Question

If the Kb of a weak base is 8.1 × 10-6, what is the pH of a 0.12 M solution of this base?

Answer 1

Lets write the dissociation equation of BOH

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

0.12 0 0

0.12-x x x

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

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 ((8.1*10^-6)*0.12) = 9.859*10^-4

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

so, x = 9.859*10^-4 M

so.[OH-] = x = 9.859*10^-4 M

we have below equation to be used:

pOH = -log [OH-]

= -log (9.859*10^-4)

= 3.01

we have below equation to be used:

PH = 14 - pOH

= 14 - 3.01

= 10.99

Answer: 11