Find the percent ionization of : a 1.35 M solution of generic base B. Kb = 6.62 x 10-3

Answer 1

We must look at the hydrolysis of the base, B:

B + H2O ==> BH⁺ + OH-

Kb = [BH⁺][OH-] / [B]

6.62x10^-3 = (x)(x) / 1.35 - x

6.62x10^-3 = x² / 1.35 - x

x² + 6.62x10^-3 - 8.94x10^-3 = 0

x = 0.0913 = [OH-]

Percent ionization = [OH-] / [B]

% = 0.0913 M / 1.35 M (x100%) = 6.76% ionized