Question

What is the pH of 0.30 M ammonia solution? Kb of Ammonia is 1.8 x 10 NH3 (aq) + H20 (l) NH; (aq) + OH-(aq)

Answer 1

Lets write the dissociation equation of NH3

NH3 +H2O -----> NH4+ + OH-

0.3 0 0

0.3-x x x

Kb = [NH4+][OH-]/[NH3]

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 ((1.8*10^-5)*0.3) = 2.324*10^-3

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

so, x = 2.324*10^-3 M

So, [OH-] = x = 2.324*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.324*10^-3)

= 2.63

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.63

= 11.37

Answer: 11.37