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Topic: pKa's and Ve's  (Read 4238 times)

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Offline Timothy

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pKa's and Ve's
« on: April 02, 2011, 04:36:25 PM »
I have got to answer a question in my laboratory manual for a report I'm doing on "Dissociation constants of dibasic acids by pH - metric titration" and the questions asks, " Given that pKa1 = pH at V = 0.5Ve1, i.e. at half -equivalence point, prove that pKa2 = pH at V = 3/2 Ve1, i.e. at 3/2 - equivalence point". I have got stuck on this question but i'm not wanting an answer but a guidance on how to reach it, Thanks


Offline Borek

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Re: pKa's and Ve's
« Reply #1 on: April 02, 2011, 04:50:26 PM »
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Offline Timothy

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Re: pKa's and Ve's
« Reply #2 on: April 03, 2011, 11:42:03 AM »
Ok, I'm still stuck but I do know that at the equivalence point volume, pH = pKa but what is the relation to the volume in the H-H equation?
« Last Edit: April 03, 2011, 12:40:53 PM by Timothy »

Offline Borek

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Re: pKa's and Ve's
« Reply #3 on: April 03, 2011, 03:06:52 PM »
Ok, I'm still stuck but I do know that at the equivalence point volume, pH = pKa

That's wrong. pH=pKa when concentrations of acid and conjugate base are identical.

You have two separate parts of the titration curve, each described by different pKa value.
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Offline Timothy

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Re: pKa's and Ve's
« Reply #4 on: April 03, 2011, 04:57:50 PM »
Sorry, I meant to say pH = pKa at half the equivalence point. Even so, when the second pKa = pH, how do you mathematically prove that pKa2 = pH at 3/2 times the first equivalence point apart from reading off the pH vs volume of base added graph and seeing at what pH the pKa2 occurs and then comparing that pH to a delta pH / delta V vs V' graph (where V' = Vi +0.5(Vi+1 -Vi) to measure the accurate volume of pKa2.

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