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## Chemistry library

### Course: Chemistry library > Unit 14

Lesson 2: Titrations- Acid–base titrations
- Worked example: Determining solute concentration by acid–base titration
- Titration of a strong acid with a strong base
- Titration of a strong acid with a strong base (continued)
- Titration of a weak acid with a strong base
- Titration of a weak acid with a strong base (continued)
- Titration of a weak base with a strong acid
- Titration of a weak base with a strong acid (continued)
- 2015 AP Chemistry free response 3b
- 2015 AP Chemistry free response 3c
- 2015 AP Chemistry free response 3d
- 2015 AP Chemistry free response 3e
- 2015 AP Chemistry free response 3f
- Titration curves and acid-base indicators
- Redox titrations
- Introduction to titration

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# 2015 AP Chemistry free response 3d

Calculating the pH at the half-equivalence point for the titration of a weak base with HCl. From 2015 AP Chemistry free response 3d.

## Want to join the conversation?

- At1:58, the video mentioned that at the half-equivalence point, the [A-] and [HA] are equivalent. Is this only true when one mixes a weak base with a strong acid? (Thanks for these excellent videos, by the way.)(6 votes)
- It is true when one mixes a weak base with a strong acid & also a weak acid with a strong base.(3 votes)

- At00:29he says that at the half-equivalence point [A-]=[HA].

I thought that was true for the equivalence point, no ?(2 votes)- No that’s correct.

At the half equivalence point half of HA has been neutralised, so half of it has been turned into A-

At the equivalence point all of HA has been neutralised and it’s all been turned into A-(6 votes)

- At1:55, why is the concentration of the acid and base at the half equivalence point equal? I thought they were equal at the equivalence point.(1 vote)
- Careful here because there are multiple acids involved in this titration. At the half-equivalence point the concentration of the base (the sorbate) and the conjugate acid to sorbate (sorbic acid) are equal. These are the concentrations being used for the Henderson–Hasselbalch equation Sal listed to calculate the pH.

At the equivalence point the concentration of the base (again the sorbate) is equal to the concentration of the acid being added to titrate the sorbate (otherwise known as the titrant). The titrant here is hydrochloric acid.

Hope that helps.(2 votes)

- What is A- in this rxn, and is Pottasium Sorbate a Weak Base?(1 vote)
- When I am doing these types of questions, is it necessary to write out all of the equation for each question or do I just write the answer without all the math?

Thanks in advance.(1 vote) - Is it the same with the equivalence point? because i thought that when ph=pka applies and not the half- equivalence point(1 vote)
- Is what the same?

At the half equivalence point pH = pKa, not the equivalence point.(1 vote)

- Wouldn't it be 4.8 because the least number of important numbers is 2. But you wrote 4.77. That's isn't in scientific notation.(0 votes)
- the number of sig figs for the concentration is the number of decimal places after the pH(4 votes)

## Video transcript

- [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.