Question

A 1.70 M solution of a weak base has a pH of 9.7. Calculate the Kb.

Answer 1

we have below equation to be used:

pH = -log [H+]

9.70 = -log [H+]

log [H+] = -9.70

[H+] = 10^(-9.70)

[H+] = 1.995*10^-10 M

we have below equation to be used:

[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)/(1.995*10^-10)

[OH-] = 5.012*10^-5 M

Lets write the dissociation equation of BOH

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

1.70 0 0

1.70-x x x

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

Kb = x*x/(c-x)

Kb = 5.012*10^-5*5.012*10^-5/(1.7 - 5.012*10^-5)

Kb = 1.48*10^-9

Answer: Kb = 1.48*10^-9