Choose the correct answer

Question

A chemist mixes 0.20 mol/l solution of a weak base where kb=3.82*10^-10. The PH of this solution is

a)between 9 and 12

b)Between 5 and 7

c)between 7 and 9

d)between 12 and 14

Answer 1

Let the weak base be represented by B. When B is put in water, you have...

B + H2O ==> BH⁺ + OH^-

Kb = [BH+][OH-]/[B] = (x)(x)/0.2 - x and assuming x is small relative to 0.2 M, we can ignore it in denominator

3.82x10^-10 = x²/0.2

x² = 7.64x10^-11

x = 8.7x10^-6 M = [OH-] = [BH+]

pOH = -log [OH-] = 5.1

pH = 14 - pOH

pH = 8.9 (answer c)