Question

What is the equilibrium concentration of H2C2O4 in a 0.542 M oxalic acid, H2C2O4, solution? For oxalic acid, Ka1 = 6.1 x 10–2 and Ka2 = 5.1 x 10–5.

Answer 1

H2C2O4 + H2O ⇋ [HC2O4]^- + H3O^+
Ka1 = [H3O^+][[HC2O4]^-]/[H2C2O4] = 6.1×10^-2
[HC2O4]^- + H3O^+ ⇋ [C2O4]^2- + H3O^+
Ka2 = [H3O^+][[C2O4]^2-]/[[HC2O4]^- ] = 5.1×10^-5
But we want:
H2C2O4 + 2H2O ⇋ [C2O4]^2- + 2H3O^+
K = [H3O^+]^2[[C2O4]^2-]/[[H2C2O4]] = Ka1× Ka2 = 3.1×10^-6
At equilibrium:
[H3O^+]^2[[C2O4]^2-]/(0.542) = 3.1×10^-6
[[C2O4]^2-] = 0,0119 M

[H3O^+] = 0,0119 M

[H2C2O4] = 0.542 – 0.0119 = 0.53 M