What is the base dissociation constant for a 0.024 M basic solution that has a pH of 11.2 ?

Answer 1

Let B represent the weak base.

B + H2O <==> BH⁺ + OH^-

Kb = [BH⁺][OH^-]/[B]

From the pH = 11.2, we can find [OH^-] which will also be [BH⁺]

pH = 11.2 so pOH = 14 - 11.2 = 2.8 and [OH^-] = 1x10^-2.8 = 1.58x10^-3 M

Kb = (1.58x10^-3)(1.58x10^-3)/0.024 - 1.58x10^-3

Kb = 2.5x10^-6/0.0238

Kb = 1.05x10^-4