Question

What is the pK_b of a weak base KA if the pH of a 0.094-M solution of KA is 10.01?

Answer 1

use:

pH = -log [H+]

10.01 = -log [H+]

[H+] = 9.772*10^-11 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)/(9.772*10^-11)

[OH-] = 1.023*10^-4 M

A- dissociates as:

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

9.4*10^-2 0 0

9.4*10^-2-x x x

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

Kb = x*x/(c-x)

Kb = 1.023*10^-4*1.023*10^-4/(0.094-1.023*10^-4)

Kb = 1.115*10^-7

use:

pKb = -log Kb

= -log (1.115*10^-7)

= 6.9527

Answer: 6.95