Question

If the Kb of a weak base is 7.5 × 10-6, what is the pH of a 0.33 M solution of this base?

Answer 1

Lets write the dissociation equation of B

B +H2O -----> BH+ + OH-

0.33 0 0

0.33-x x x

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

Kb = x*x/(c-x)

Assuming small x approximation, that is lets assume that x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((7.5*10^-6)*0.33) = 1.573*10^-3

since c is much greater than x, our assumption is correct

so, x = 1.573*10^-3 M

So, [OH-] = x = 1.573*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (1.573*10^-3)

= 2.80

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.80

= 11.20

Answer: 11.20