pH's and pKa's of polyprotic acid?

[DeathandTaxes]

Can anyone help me understand how to start this problem? The book says that mixing H2PO4 and -HPO4 means the pH should be around pKa 2... thus we would use the equation pH = pKa2 + log(A-/HA).. but why pKa2?

Secondly, it says that if we had an equal mixture of H2PO4 and -HPO4 the pH would be equal to ( pKA1 + pKA2) / 2, what is the rationale behind this?

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[Teleologist]

Use the HH equation.

 

[DeathandTaxes]

Elaborate? I'm using it in the first equation but I don't know why you'd use pKa2 instead of say pKa 1.


Also How does that explain the second question?

 

[DeathandTaxes]

Bump anyone?

 

[sillyjoe]

(pKa2 + pKa3 / 2) gives you the pH at equivalence of this buffer mixture (H2PO4 to HPO4) if you were doing a titration.

For this buffer/titration A- is HPO4 and HA is H2PO4

The HH equation says pH = pKa + log(A-/HA)

[A-] =.2
[HA] = .1

Since there is still HA in the solution we are not yet at the equivalence point so the pH at equivalence ([pKa2 + pKa3 / 2]) is greater than the pH of this solution.

Since the HH equation gives you log(.2)/(.1) the pH of the solution will be above your pKa2 since pH = pKa at the half equivalence point.

Therefore, the answer choice is really saying the pH at equivalence for this titration > pH of the solution > pH at the half equivalence point where pH = pKa2

Essentially, this solution is in between the half equivalence point and equivalence point.

You had also asked why you use pKa2 and not pKa1. The reason for this is you have a triprotic acid so you must use the relevant pKa for the reaction.

H3Po4 --> H2PO4 = pKa1

H2PO4 --> HPO4 = pKa2

HPO4 --> PO4 = pKa3

In this question we are talking about H2PO4 --> HPO4 so we must use the pKa2