Question

1. Calculate the concentration of H3O+, HC2O4-1, and oxalate ion (C2O42-) in a 0.175 M solution of oxalic acid (C2H2O4).
[For oxalic acid, Ka1 = 6.5 ´ 10-2, Ka2 = 6.1 ´ 10-5.]

Answer 1

.......H2C2O4 + H2O --> H3O^+ + HC2O4^-
I......0.175.............0.......0
C........-x..............x.......x
E.....0.175-x............x.......x

Substitute the E line into Ka1 expression and solve for x = (H3O^+) = (HC2O4^-).

That gives you H3O^+ and HCO4^- and note they are equal. It also gives (H2C2O4). That leaves C2O4^2-


.......HC2O4^- + H2O ==> H3O^+ + C2O4^2-
Ka2 = (H3O^+)(C2O4^2-)/(HC2O4^-)
Since (H3O^+) = (HC2O4^-), then (C2O4^2-) = ka2.