calculating pH of a weak base

What is the pH of a 0.1 molL-1 solution of NH3? Given Pka(NH4+) = 9.25

Ka = inv log - 9.25
Ka = 5.62x 10^-10

Kb = Kw/ Ka
Kb = 1 x 10^-14 / 5.62 x 10^-10
Kb = 1.77 x 10^-5

NH3 + H2O -----> NH4+ + OH-
[NH3] = 0.1
[NH4+] [OH-] = X

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

Kb = [x][x] / [0.1]

[x] = square root of [1.77 x 10^-5] x [0.1]

[x] = [OH-] = 1.33 x 10^-3

pOH = -log[OH-]

pOH = -log [1.33 x 10^-3]

pOH = 2.876

pH = 14 - 2.876

pH = 11.12