Chemistry-acid base equilibria

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p-Hydroxybenzoic acid is a weak diprotic acid, and during titration with 0.0200 M NaOH, the first equivalence point occurs at 16.24 mL, with pH values of 4.57 and 7.02 at 8.12 and 16.24 mL, respectively. The calculation for Ka1 was derived using the ICE method, resulting in Ka1 = 2.69 x 10^-5. The participant is struggling with calculating Ka2 and has found a value of 3.39 x 10^-10 but is uncertain about its accuracy. To find the pH at the beginning of the titration and at the second equivalence point, further analysis of the concentrations and equilibrium expressions is necessary. Understanding the amphoteric nature of HA- is crucial for these calculations.
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p-Hydroxybenzoic acid, HOC6H4COOH, is a weak diprotic acid. A 25.00 mL aliquot of a dilute solution of HOC6H4COOH is titrated with a 0.0200 M NaOH solution. The first equivalence point was reached after 16.24 mL of the NaOH solution was added.
a. If the values of the pH after 8.12 and 16.24 mL of base added were 4.57 and 7.02, respectively determine Ka1 and Ka2 for p-hydroxybenzoic acid.
b. Calculate the pH of the solution at the beginning of the titration.
c. Calculate the pH of the solution at the second equivalent point of the titration.
 
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What things have you tried so far?
 
Here's what I've tried- the moles of H2A at the beginning is .01624L times 0.0200M. You also need 32.48mL of base to reach 2nd equiv point. the part that trips me up is that HA-is an amphoteric species.
I used ICE with HA- + H20 -> A2- + H3O+, finding the conc. of [H3O+] as 10^-7.02 with .012992M as conc. of HA-. I ended up with 7.02 * 10 ^-13 as my answer for Ka1, but I don't think that's right. I also don't know what to do for Ka2.
 
i figured out how to get ka1, but I'm still having trouble with ka2.

how i got ka1-
use ICE with H2A + H2O -> HA- + H3O+, using the values for halfway to the first equivalence point.
Ka1=([HA-][H3O+]/[H2A])
[H2A]=[HA-]=.0049. (cancels out), leaving Ka1 = [H3O+] = 10^-4.57=2.69*10^-5.

Any tips for solving Ka2?
 
found ka2 to be 3.39*10^-10.
 
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