Question

What is the pKb of a weak base KA if the pH of a 0.075-M solution of KA is 9.69?

Answer 1

use:

pH = -log [H+]

9.69 = -log [H+]

[H+] = 2.042*10^-10 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)/(2.042*10^-10)

[OH-] = 4.898*10^-5 M

A- dissociates as:

A- +H2O -----> HA + OH-

7.5*10^-2 0 0

7.5*10^-2-x x x

Kb = [HA][OH-]/[A-]

Kb = x*x/(c-x)

Kb = 4.898*10^-5*4.898*10^-5/(0.075-4.898*10^-5)

Kb = 3.201*10^-8

use:

pKb = -log Kb

= -log (3.201*10^-8)

= 7.4948

Answer: 7.49