Question

If the Kb of a weak base is 4.1 × 10-6, what is the pH of a 0.49 M solution of this base?

Answer 1

B + H2O <----------------> BH+    + OH-

0.49                                0             0 -----------------> initial

0.49-x                             x             x -------------------> equilibrium

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

Kb = x^2 / 0.49 -x

4.1 x 10^-6 = x^2 / 0.49 -x

x^2   + 4.1 x 10^-6 x - 2 x 10^-6 = 0

by solving this

x = 1.41 x 10^-3

x = [OH-] =1.41 x 10^-3 M

pOH = -log [1.41 x 10^-3 ]

pOH = 2.85

pH + pOH = 14

pH = 14-pOH

pH = 11.15