2015 AP Chemistry free response 3d | Chemistry | Khan Academy

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. 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? A good thing to do whenever you feel a little bit stuck here is to say, "Well, what other information have they given us?" Well, they gave us the Ka of sorbic acid as being 1.7 * 10^5.

So somehow, can we connect the Ka of sorbic acid to the pH at the half equivalence point? Well, the other thing that they give you is a whole series of formulas. In fact, they give you a lot—all the formulas that I'm using here on 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.

The one that might show up that looks interesting is the Henderson-Hasselbalch equation. It's actually not hard to prove; it comes straight out of the definition of Ka and then rearranging things, 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 to pKa 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 and the conjugate base are going to—their concentrations are going to be equivalent.

So this and this are going to cancel out; you're just going to get one, and the log of one is just going to be zero. So at the half equivalence point, the pH is going to be equal to the pKa.

And so what is the pKa here? Well, they told us that the Ka is equal to—and this is the Ka of sorbic acid. Ka of sorbic acid is 1.7 * 10^5. Once again, when we're thinking about Ka, we're thinking about the dissociation constant for the acid, and that's why we used sorbic acid there.

Either way, these two concentrations are going to be the same. If I want to find the pKa, I take the negative log of this. So pKa is equal to the negative log base 10 of the Ka. They actually give you all of these formulas on the first page.

This is going to be equal to the negative log base 10 of 1.7 * 10^5. What is that going to be? Let me get a calculator out. So, let's see, I'll write 1.7, these two capital e—that's times 10 to the—you know, not the fifth, the negative fifth power.

Now, I want to take log base 10—that's this button here. If your calculator just has a log button, that’s going to default to base 10. So, I take the log base 10 of that, and then I want to take the negative of that.

So this is going to be approximately 4.77. So, the pH is equal to pKa, which is equal to that right over there.