How to find pH after the second equivalence point?

Question

I have the following information to sketch a titration curve: 10 mL aliquot of 0.100MNa3AsO4 titrated with 0.100M HCL with pkas pK1=2.25, pK2=6.77, pK3=11.60.

The hint was to use the volumes in mL: 0, 5, 10, 15, 20, 25, 30, and 40.

Where I get stuck is in finding the pH at 30 mL. I heard I have to use the quadratic formula but I don't know how to find the kb with the following information. Any help would be much appreciated!

Answer 1

upon adding  the first 10 ml of HCl to the Na3ASO4, you are at the first equiv point having prorduced the acid HASO4)-2 whose pKa is 11.60

upon adding  the second 10 ml of HCl to the Na3ASO4, you are at the second equiv point having prorduced the acid H2ASO4)-1 whose pKa is 6.77

upon adding  the third 10 ml of HCl to the Na3ASO4, you are at the third equiv point having prorduced the acid H3ASO4)-1 whose pKa is 2.25

at this point you have 0.100 Molar Na3AsO4 converted to H3ASO4 but diluted:

0.100 Molar  @ 10 ml / 40 ml total = 0.025 Molar H3AsO4

and this acid dissociates:

H3AsO4 --> H+  & AsO4)-2

0.025 -x   -->  x  &  x

Ka1 = [H+] [H2AsO4)-2] / [H3AsO4]

& for a pKa = 2.25,  the Ka = 5.62e-3

5.62e-3 = [x] [x] / [0.025-x]

X2 = ( 5.62e-3) [0.025-x]

X2 = 1.4e-4  - 5.62e-3x

X2  - 1.4e-4  +  5.62e-3x = 0

http://www.1728.com/quadratc.htm

x = [H+] = 0.0093 molar

pH = 3.85