Question

Ammonia, NH3, is a weak base with a Kb value of 1.8×10−5.

Part A

What is the pH of a 0.480 M ammonia solution?

Express your answer numerically to two decimal places.

Part B

What is the percent ionization of ammonia at this concentration?

Express your answer with the appropriate units.

Answer 1

A)

Lets write the dissociation equation of NH3

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

0.48 0 0

0.48-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)*0.48) = 2.939*10^-3

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

so, x = 2.939*10^-3 M

So, [OH-] = x = 2.939*10^-3 M

we have below equation to be used:

pOH = -log [OH-]

= -log (2.939*10^-3)

= 2.53

we have below equation to be used:

PH = 14 - pOH

= 14 - 2.53

= 11.47

Answer: 11.47

B)

  

% dissociation = (x*100)/c

= 2.939*10^-3*100/0.48

= 0.6124 %

Answer: 0.612 %