pH question?

Question

What is the pH of a 0.10 M solution of NH4NO2? Ka for HNO2 = 4.6 10-4, Kb for NH3 = 1.8 10-5 (hint: consider the equilibrium NH4+ + NO2- NH3 + HNO2)

a)6.30
b)2.26
c)9.26
d)2.17

Answer 1

NH3 + HNO2 <==> NH4NO2 <==> NH4+ + NO2-
It is very obvious that [NH3] = [HNO2], and [NH4+] = [NO2-]
We are given:
Ka = 4.6x10-4 = [H+]*[NO2-] /[HNO2]
Kb = 1.8x10-5 = [NH4+]*[OH-] /[NH3]
We also know:
[H+]*[OH-] = Kw = 1.0x10^-14
Therefore:
Ka*Kw /Kb = [H+]*[NO2-] *[H+]*[OH-] *[NH3] /{[HNO2]*[NH4+]*[OH-] }
= [H+]^2*[NO2-]*[NH3] /{[HNO2]*[NH4+]}
= [H+]^2 = Ka*Kw /Kb = 5.06x10^-7 (M)
So the answer is (a): pH = 6.30