Hard: what is the pH of a buffer solution produced when Oxalic acid and NaOH are used?

Question

I have titration curve Weak acid x Strong Base
Oxalic acid! x NaOH
acid conc = 0.100 mol dm-3
base conc = 0.101 mol dm-3
Th alkali was added to the acid (25ml)
The whole titration was 40ml
1st equivalence point is at 12.3 ml (pH 2.6) the second equivalence point was at 24.6ml. (pH 8.2)
But i do not know how to find the pH of the buffer solution produced, i was told to use this formula : pH=Pka + lg [A-] / [HA]. The equation is
H2C2O4 + 2NaOH ==> Na2C2O4 + 2 H2O
how you find the pKa? because i was told that there are two, bcos there are the 2 equivalence points . But then how do you work out the buffer?
I have managed to work out the pH of other buffer solutions produced, e.g, KHP x NaOH
But this one is a lot more challenging ad i keep onfusing myself

Do you find the pKa, by using the half equivalence point, if so would you use both to work out the pH of the buffer?
=S How is the buffer calculated ?
Please help

Answer 1

You are almost there. Just look at the pH at 6.15 ml (half way to the first equivalence point) and 18.45 ml (half way between the two). These pH values should correspond to the two pKa values of (COOH)2.

Answer 2

Using your calculator -log[.045] = pH