Question

A 0.0100 M solution of a weak base has pH = 9.78 at T = 25 C. What is the numerical value for Kb for the weak base?

Answer 1

we have below equation to be used:

pH = -log [H+]

9.78 = -log [H+]

log [H+] = -9.78

[H+] = 10^(-9.78)

[H+] = 1.66*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)/(1.66*10^-10)

[OH-] = 6.026*10^-5 M

Lets write the dissociation equation of BOH

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

1*10^-2 0 0

1*10^-2-x x x

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

Kb = x*x/(c-x)

Kb = 6.026*10^-5*6.026*10^-5/(0.01-6.026*10^-5)

Kb = 3.65*10^-7

Answer: 3.65*10^-7