Question

If the Kb of a weak base is 3.8 × 10-6, what is the pH of a 0.39 M solution of this base?

Answer 1

lets B is the base

construct ICE table

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

I     0.39                                  0                  0

C    -x                                      +x               +x

E   0.39-x                                +x                +x

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

3.8 x 10^-6 = [x] [x] / [0.39-x]

x² + x 3.8 * 10^-6 - 1.482 * 10^-6 = 0

solve the quadratic equation

x = 0.00121 M = [OH-]

pOH = -log[OH-] = -log[0.00121] = 2.91

pH + pOH = 14

pH = 14-pOH = 14 - 2.91 = 11.01