Posted by Chay on .
Consider a 0.10 M solution of a weak polyprotic acid (H2A) with the possible values of Ka1 and Ka2 given below, calculate the contributions to [H3O+] from each ionization step.
Part 1: Ka1=1.0*10^4 Ka2=1.0*10^5
Part 2: Ka1=1.0*10^4 Ka2=1.0*10^6

Chemistry 
DrBob222,
I don't know how accurate these are to be but as a first approximation they are as follows:
Note that
........H2A ==> H^+ + HA^ and
........HA^ ==> H^+ + A^2
If you calculate H^+ from k1 (both acids) you obtain about 3.16E4. That comes from k1 = (x)(x)/(0.1x).
Since (H^+) = (HA^), then those cancel each other when you substitute into k2 so (A^2 ) = k2 for each and that is the contribution of H^+ from the second k. You can see that it is almost negligible for #1 but has a greater impact on #2.