Question

The Ka of a monoprotic weak acid is 3.98 × 10-3. What is the percent ionization of a 0.146 M solution of this acid? Solve this using the quadratic equation.

Answer 1

HA   ---------------> H+    +   A-

0.146                      0           0

0.146 - x                 x            x

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

3.98 x 10^-3 = x^2 / 0.146 - x

x^2 + 3.98 x 10^-3 x - 5.81 x 10^-4 = 0

x = 0.0222

percent ionization = x / C x 100

                             = (0.0222 / 0.146 ) x 100

                            = 15.2 %