Question

The Ka of a monoprotic weak acid is 6.55 × 10-3. What is the percent ionization of a 0.176 M solution of this acid?

Answer 1

Ka of a monoprotic weak acid = 6.55 × 10^-3

concentration of acid = 0.176 M

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

0.176                         0         0

0.176 - x                    x          x

Ka = x^2 / 0.176 - x

6.55 × 10^-3 = x^2 / 0.176 - x

x = 0.0308

[H+] = 0.0308 M

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

                                 = (0.0308 / 0.176) x 100

                                 = 17.5%

percent ionization % = 17.5%