How can I calculate the Kb of a weak base when given the concentration, and the pH at equilibrium?

Question

The problem:A weak base NH3 with a concentration of 0.520 has a pH of 9.00 at equilibrium. Calculate the Kb for this base.

( Equation: NH3 H2O <----> NH4 OH- )

Any help is GREATLY appreciated!

(Summer Assignment! Due tomorrow!)

Answer 1

pH + pOH = 14  given rule

Therefore if pH = 9, pOH = 5, which means OH = 1x10^-5

Kb = {nh4}{OH}/NH3  for simplification we can ignore water

OH must = NH4

by substitution

Kb = (10^-5)(10^-5)/0.52

=5.2E-11

Answer 2

0.520 M (I assume!) NH3 will partially dissociate in water to form x M NH4+ and an equal amount of OH-. The [NH3] at equilibrium = 0.520 - x M.

But we were told that the pH = 9.00, so

pOH = 14 - 9.00 = 5.00.

[OH-] = 10^-pOH = 10^-5.00 = 1 x 10^-5 M = x

Kb = [NH4+][OH-] / [NH3] = (x)(x) / 0.520 - x

= (1 x 10^-5)^2 / (0.520 - 1 x 10^-5) =

1.9 x 10^-9

The real Kb for NH3 is 1.8 x 10^-5.