Question

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.What is the pH of a 0.335 M ammonia solution?What is the percent ionization of ammonia at this concentration?

Answer 1

1)

Lets write the dissociation equation of NH3

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

0.335 0 0

0.335-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)*0.335) = 2.456*10^-3

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

so, x = 2.456*10^-3 M

So, [OH-] = x = 2.456*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.456*10^-3)

= 2.61

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.61

= 11.39

Answer: 11.39

2)

% dissociation = (x*100)/c

= 2.456*10^-3*100/0.335

= 0.7331 %

Answer: 0.7331 %