Question

The Ka of a monoprotic weak acid is 4.32*10^-3 what is the percent ionization of a 0.102 M solution of this acid?

Answer 1

HA --------------> H+ +   A-

0.102 M                      0        0

0.102 - x                      x         x

Ka = [H+][A-] / [HA]

4.32 x 10^-3 = x^2 / 0.102 - x

x = 0.0189

[H+] = 0.0189 M

% ionization = ([H+] / initial concentration ) x 100

                    = (0.0189 / 0.102 ) x 100

% ionization = 18.5