts to a strong base to give you conjugate base and water.  Now this reaction, remember in titrations the reactions have to be clean, quantitative  and rapid.  And so this is a reaction that takes place completely, it takes place cleanly and it  takes place quantitative.  That means if you're looking at a limiting reactant and excess reactant and so on, this  reaction would take place completely and what you end up with is the excess reactant left  over.  Now usually up to this point you would have seen the general reaction.  And the general reaction is quite misleading.  So if you look at the general reaction, you have to kind of read between the lines.  If you look at the general reaction, reactant is the weak acid, base is the strong base  sodium hydroxide giving you the salt, which is this, plus water.  All right?  But what the general reaction doesn't show you is that the strong base actually dissociates  completely.  The strong base doesn't stay as an AOH.  We know that once the products are formed you end up with water and salt.  So usually acid-base reactions result in the formation of water and salt.  The salt that you form is the sodium salt of the acetate anion, but because it's a  soluble salt you know that it dissociates completely.  So even though you write it like that, we know that it has dissociated completely.  All right?  So this is what the reactions look like and overall the net ionic equation is what summarizes  exactly what's going on in that solution.  So now we want to do the same thing.  We're going to calculate the pH when you add certain amounts of sodium hydroxide to it.  But before that I want to show you what the titration curve looks like.  Now to compare the two, we'll start by looking at the strong acid.  So let me just make this smaller.  So this is the strong acid versus strong base.  This is what we've seen before for hydrochloric acid and NaOH.  And we said that has an S-shaped curve and that's shown here.  Now if instead of having the acid in the flask, if I have the base in the flask and I'm adding  the strong acid, you can see now that the curve is the mirror image of this.  So here to begin with, before you add any of the titrant, the pH is 1.  If instead in the flask if you have a strong base, you can see now before any titrant is  added all you have is 0.1 molar NaOH.  0.1 molar NaOH will have a pH of 13 because hydroxide ion is 10 to the negative 1.  Therefore, a hydronium ion would be 10 to the negative 13 and if that would give you  a pH of 13.  You see that?  So if you just look at the pH before you add any of the titrant, you know whether you have  a strong acid or a strong base in the flask and you can see that this curve is actually  a mirror image of that.  And so the whole process is now reversed.  Now we're going to move on to looking at a weak acid.  Now that titration curve looks like this.  So what you have here is this is the pH and over here the volume of base added and remember  we said the shape of this curve, if it's a strong acid strong base, is an S-shaped curve  of sigmoidal.  So you see that for a strong acid you see that it is S-shaped.  But if you have a weak acid you actually have two S's.  See there's one S here and the second S here.  So if you have a curve that has two S's in there, that means that now you're looking  at a weak acid versus a strong base.  So that's the difference between the plot for a strong acid versus a weak acid.  In fact, it depends on, this curvature depends on the K of the weak acid.  So here I have a picture that kind of shows you what this looks like.  So way down here this represents the strong acid, that's the typical S-shaped curve.  Now this is another strong acid over here but now you can see as the acids get weaker  and weaker and weaker you can see that that curve here becomes more pronounced.  And so the end result is that what I want you to remember is that if you take a weak  acid like acidic acid and sodium hydroxide this is what that curve looks like.  Okay, not very good.  And if you want to see a picture of what this plot looks like if you were to collect the  data in the laboratory this is what it looks like.  So here what we have is once again a curve and these are actual data points that were  collected.  So you can see that as you add alkyl-tosodium hydroxide the dots represent the actual titration  and you can see that there is a first curve here and then one like that.  If it were like this that would be a strong acid, strong base.  So this is a titration curve that corresponds to a weak acid versus a strong base.  Now if you look at the equivalence point so this is the equivalence point and remember  one of the things that we said was if you take a strong acid, strong base at the equivalence  point the solution will be neutral because all the strong acid has been neutralized by  all the strong base and all you have in that solution is salt and water.  The salt is sodium chloride, the pH is neutral, water the pH is neutral therefore all you  have in there is neutral solutions and therefore the way to tell where the equivalence point  is you look for pH 7 and that will tell you the equivalence point.  Now in this instance the equivalence point will always be greater than 7.  The solution is going to be basic and the reason is that if you look at the net ionic  equation so here we are if you look at this net ionic equation you know that the weak  acid reacts with hydroxide to give you acetate ion.  So can you see that here what we have is the conjugate base of a weak acid.  At the equivalence point you have water which is neutral and you have the conjugate base  of a weak acid.  The conjugate base of a weak acid is acidic, basic or neutral.  It's going to be basic and therefore the pH has to be greater than 7.  Do you see that?  So for weak acid strong base titrations at the equivalence point the pH will be basic.  It has to be above 7 and you can see in this plot it's around 8.  So this would be the equivalence point right here which is above 8.  Now in this region you can see that it's almost flat.  Can everybody see that that plot the pH hardly changes in this region.  The reason that the pH hardly changes in this region is because as you add strong base to  acidic acid what you're forming is acetate anion.  So in this region what we have is a weak acid and it's conjugate base.  Whenever you have a mixture of a weak acid and it's conjugate base in substantial amounts  in approximately equal amounts what does it act like?  A buffer.  So can everybody see that in that region it's acting like a buffer because you have both  the weak acid and you have the weak base and because both of them are coexisting that region  is going to act like a buffer and if it acts like a buffer you can see that the pH hardly  changes and so that's why we see this flat region here.  This is a flat region where the pH hardly changes.  Here is another example of that flat region where the pH hardly changes.  So what we're going to be doing is now we're going to do these same calculations that we  did with strong acids and strong bases.  Now we're going to look at it in terms of weak acid strong base.  And so we're going to go on the presumption that we have.1 molar sodium hydroxide in  there.  We're going to have 100 milliliters of.1 molar acidic acid in the flask.  So let's start calculating this.  So once again we have to go through the whole process.  So we start by looking at the first scenario where we calculate the pH before any base  is added.  Before any base is added what we have is.1 molar acidic acid and we know that the KA  for the acid is 1.76 times 10 to the negative 5.  Now all of you know how to calculate the pH of a weak acid.  How do we do that?  Remember that's the first kind of problems that we work.  We have acidic acid.  You can set up the equilibrium.  So what we have is we have acidic acid aqueous in water giving me the... so this is the  acid, this is the base, this is the proton donor, this is the proton acceptor.  So what you end up with is the conjugate base and hydronium ion.  All of you remember that.  Okay and we know KA is 1.76 times 10 to the negative 5.  We say at the beginning the concentration of acidic acid is.1 molar.  We're not interested in water.  Zero, zero.  The change would be minus x, not interested in water, plus x, plus x.  At equilibrium we know this would be.1 minus x, we know this would be x, we know that KA  equals 1.76 times 10 to the negative 5 which equals the concentration of the acetate ions  times the concentration of the hydronium ion divided by the concentration of acidic acid  which is x squared over.1 minus x.  We can approximate that to x squared divided by.1.  You guys know the process.  So if you solve for this and validate your approximation and so on, I'm not going to  go through the whole process because we've worked problems like this so many times and  you know how to calculate the pH of a weak acid.  But we know if you calculate x, x gives us the hydronium ion concentration which turns  out to be 1.33 times 10 to the negative 3 molar and therefore if you calculate the pH of that  solution it comes out to be 2.877.  So I want you to try this at home and make sure you know the process.  So all we have is a weak acid in the flask and we know how to calculate the pH of a weak  acid once you know the concentration.  So that's the pH before we add any base.  Now we're going to go on to the second scenario where we look at what happens when we add  30 milliliters.  So the second scenario is we're going to calculate pH after 30 milliliters of NaOH is added.  Now we said in these titrations because the volume is constantly changing because we're  calculating the pH in the flask and because we're incrementally adding sodium hydroxide  and because the volume is changing.  In other words, you're diluting that solution we need to, we can't think of everything in  concentration because concentration is moles divided by the volume and if the volume keeps  changing that means the concentration is changing.  And therefore in these calculations we always calculate the number of moles because the  number of moles of the acid does not change.  As the reaction takes place you can figure out the number of moles of the acid that's  consumed and then we can figure out the number of moles that's left over and if we can figure  out the number of moles then we can calculate the new concentration.  So we're going to use the same strategy that we used last time.  So we're going to start by calculating the initial number of moles of acidic acid.  So we want to know how many moles of the acid are actually in the flask before any reaction  takes place.  And that is 0.100 moles per liter times we had 100 milliliters which is 0.100 liters.  We're taking everything in three significant figures.  So that comes out to 1 times 10 to the negative 2 moles of acidic acid.  So before any reaction takes place this is the number of moles of acidic acid that you  had.  Okay, likewise we're going to calculate the initial number of moles of hydroxide ion that  we added.  So this is the number of moles of hydroxide that's in the 30 milliliters that you determined.  Now how do I convert concentration to moles?  Remember we added 30 milliliters of NaOH to the flask.  So if I want to calculate the number of moles of hydroxide in the 30 milliliters what do  I need to know?  How do I calculate moles if I have the concentration?  Times the volume.  Okay, so I'm going to take concentration times volume.  So I have 0.1 moles per liter of sodium hydroxide times I added 30 milliliters so that becomes  it goes to three significant figures.  So it becomes 0.03 liters and that gives me 3 times 10 to the negative 3 moles of hydroxide.  So what I've calculated is the number of moles of each of the reactants before any reaction  took place.  Now I need to figure out after this reaction is completed how much each component do I  have.  So in order to do this I need to write the net ionic equation.  What was the net ionic equation for this?  We know that it's a weak acid so we start with a weak acid.  So we have acidic acid which is the weak acid.  Okay, the other reactant was hydroxide ion.  It reacts cleanly, rapidly and quantitatively to give me the conjugate base which is the  acetate anion plus water.  So we're going to start by saying before any reaction takes place what I have is I just  calculated that I have 1 times 10 to the negative 2 moles of acidic acid and that I have 3 times  10 to the negative 3 moles of hydroxide and I have none of that.  Okay, I'm not concerned about water because water is the solvent.  So now we're going to let the reaction takes place.  So what we have is after reaction occurs.  Now this is a limiting reactant problem.  You can see that this is excess, this is limiting.  So that means that if I subtract this from this what I end up with is 7 times 10 to the  negative 3 moles.  All of this is consumed and the reason it's consumed is because this is the limiting reactant  and this is the excess reactant.  So if all of this reacts you can see on the product side I end up with 3 times 10 to the  negative 3 moles.  So once this reaction has taken place we know that we have some unreacted acidic acid left  over and we formed the conjugate base.  So now what do I have there?  If I have a weak acid in this conjugate base how is that system going to act?  If you have a weak acid and it's conjugate base together in approximately equal amounts  what do we know?  It's going to act as a buffer.  If it's a buffer how do you calculate the pH of a buffer?  You can use the Henderson-Hasselbalch equation.  So now I know the moles of the acidic acid.  I know the moles of its conjugate base.  If I'm going to apply the Henderson-Hasselbalch equation I need to know concentration.  So how do I convert moles to concentration?  Divide by the volume.  What is my volume?  100 milliliters I added.  30.  So now my total volume is what?  130.  So now I'm going to go calculate the new concentration.  So I'm going to start by calculating the concentration of the excess reactant that's left over which  is the moles divided by the new volume which is 130 liters of 0.13 liters or 130 milliliters  and that comes out to be 5.38 times 10 to the negative 3, 10 to the negative 2 molar  and then I'm going to calculate its conjugate base which is 3 times 10 to the negative 3  moles divided by 0.13 liters which gives me 2.31 times 10 to the negative 2 molar.  So now I know the concentration of the weak acid and its conjugate base.  Now I can say pH equals pKa minus log the concentration of the acid form divided by  the concentration of the acetate anion after the titration is completed.  So we're going to assume that that's our new initial and so this is acidic acid so this  is 4.754 minus log concentration of the acid is 5.38 times 10 to the negative 2 divided  by the concentration of the acetate which is 2.31 times 10 to the negative 2 and so  that gives me a pH of 4.387.  So now we've calculated what the pH of that solution is in that flask.  Can everybody see that what we're doing is the same stuff that we've done before except  that now we have a lot more things going on and I want you to be able to understand this  at the same level.  There's a lot of detail there and you need to kind of understand that in this instance  what's happening is there's a reaction taking place after that reaction is complete we want  to calculate what are the species in that solution in the flask and then we need to  calculate the concentration of those and from that we can figure out what the pH is.  Now as homework can I ask you guys to calculate what the pH of that solution would be if you've  added let's say 45 milliliters.  If you added 30 milliliters now I want you to try to calculate what would the pH of that  solution turn out to be if you have 45 milliliters.  Now we're going to look at the next scenario which is at the equivalence point.  The next one would be so we looked at after you add 30 milliliters now let's look at pH  at the equivalence point.  So we need to calculate what the equivalence point is remember what is the equivalence  point we call this equivalence point or the stoichiometric point and what does that represent?  Somebody tell me what happens at the equivalence point?  Equal amount in other words the number of moles of the acid should equal the number of  moles of the base in other words all of the acid has been completely neutralized by the  base and usually when you calculate the equivalence point the underlying principle is that the  number of moles in this case the number of moles of acidic acid should equal the number  of moles of hydroxide that you have in that solution.  They have both reacted in a one to one stoichiometry because the balanced equation shows us that  they react in a one to one ratio.  So if all of the weak acid has been neutralized by the strong base then you have to have the  same number of moles of both and we know that moles is what?  Concentration times volume.  So on this side we have concentration of acidic acid times the volume of acidic acid and over  here you have concentration of hydroxide which is sodium hydroxide times the volume of hydroxide  which is the volume of sodium hydroxide that you added.  So at the equivalence point if I want to calculate the volume of hydroxide that's there that  would be the concentration of acidic acid times the volume of acidic acid divided by  the concentration of hydroxide which is we know concentration of acidic acid is 0.1 moles  per liter times the volume was that we had 100 milliliters and we know that we had 0.1  moles per liter of base so can you see that your volume of the base should be 0.1 milliliters.  Now sometimes in the laboratory you guys remember sometimes in the lab they give you this equation  M1 V1 equals M2 V2 this is the same thing.  M is the molarity, volume is the volume of acid that you added and then M2 V2 would be  the molarity of the base times the volume of the base that you added.  You have to be careful because you can only apply that if the acid and the base react in  a one to one ratio alright so that you have one acidic proton alright and therefore the  hydroxide will react with that one acidic proton and therefore it's a one to one ratio  alright so what do we know?  How much base do I need to add at the equivalent point?  You can see that I have to add 100 milliliters alright so if I add when I've added 100 milliliters  base I would be at the equivalent point alright so if I want to calculate pH now I've got  to look at this so pH at the equivalent point is remember the reaction we're looking at  is the weak acid reacting with the base to give you the conjugate base plus liquid water  okay and we said before reaction we had one times 10 to the negative 2 moles remember  in the flask when you have 100 milliliters of.1 molar that's the number of moles of  acidic acid that you had to start with so that was 100 milliliters times.1 molar.  Now we had at the equivalent point we have to have 100 milliliters so can somebody tell  me how many moles of hydroxide did I add?  We had 100 milliliters.1 molar in AOH so what would be the number of moles of NaOH?  What do you think?  Louder it's the same thing do you understand that because at the equivalent point the number  of moles have to be the same you see the moles of acid have to be the same as the moles of  base so good so you should see that that you're at the equivalent point so you're going to  have the same number of moles so can everybody see that this would be 10 to the negative  2 moles and we're going to have 0 of this alright now after reaction so now the reaction  is going to take place this is the titration you've added the base to the acid they cannot  coexist they're going to react cleanly rapidly and quantitatively so the reaction is going  to go to completion so now can you see these are in the same amount so if all of this reacts  with all of that you're going to end up with 1 times 10 to the negative 2 moles of the  product so remember we said at the equivalent point all the acid has been used up all the  base has been used up so what are we left in solution at the end of the titration the  conjugate base so now we need to calculate the pH of the conjugate base how do you calculate  the pH of a weak base alright so I need the concentration now alright so I'm going to  take the number of moles and calculate the concentration and therefore that would be  the concentration of the conjugate base which is 1 times 10 to the negative 2 moles divided  by the volume what is my volume I had 100 millilitres in the flask what's the volume  of base that I added 100 so what is my total volume 200 okay so I'm going to divide by  the total volume which is 0.200 liters which gives me 0.5 00 moles per liter so now we  go back to working remember this is a conjugate base how do I calculate the pH of this is  actually a salt can you see that it's the it's the salt of the conjugate base of the  weak acid acidic acid I know the initial concentration so can everybody write down how you would  calculate the pH of that solution this is applying what we did like that the first few weeks  of class what so if you're going to calculate the pH of a weak base what is the starting  point what do you need to write down what do you start with what do you need to write  down how do you calculate the pH of a weak base all of you did really well on the test  so you need to know this you have to draw the equilibrium and use the ice chart I see  that so so we're going to write down the equilibrium first so this is acidic acetate anion which  is the salt of the conjugate base in water giving me the weak acid what would I put here  OH minus great and this represents KB did they give me KB no they give me all the KA  so I need to convert KA to KB therefore KB equals one times 10 to negative 14 divided  by 1.76 times 10 to negative 5 which gives me the KB which is 5.68 times 10 to negative  10 you guys understand that I can move really fast because we've done many problems of  this type once again we're just applying what we've learned before okay so if I want to  write the ice chart I start with my initial concentration would be point zero five zero  zero I'm not concerned about water I have zero zero to begin with but I want to keep  figure out what the changes would be negative x plus x plus x not concerned about water  at equilibrium I know this will be point zero five zero minus x x x okay so we're looking  at KB because this is the formation of hydroxide from the weak conjugate base therefore KB  equals 5.68 times 10 to negative 10 which is concentration of the acidic acid times the  concentration of hydroxide divided by the concentration of the weak base alright so  that gives me x squared over point zero five zero zero minus x alright and as all of you  know we can approximate this to x squared over point zero five zero now since you guys  know how to solve for this I would like you to try this at home you have to check the  validity of your approximation but if you calculate x after you check for a validity  of approximation and so on x will come out to be 5.33 times 10 to the negative 6 molar  so we know that the hydroxide ion concentration equals x which is 5.33 times 10 to the negative  6 molar remember we need to calculate pH so I need to convert that to hydronium ion concentration  which is Kw divided by the hydroxide ion concentration which is 1 times 10 to the negative 14 divided  by 5.33 times 10 to the negative 6 molar this will be molar squared and therefore if you  work that out it comes out to be 1.88 times 10 to the negative 9 molar therefore pH equals  8.727 remember we said at the equivalence point the pH has to be greater than 7 because  that solution is basic alright and so we figured out what the pH of that solution is at the  equivalence point okay now I need you to be able to calculate so if you look at the  fourth scenario you need to be able to calculate pH after 100.05 milliliters of NaOH is added  okay now the solution to this is very similar to the strong acid because once we pass the  equivalence point now all of the acid has been consumed completely so now all you have  is a strong base alright and so the solution for this part will be exactly the same as  the solution for the previous part because what you have is now at the equivalence point  you have added 100 milliliters of the base now you have an excess of 0.05 alright so  in that solution now you have a strong base and a weak base and the strong base is much  stronger than the weak base you see that so the major contribution is going to come from  the strong base so if you have 0.05 milliliters of excess hydroxide all you have to do is  calculate the number of moles in 0.05 milliliters of hydroxide divide by the new total volume  which is 200.05 and that gives you the concentration of the strong base and because the strong  base dissociates completely you know what the hydroxide concentration is and so you  know the hydroxide concentration that comes from that extra 0.05 that was not neutralized  alright and you can calculate what the pH of that solution is okay so you should be  able to solve these kinds of problems and in discussion for next week actually it's  this week isn't it the discussions actually get you to solve some of these problems I  need you to be able to look at these different types of problems and to be able to solve  them okay now a couple of things I want to point out so we're going to go back to this  graph and I want to point out some things on this graph so if you look at this titration  curve this is the point at which there is no base added so you should be able to calculate  the pH of that weak acid okay we know this region where it flattens out is called the  buffer region okay so in this region where it's flat we call that the buffer region because  that's the region where it's going to act as a buffer so whenever you have mixtures  of the weak acid and it's conjugate base it's going to act as a buffer okay now this is  the equivalence point alright and we know that the equivalence point the pH will be  greater than 7 okay now if you take we already calculated that you have to add 100 milliliters  to reach the equivalence point so exactly half of that is called the half equivalence  point alright and at the half equivalence point what you realize is so at the half equivalence  point what you realize is that the concentration of the acidic acid that's left over equals  the concentration sorry CH3CO2 minus can I just rewrite that okay because it's not very  clear and I don't want to rush that so at the half equivalence point  we know that the concentration of the acidic acid that's left over in the flask the excess  amount should equal the concentration of the acetate ion alright so you know that pH equals  pKa minus log of the acid over the base and these two are equal to each other what do  we know pH equals pKa minus log 1 which is pKa minus 0 so we know at the half equivalence  point pH should equal pKa so if you're at the half equivalence point so you figured  out that the equivalence point is 100 milliliters if you're at the half equivalence point that's  the perfect buffer you have equal amounts of both alright and therefore the pH should  equal pKa so if that's the case you don't need to do any calculations if you get a question  saying you know the point in the equivalence point they ask you what is the pH of the solution  at this point you don't have to do any math because you know pH equals pKa and this is  acidic acid so pKa acidic acid is what 4.75 so you know that the pH at that point will  be 4.75 okay now I want to finish up by just quickly going through this question so this  is an old question from an older exam so I just pulled it this morning and I want you  guys to take a look at this and I'll put a copy of this on the class website okay so  why is that here we are so the question is consider the titration of a generic weak acid  with a strong base that gives the following titration curve so you have already seen this  curve you know that it's a weak acid strong base curve okay on the curve indicate the  points that correspond to the following the first is the equivalence point so all of you  know that the equivalence point of the stoichiometric point has to be this got it so that's the  equivalence point the region with maximum buffering so you want to show an arrow that  kind of shows where this is plateaus out and flattens out this is the region of maximum  buffering somewhere between this and this that region got it pH equals pKa where would  that be you look at the equivalence point it's a little above 25 all right and it's 30 here  so it's somewhere between so it's it's actually somewhere here right in the middle so you  have to look at this value and divide that exactly by half and that will give you where  pH equals pKa and that will be that point there okay pH depends only on the initial concentration  so this should be a subscript okay and so if the pH depends on only the initial concentration  of the weak acid where would that be zero before you add any base that will be the point  at which okay pH depends only on the conjugate base at the equivalence point because the  equivalence point is the only point at which you have exclusively just the conjugate base  lastly pH depends only on the amount of excess strong base added where would that be where  does the pH depend on the amount of strong base anything past the equivalence point so  once you pass this all of these points once you pass this anything above that will depend  on the concentration of the hydroxide okay so we'll stop there for today and next time  we'll move on to the next topic.
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