# 17.3 pH Calculations Involving Titrations

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##### Video Transcript

Okay. So here the question doesn't actually use the word titration in it. The question just says 100 milliliters of 0.125 molar NaOH, is added to 150 milliliters of 0.1 molar HCl, what is the pH of the resulting solution? How would you know this is a titration question? You've mixed an acid and a base and at least one of them is strong. In this case they're both strong, but at least one of them is strong. That's a titration. And for our titration, you should run the titration reaction. In this case HCl reacts with NaOH, because a titration will always have at least either the acid or the base, or in this case both, being strong. And how much do strong acids and bases dissociate? Completely. Which means these are not equilibrium reactions. They're going to go completely. And so what products do I get here? Will water affect pH? No. So I will form some water here. I'm going to ignore it. Doesn't affect the pH of the solution. Does NaCl affect the pH of a solution? No. He's a negligible salt. Negligible cation. Negligible anion. I'll ignore him as well. So I just figured out that when I worked this, I don't have to worry about any of these products that I form. They're not going to check the pH of the solution at all. Cool. If a reaction is an equilibrium reaction, we always deal with molarities. Because we're going to plug this into like a Keq, where we have molarities of products over molarities of reactants. But this is not an equilibrium reaction. It goes to completion. And for reactions that go to completion, we don't deal with molarities usually. We need to know exactly how many moles of every reactant we have. Not molarities. So, if you look, definition of molarity is moles over the volume in liters. If you rearrange this, then moles is equal to molarity times the liters. Have you guys been using moles or millimoles, by the way? Moles. Good. We'll use moles. I'll ignore the millimoles. It's usually a little bit easier when you have a calculator handy, but I like moles. So somebody give me the number of moles of NaOH, that we started with here. Awesome. 0.1 liters, times 0.125 moles per liter is 0.0125 mols. How many moles of HCl? 0.015 moles. Since this reaction is going to completion, these react one two one. Which one is the limiting reagent? Yeah I got less NaOH. And so I lose all of it. It's gone. And there's none left. This is not technically an ice chart we're setting up, because ice charts deal with molarities. We're dealing with moles. There's going to be no X's here. Because- since it goes to completion, I'll know exactly how much limiting reagent I lose. All of it. Since these react one to one, if I lose that much NaOH how much HCl do I need to lose as well? And so how many moles of HCl are left? Cool. And we already said that we're going to ignore the products, but how many moles of water actually did form in this process? 0.0125 formed. How many moles of NaCl formed? Say I'm out, do I really care? Not at all. They don't affect pH, so we ignored them. So when I look at what still can affect pH and it's in my solution, what is in my solution that affects pH? Hust HCl, which is a what? What kind of acid? Strong acid. So from this point on, what kind of calculations are we doing? Which of the five types? A strong acid calculation. How do you do a strong acid calculation? It dissociates completely. So all you got to do is take the negative log of the concentration. Is that a concentration? No. How do I make it a concentration? Divide by the total volume in liters. Take the moles and divide by the total liters. So how many total liters do we have right now? Yeah, 250 milliliters is 0.25 liters. And so we find out that our concentration of HCl, and because it dissociates completely, the concentration of H+, is equal to- somebody get that for me- 0.01 molar. And we can just take the negative log because it dissociated completely. And that'll get us a pH of? Two. This is how every titration problem works. You run the titration to completion. They're always going to have at least one of these being strong, and you know it goes completion. And then you look at what's left that still affects the pH. And based on what's left, you decide which of the five types of problems you're doing. Cool. All right, we got one more set of problems left to go here. We got 100 milliliters of 0.1 molar HF. The Ka of which is 6.8*10^-4. And the question, or questions, are calculate the pH after the addition of 25, 50, and 100 milliliters of solution. 25, 50, and 100 milliliters of solutions. Let's stop for a minute before we do 25 milliliters of adding NaOH here. How would I calculate just the pH of the original solution? Before I did anything? I'm not going to use an ice chart because I know I don't have to use the quadratic. Which of the five types of problems is this? Weak acid. And use the appropriate shortcut for that. Cool. Ka times Ha concentration. Take the square root, and then take the negative log. Great. We can do that pretty quickly. So it wasn't asked, but we could have done it. But let's move on. Now the first problem we're going to do is adding 25 milliliters of- does it say concentration- yeah, 0.1 molar NaOH. Cool. How do I know this is a titration? Well the question said titration. But how would I know otherwise? NaOH is strong. I've mixed an acid to base and one of them strong. So in this case you'll set up your titration to go to completion. What's my other product? Good. Will water affect the pH? No. Will NaF affect the pH? Yes, he's a base. I can't ignore the amount of this I form. It affects pH. F- is not negligible. It really is one of those basic anions. Okay. Because this goes to completion, I can't deal with molarities, I need to deal with moles. How many moles of HF did we start with here? Good. How many moles of NaOH have we started with this 25 milliliters? So if you take 0.025 liters times 0.1 moles per liter, you'll get 0.025 moles. Who's my limiting reagent? NaOH. So how much do I lose? All of it. And I have none left. How much HF has reacted and I've lost as well? Same. It's a one to one ratio. So if I lost 0.025 of NaOH, I'll lose 0.025 of HF. But I should also keep in mind that how much NaF did we form? The same. It's a one to one to one ratio. So plus 0.025 moles. How much of that did we start with? We didn't have any. And so in this case I'm going to have 0.0025 moles of NaF. How many moles of HF are left? 0.075 moles. And so if we look at this now, if we look at what's left in our solution, which of the five types of problems- of calculation problems- is this going to be? I got a weak what? Weak acid, and? And it's conjugate base. What kind of problem is this? It's a buffer. And what do I use to do a calculation for a buffer? Henderson-Hasselbalch. Awesome. So we'll go pH equals pKA, plus log A- over HA. In this case the pKa, well the Ka, is given right there. And pH will equal the negative log of that, which happens to be 3.17 by the way, plus the log- in this case the A- concentration is the 0.0025. And you might be like "Wait a minute Chad, you can't do that." Are these molarities? Awesome. It doesn't matter if it's a ratio. Because if I do moles per liter over moles per liter, well it's the same solutions. The same liters. And the liters could cancel anyway, so I can just do it as a ratio as moles over moles. Only because it's a ratio. Cool. Can somebody get me the pH here? 2 point- what? Awesome. Cool. First one is done. You guys ready for the next one? Okay. So we got 100 milliliters now of 0.1 molar HF again; we started with. But now instead of 25 milliliters, we've now added 50 milliliters of 0.1 molar NaOH. You guys want to see how we do this one? I'm done. Good. If you notice, they're the same molarity. How many milliliters of NaOH would I have to add to get to the equivalence point, if they're the same molarity? No. But same amount. It's a 1:1 ratio so I'd have to- it's the same molarity, I'd have to add a hundred milliliters of NaOH to neutralize it completely. Which means I have added exactly half. Where am I at in the titration? The half equivalence point. What's true at the half equivalence point? pH equals pKa, which is 3.17. Now let's actually set this up and verify that. But if you recognize you're at the half equivalence point you can short-circuit it really quick. pH equals 3.17 because that's the pKa value. Done. But let's verify that we actually get that if we worked it all the way out. We set this up just like the last one. We're still going to ignore water. In this case you've still started with 0.01 moles of HF. But now you've added 0.005 moles of NaOH. Who's your limiting reagent? NaOH. And how much do you lose? 0.005. You lose the same amount of HF and gain the same amount of NaF. You got no NaOH left. 0.01 minus 0.05 will be 0.005 moles of HF left. You formed 0.005 moles of NaF, and you can see that yeah we have equal amounts of weak acid and conjugate base. Whether you saw that or not, you should still recognize that this is what type of problem? Buffer calculation again. So the fifth type we learned. So we're still going to plug this into Henderson- Hasselbach, but in this case because the moles of A- and the moles of Ha are equal, then the log of one is zero and goes away. And we're left with the pH just equaling the pKa, which was our 3.17. The negative log of 6.8*10^-4. So it works out that way anyways, even if you don't recognize it. But if you recognize it from the get-go, you just saved yourself a ton of time. Because it's a calculation that students can answer quickly if they've worked through a lot of examples and get this familiar with it, professors love asking half equivalence point calculations. Because the students that are really on the ball, can do it in five seconds. Cool. Half-Equivalence? Again the key is, they're both 0.1 molar. Had they not been both 0.1 molar, it would been much harder to recognize. But because they're both 0.1 molar, if I start with 100 milliliters of the weak acid, how much NaOH- how many milliliters of NaOH, would it take to get all the way to the equivalence point? 100. That's exactly half of a hundred. Correct. Because truth be told, the key is I have to add half as many moles of the strong, in this case base, as of the weak acid. Yeah. So it was just easy to recognize because the molarities are the same. If the molarities weren't the same, I still probably- if the numbers would have worked out moles to these two numbers- I still would have recognized it at that point. But at some point, even if you never recognize it, you're still going to get the right answer. Question. So what's the pKa? So, you cool with the A- and the HA being the same, and so the log of one goes away. If you take the negative log of the Ka, it is 3.17. Nope. Next thing- next one we're about to do is not a buffer problem. So notice we did a strong acid, strong base titration. That wasn't a buffer one. We've done a couple weak/strong, they both turn into buffer problems. With this last one, it's not going to be a buffer problem. Cool. So here, this is the worst of all titration calculations right here. Notice what point in the titration am I at now? I'm at the equivalence point. The equivalence point is either the easiest of all calculations or the hardest of all calculations. If I would have had a strong acid and strong base, it would have been the easiest of all calculations and the pH would have simply been exactly 7. But in this case it's not going to be 7. In fact I should know which side of 7 it's on. Which side of 7 is this equivalence point pH going to be on? Yeah, it's going to be higher than 7 because it's the base that's strong. My way of remembering it. So in this case, if we set this up, again you still got 0.01 moles of HF to start. But now you've added also 0.01 moles of NaOH. Which one is the limiting reagent? Well they both are. They both run out completely. You lose all of both of them in this case. They're equal. And how much NaF do you form? The same amount. So in this case you don't have any HF left. You don't have any NaOH left. But now you form 0.01 moles of NaF, bless you. What's left in your solution that affects the pH? Base. What kind of base? A weak base. This is a weak base problem. How do you solve a weak base problem? Not- not a buffer problem. How do you solve just a weak base? Shortcut. What's the shortcut? Shortcut for weak base. Well it's a base so you're solving for hydroxide not hydrogen ions. And it's the square root of Kb times the actual base concentration. This is the hardest point because there's two things. Do I know the Kb right now? If I solve for it. But I don't know it actually directly right now. I've got the Ka, but I'm going to have to use Ka times Kb equals Kw, to get the Kb. So in this case if I divide through by the Kb I'll get 10 to the negative 14 all over the Ka, 6.8 times 10 to the negative 4, to get the Kb. So that's the first part I got to get a Kb to plug in. What is that? Anybody got an answer for me? For the Kb? 1.47 times ten to the? You said 10 to the minus 11 right? Awesome. But I also need a base concentration. Do I know the molarity of the base right now? No. I just know the moles. How do I turn that back into a molarity? Divide by what is the total liters now? 200 milliliters is 0.2 liters. And so this is going to get me my concentration of base here. In this case the concentration of F-. And what does that come out to? It's 0.05 molar. That is my concentration of base that gets plugged back into the shortcut here. So can anybody get me the hydroxide concentration then? And what will I get from here? So what's the pOH? And so what's the pA. Oop, 6.1? My bad, 6.1. Then what's the pH? Awesome. Whoa. 7.9? Does that make sense? Yes it does. With a weak acid/strong base at the equivalence point, it should be on the basic side of seven. Awesome. Yeah like I said that's the worst of all calculations. If they give you the equivalence point in a weak/strong titration, it doesn't get any worse than this. But, again once you run the titration to completion, you should realize that "Oh this is just a weak base, and there's my shortcut!" And getting there is the hardest part. Correct. So I didn't give you one where we went beyond the equivalence point. But let's say we went past it. Let's say we had 125 milliliters NaOH. Then you'd have leftover NaOH. You have a weak base, still. But you'd also have strong base. And the easiest way to do that is to ignore the weak one. If I do a tug-of-war and it's me and a mosquito on the same team. Is the mosquito really doing much? No, ignore it. Same thing here. If we would have gone past the equivalence point, you'd have strong base and weak base just ignore the weak one. Just use the strong base, divide by the total volume to get it to a molarity again, take the negative log to get the pOH, and subtract from 14. Well weak acid/weak base you're never going to deal with calculation wise. But again your titrations will always have at least one of the two, either the acid or the base being strong. But, but it would get totally ridiculous. Question. Do I think she'll give one like this? These show up about half the time on this exam for the average chem professor. Unfortunately. It can't get harder than this. This is as hard as it gets. If you can do this, it doesn't get any harder. This is the absolute hardest calculation in this chapter. Probably wouldn't go for all three. Especially if you only have five calculations. Notice you have five different main principle types of calculations, but then every titration turns into one of those. As well. Which makes them harder because it's a little more plus that. So I doubt she'd ask you. Especially if you guys are only getting five calculations I doubt three of them would all be titration calculations. So. True. If you got three seconds left and you don't have time to work this, at the very least if you recognize it was equivalence point you could be like how many answers are higher than seven, a little higher than seven, and at least narrow it down.