What is the pH of a 0.49 M solution of sodium oxalate, Na2C2O4?

Question

Acid ionization constants for oxalic acid are Ka1 = 5.6 x 10-2 and Ka2 = 5.4 x 10-5.

Answer 1

C2O4 2-  in water --> HC2O4-  &  OH-

0.49 molar -x -->  -->  -->  x      &    x

Kb = [HC2O4-]  [OH-] /  [C2O4-2]

Kb = Kw / Ka2 = 1e-14 / 5.4e-5 = 1.85e-10

1.85e-10 = [HC2O4-]  [OH-] /  [C2O4-2]

1.85e-10 = [x]  [x] /  [0.49]

x2 = 9.07e-11

x = [OH-] = 9.526 e-6

pOH = 5.02

pH = 14 - pOH

your answer is : pH = 8.98

the pH comes from the hydrolysis with C2O4-2, though some HC2O4- is made its further hydrolysis  is very insignificant

Answer 2

7.33 ? Saturated solution,  3.4 or 0.25 m calculated. Do not carve this in stone!!! You will want to research more or ask a more intelligent source than us nimrods online! Good luck!