Question

A 0.120 M solution of a weak base has a pH of 11.27.

Part A

Determine Kb for the base.

Answer 1

use:

pH = -log [H3O+]

11.27 = -log [H3O+]

[H3O+] = 5.37*10^-12 M

use:

[OH-] = (1.0*10^-14)/[H3O+]

[OH-] = (1.0*10^-14)/(5.37*10^-12)

[OH-] = 1.862*10^-3 M

Let the base be written as BOH

BOH   <—> B+   +   OH-

0.120   0       0   (initial)

0.120-x   x       x   (at equilibrium)

[OH-] = x = 1.862*10^-3 M

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

Kb = x*x / (0.120-x)

Kb = (1.862*10^-3)*(1.862*10^-3)/(0.120 - 1.862*10^-3)

Kb = 2.93*10^-5

Answer: 2.93*10^-5