Question

If the Kb of a weak base is 1.2 × 10-6, what is the pH of a 0.32 M solution of this base?

Answer #1

Let the base be written as BOH

Lets write the dissociation equation of BOH

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

0.32 0 0

0.32-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 ((1.2*10^-6)*0.32) = 6.197*10^-4

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

so, x = 6.197*10^-4 M

so.[OH-] = x = 6.197*10^-4 M

we have below equation to be used:

pOH = -log [OH-]

= -log (6.197*10^-4)

= 3.21

we have below equation to be used:

PH = 14 - pOH

= 14 - 3.21

= 10.79

Answer: 10.79

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