Question

The pH of a 0.032 M solution of a weak base is 9.83. What is the pKb for this base? That is the -log(Kb).

Answer 1

Let the weak base be written as BOH

we have below equation to be used

pH = -log [H3O+]

9.83 = -log [H3O+]

[H3O+] = 1.479*10^-10 M

we have below equation to be used

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

[OH-] = (1.0*10^-14)/(1.479*10^-10)

[OH-] = 6.761*10^-5 M

BOH dissociates as:

BOH +H2O -----> B+ + OH-

3.2*10^-2 0 0

3.2*10^-2-x x x

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

Kb = x*x/(c-x)

Kb = 6.761*10^-5*6.761*10^-5/(0.032-6.761*10^-5)

Kb = 1.431*10^-7

we have below equation to be used

pKb = -log Kb

= -log (1.431*10^-7)

= 6.84

Answer: 6.84