Question

The Ka of a monoprotic weak acid is 8.87 x 10^-3. What is the percent ionization of a 0.121 M solution of this acid?

Answer 1

Ka = 8.87 x 10^-3

concentration = C = 0.121 M

HA    -------------->   H+    +      A-

0.121                       0              0

0.121 - x                  x                x

Ka = x^2 / 0.121 - x

8.87 x 10^-3 = x^2 / 0.121 - x

x = 0.0286

[H+] = 0.0286 M

% ionization = [H+] / C ) x 100

                    = 0.0286 / 0.121 ) x 100

% ionization = 23.6 %