Question

A 0.250 M solution of a weak base has a pH of 9.30. What is the base hydrolysis constant, Kb, for the weak base?

Answer 1

from pH find the pOH

pH + pOH = 14

pOH = 14 - pH = 14 - 9.30 = 4.7

[OH-] = 10^-pOH = 10^-4.7 = 2 x 10^-5 M

let us week base is called B

construct ICE table

B (aq) + H2O (l) <-----> BH (aq) + OH- (aq)

I 0.25 0 0

C -x +x +x

E 0.25-x +x +x

but we have alredy concentration of OH- = x = 2 x 10^-5 M

if 1 mol of B hydrolyse will give 1 mol of BH and 1 mole of OH_

so concentration of [OH-] = [BH] = 2 x 10^-5   

concentration of [B] = 0.25 - x = 0.25 - 2 x 10^-5 = 0.24998 M

now expression for Kb is

Kb = [OH-] [BH] / [B]

Kb = [2 x 10^-5 ] [2 x 10^-5 ] / [0.24998]

Kb = 1.6 x 10^-9