Question

What must be the molarity of an aqueous solution of trimethylamine, (CH3)3N, if it has a pH = 11.04? (CH3)3N+H2O⇌(CH3)3NH++OH−Kb=6.3×10−5

Answer 1

Let the concentration of (CH3)3N c molar

use:

pH = -log [H+]

11.04 = -log [H+]

[H+] = 9.12*10^-12 M

use:

[OH-] = Kw/[H+]

Kw is dissociation constant of water whose value is 1.0*10^-14 at 25 oC

[OH-] = (1.0*10^-14)/[H+]

[OH-] = (1.0*10^-14)/(9.12*10^-12)

[OH-] = 1.096*10^-3 M

(CH3)3N dissociates as:

(CH3)3N +H2O -----> (CH3)3NH+ + OH-

c 0 0

c-x x x

Kb = [(CH3)3NH+][OH-]/[(CH3)3N]

Kb = x*x/(c-x)

6.3*10^-5 = 1.096*10^-3*1.096*10^-3/(c-1.096*10^-3)

c-1.096*10^-3 = 1.908*10^-2

c=2.018*10^-2

Answer: 2.02*10^-2 M