Question

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

A) What is the pH of a 0.280 M ammonia solution?

Express your answer numerically to two decimal places.

B) What is the percent ionization of ammonia at this concentration?

Express your answer with the appropriate units.

Answer 1

Ammonia solution is a weak base and will dissociates as:     

                                  NH3 + H2O <----> NH4+ + OH-
initial concentration    0.28                    0               0
change                         -x                   x                x
at equilibrium                0.28-x              x              x

Now
Kb = 1.8 x 10^-5 = (x)(x)/ 0.28-x

1.8 X 10^-5 = x² / 0.28-x

As it is weak base we can ignore x in denominator

So 0.504 X 10^-5 = x²
x = [OH-]= 2.24 X 10^-3 M
pOH = - log (2.24 X 10^-3) = 2.65
pH = 14 - pOH = 14 - 2.65 =11.35

We can calculate percentage ionizaition as:

% ionization = x X 100 / Initial concentration

= 0.00224 x 100/ 0.280=0.80 %