Question

The Ka of a monoprotic weak acid is 2.50 × 10-3. What is the percent ionization of a 0.176 M solution of this acid?

Answer 1

Ka of monoprotic acid = 2.50 × 10^-3

   HA     ---------------> H+ + A-

0.176                          0         0

0.176 - x                     x          x

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

2.5 x 10^-3 = x^2 / 0.176 - x

x^2 + 2.5 x 10^-3 x - 4.4 x 10^-4 = 0

x = 0.0197

[H+] = 0.0197 M

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

                    = (0.0197 / 0.176 ) x 100

                    = 11.19%