Question

Calculate the concentration of H+ in a 0.250 M solution of oxalic acid. Consider only [H+] from the first ionization. (Ka = 5.37 x 10^-2) Then calculate [H+] of the same solution, including the H+ ions resulting from the second ionization. (Ka = 5.37 x 10^-5)

Answer 1

                     (COOH)2 ----------> HOOC- COO- + H+

initially           0.25                             0                      0

at equi       (0.25 - x)                         x                      x

Ka = [H+] [HOOC-COO-] / [(COOH)2]

5.37 * 10^-2 = x*x / (0.25-x)

x = [H+] = 0.0921

                     HOOC-COO- -------> (COO)2 2- + H+

initially          0.0921                             0                0

at equili       (0.0921 - x)                     x                 x

Ka = [H+] [(COO)22-] / [HOOC-COO-]

5.37 * 10^-5 = x * x / (0.0921 - x)

x = [H+] = 0.002197