Titration question

[MedGrl@2022]

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If the pH of the half equivalence points do not depend on concentration of the acid or its conjugate base, how can the pH of the equivalence point depend on the concentrations?

Please let me know how I can understand this and/or where I can find more information to help me understand this better.

Thank you,

Verónica

 

[slz1900]

Are you talking about weak acid/base titrations? If so, the pH at the half equivalence point does depend on concentrations. You can see this with the Henderson Hasselbach equation. If you are talking about strong acid titrated with a strong base, you are right that pH at the half equivalence point does not depend on concentrations of conj. acid/base. You calculate pH at any given point with a strong acid/base titration by calculating the concentration of strong acid or base in solution that hasn't been neutralized. With a strong acid/base titration you don't have a conjugate acid/base in the same sense; instead, you have salt and water. When all of your strong acid or base has been neutralized, you don't have any acid or base in solution. All you have is water and an ionic salt, so your pH is 7.

 

[MedGrl@2022]

Are you talking about weak acid/base titrations? If so, the pH at the half equivalence point does depend on concentrations. You can see this with the Henderson Hasselbach equation. If you are talking about strong acid titrated with a strong base, you are right that pH at the half equivalence point does not depend on concentrations. This is because you don't have a conjugate acid/base in the same sense; instead, you have salt and water. When all of your strong acid or base has been neutralized, you don't have any acid or base in solution. All you have is water and an ionic salt, so your pH is 7.

860. The Ka1 for H2C03 is 4.3x10^-7. The Ka2 for the H2C03 is 5.6x10^-11. Which of the following depends upon the concentration of acid and conjugate base?

A. The pH of the first equivalence point
B. The pH of the first half equivalence point
C. The pH of the second half equivalence point
D. The pKal

So I reasoned that pKa=-logKa so the concentration does not matter and we can solve for the pH of the half equivalence points using Henderson-Hasselbalch equation: pH=pKa+log([A-]/[HA]) since I already know the Ka's I can solve for the pH's of the half equivalence points. Thus that leaves "A. The pH of the first equivalence point". Which according to EK is the correct answer? But why? If you can solve for the pH of the half equivalence points then you should be able to solve for the equivalence point, right? It should be half way between the first and second equivalence point.

Maybe I am just confused. If the solution is diluted to begin, the pH would already be affected from the beginning right? So couldn't everything be somewhat effected?

Let me know what you think.

Your other answer on the pH of water really helped.

Thank you,

Verónica

 

[slz1900]

OK, I see what you mean by the halfway equivalence pt not depending on concentration now. My interpretation is, for the statement pH=pKa to be true, your concentrations of conjugate acid and conjugate base have to be equal, so concentration matters but I get what you're saying.

wrt your question, I don't know why ek's answer would be correct. You are right that (pKa1+pKa2)/2 would be equal to the equivalence point. Maybe they mean a direct derivation of the equivalence point would require you to know the concentration, although I'm not sure how.

 

[MedGrl@2022]

OK, I see what you mean by the halfway equivalence pt not depending on concentration now. My interpretation is, for the statement pH=pKa to be true, your concentrations of conjugate acid and conjugate base have to be equal, so concentration matters but I get what you're saying.

wrt your question, I don't know why ek's answer would be correct. You are right that (pKa1+pKa2)/2 would be equal to the equivalence point. Maybe they mean a direct derivation of the equivalence point would require you to know the concentration, although I'm not sure how.

Okay cool. EK is known for Errata. I think EK also had a similar problem elsewhere in their 1001 Chem book. Let me see if I can find it and try to figure it out too.

Thank you once again for your help!

Best,

Verónica 🙂

 

[Postictal Raiden]

860. The Ka1 for H2C03 is 4.3x10^-7. The Ka2 for the H2C03 is 5.6x10^-11. Which of the following depends upon the concentration of acid and conjugate base?

A. The pH of the first equivalence point
B. The pH of the first half equivalence point
C. The pH of the second half equivalence point
D. The pKal

So I reasoned that pKa=-logKa so the concentration does not matter and we can solve for the pH of the half equivalence points using Henderson-Hasselbalch equation: pH=pKa+log([A-]/[HA]) since I already know the Ka's I can solve for the pH's of the half equivalence points. Thus that leaves "A. The pH of the first equivalence point". Which according to EK is the correct answer? But why? If you can solve for the pH of the half equivalence points then you should be able to solve for the equivalence point, right? It should be half way between the first and second equivalence point.

Maybe I am just confused. If the solution is diluted to begin, the pH would already be affected from the beginning right? So couldn't everything be somewhat effected?

Let me know what you think.

Your other answer on the pH of water really helped.

Thank you,

Verónica

This is what I think:

At 1/2 equivalent point, [acid] = [conj. base], so pka = pH. This eliminates options B and C. Option D is the same as option B, because pKa1 = pH at the first 1/2 equivalent point. This leaves you with option A.

 

[slz1900]

This is what I think:

At 1/2 equivalent point, [acid] = [conj. base], so pka = pH. This eliminates options B and C. Option D is the same as option B, because pKa1 = pH at the first 1/2 equivalent point. This leaves you with option A.

You could choose the answer this way, but tbh I really dislike this kind of logic. If the question doesn't make sense or is wrong (as I'm pretty sure it is in this case), you really don't gain anything by fishing for a situation that would lead you to choose an incorrect answer. I don't believe that jumping through hoops like that is a good test taking strategy, especially since there won't be errata on the real deal.

 

[mehc012]

You could choose the answer this way, but tbh I really dislike this kind of logic. If the question doesn't make sense or is wrong (as I'm pretty sure it is in this case), you really don't gain anything by fishing for a situation that would lead you to choose an incorrect answer. I don't believe that jumping through hoops like that is a good test taking strategy, especially since there won't be errata on the real deal.

It gives you practice at figuring out the answer to a question you can't get from your background knowledge. Personally, I think that learning to sort through the answer options and choose the best one without a clearcut knowledge-based answer is invaluable for standardized tests; it gives you a much better shot at getting the right answer when you happen not to know something/are running low on time.

 

[brood910]

This is what I think:

At 1/2 equivalent point, [acid] = [conj. base], so pka = pH. This eliminates options B and C. Option D is the same as option B, because pKa1 = pH at the first 1/2 equivalent point. This leaves you with option A.

This is correct.

At half eq points, [conj] = [acid].
pH = pka + log(conj base/acid)
log1 = 0 (concentrations get removed from the eq).
Thus, concentrations do not matter.