2015 AP Chemistry free response questions
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2015 AP Chemistry free response 3d
- [Voiceover] Calculate the pH at the half-equivalence point. So, let's just remind ourselves what the half-equivalence point even is. The equivalence point is when the titrant, in this case the hydrochloric acid, completely reacts with the potassium sorbate, the thing that we are titrating. Now the half-equivalence point is the point at which half of the potassium sorbate has been converted to the sorbic acid. Or another way of thinking about it is, the concentrations of the potassium sorbate and the sorbic acid are equivalent. Well, how do we relate that to pH, and what other information have they given us to actually solve this? And a good thing to do whenever you are, if you feel a little bit stuck here, say, "Well, what other information have they given us?" Well, they gave us the K-a of sorbic acid is being 1.7 times ten to the negative fifth. So somehow, can we connect the K-a of sorbic acid to the pH at the half-equivalence point? Well the other thing that they give you, is they give you a whole series of formulas. In fact, they give you all the formulas that I'm using here in the first couple of pages of the free-response section, and even a whole bunch of formulas on equilibria and all of these different notations they use. And the one that might show up that looks interesting is this one right over here that you might recognize as the Henderson-Hasselbalch equation. And it's actually not hard to prove it. It comes straight out of the definition of K-a, and then rearranging things, and then taking the negative log of both sides and doing things like that. I encourage you to watch those videos on Khan Academy if you are curious, but what's neat here is it connects pH, it connects pH, it connects pH, to p-K-a, and the concentrations of an acid and its conjugate base. So how do we make a relationship here? Well, at the half-equivalence point, the acid in the conjugate base are gonna, their concentrations are going to be equivalent. So this, and this are going to cancel out. You're just gonna get one. And log of one is just going to be zero. So, at the half-equivalence point, the pH is going to be equal to the p-K-a. And so what is the p-K-a here? Well, they told us that the K-a, the K-a is equal to, and this is the K-a of sorbic acid, K-a of sorbic acid is 1.7 times ten to the negative fifth. 1.7 times ten to the negative fifth. 1.7 times ten to the negative fifth. And once again, when we're thinking about K-a we're thinking about the dissociation constant for the acid, and so that's why we used sorbic acid there. And, either way, these two concentrations are going to be the same. And if I want to find the p-K-a, I take the negative log of... I take the negative log of this. So, p-K-a, which is equal to the negative log, base ten, of the K-a. And they actually give you all these formulas on the first page. This is going to be equal to the negative log, base ten, of 1.7 times ten to the negative fifth power. And what is that going to be? Let me get a calculator out. So... So let's see, I'll write 1.7... These two capital Es, that's times ten to the... So, times ten to the, not the fifth, the negative fifth power. Now I want to take log, base ten, that's this button here. If you're calculator just has a log button that defaults to base ten. So I take log, base ten, of that. And then I want to take the negative of that. And so this is going to be, approximately, 4.77. So, this is approximately, approximately, 4.77 So, the pH is equal to the p-K-a which is equal to that, right over there.