Calculate the pH and concentration of H3O+ of a 0.285M oxalic acid solution.

Answer 1

.......H2C2O4(aq) ==> H^+ + HC2O4^-
I.......285...........0......0
C.......-x............x......x
E....0.285-x..........x......x

Substitute the E line into the Ka1 expression for H2C2O4 and solve for (H^+), then pH = -log(H^+).
I assume that the (H^+) is largely due to that contributed by Ka1 and none by Ka2. Actually, some will come from Ka2 but only about 1 in 1000 molecules that dissociate so the error will not be that large by neglecting that contribution from Ka2.