Question

The Ka of a monoprotic weak acid is 6.59 × 10-3. What is the percent ionization of a 0.149 M solution of this acid?

Answer 1

HA   --------------> H+ + A-

0.149                     0       0

0.149 - x                x        x

Ka = x^2 / 0.149 - x

6.59 x 10^-3 = x^2 / 0.149 - x

x = 0.0282

% ionization =( x / initial concentration ) x 100

                    = 0.0282 / 0.149 ) x 100

% ionization = 18.9 %