Question

The Ka of a monoprotic weak acid is 4.51*10^-3. What is the percent ionization of a 0.106M solution of this acid?

Answer 1

HA   -----------------> H+ + A-

0.106                          0      0

0.106 - x                      x         x

Ka = x^2 / 0.106 - x

4.51 x 10^-3 = x^2 / 0.106 - x

x = 0.0197

[H+] = 0.0197 M

% ionization = [H+] / C x100

                    = 0.0197 / 0.106 x 100

                     = 18.6 %