What is the pH of a 2.62 x 10 -1 M NH3 solution? NB the pKa for NH4+ = 9.241

NH3 + H2O ---> NH4^+ + OH^-
pKa + pKb = 14
pKb = 14 - 9.241 = 4.759
Kb = 1.74x10^-5
Kb = 1.74x10^-5 = [NH4^+}[OH^-]/[NH3] = (x)(x)/0.262 - x (neglect x in denominator)
1.74x10^-5 x 0.262 = x^2
x^2 = 4.56x10^-6
x = 2.14x10^-3 = [OH]
pOH = 2.67
pH = 14 - 2.67
pH = 11.33