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Ch.17 - Applications of Aqueous Equilibria

Chapter 17, Problem 41c

The following plot shows two pH titration curves, each representing the titration of 50.0 mL of 0.100 M acid with 0.100 M NaOH:

. (c) What is the approximate pKa of the weak acid?

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hi everyone for this problem. It reads shown below are the tight rations curves of two acids, each with a concentration of 20.5 moller a 40 millimeter sample of each acid was tight rated with 400.5 molar potassium hydroxide, estimate the P. K. A. Of the weak acid. So we want to answer or estimate the P. K. A. Of the weak acid. So let's take a look at our titillation curve. Okay. And this is going to tell us a few things. So let's take a note. If we have a strong acid, the strong acid is going to be, or the curve for a strong acid is going to start at a low ph. So right here curve starts at low P H. So when we look at our two curves here, we can say that this curve because it has a low ph we can say this one is going to be a strong acid and for a weak acid, the curve starts at a higher ph Okay, so we'll say here that we have a weak acid here. So, right here a week acid. All right or not there? Excuse me, It's going to be here weak acid. Okay, so let's recall that at equivalence point. That is when the molds of acid equals the molds of base. Okay, so what we want to essentially solve for here is the volume of the base. That's what's going to help us solve this problem. So we have everything we need to solve for the volume of the base and the way that we're going to solve for the volume of the base is by using the malaria T times volume of the acid equals malaria T times volume of the base. And we'll just plug in what we know because what we essentially want to solve for here is the volume of the base. We know the malaria T. And volume of the acid. So let's plug that in. So for the acid, the malaria T is 0.5 molar and the volume of the acid is mL And this is equal to the mill arat e. of the base is 0.05 molar. Because we're at equivalence points so the molds of acid equals the molds of base times the volume of the base. So we don't know what the volume of the bases. So when we saw for this, then what we're going to get is the volume of the base is equal to 40 mL. Okay, so half the equivalence point, half the equivalence is equal to 20 mL. And at half the equivalence point P. H equals P. K. A. So let's go to our curve. So we're looking at 20 mL and we want to see for the weak acid what is the estimated P. H. So the region that we're looking at is right here at half the equivalence point. So our volume of the bases, 40 mL and half the equivalence is 20 mL. So looking at our weak acid curve. When we go to 20 mL, we see the region that the pE HSN and out of the answer choices given the one that is closest to reflect this is answer choice C. Okay, so we can estimate that the PKA is equal to 5.3. So let's go ahead and highlight that and this is going to be our final answer. That is it for this problem. I hope this was helpful.