Question

What is the H3O+ concentration in 1.2 M NH4+ (aq)? The Kb of NH3 is 1.8x10^(-5)

Answer 1

NH4+    -----> NH3 +   H+
1.2                          0         0   (initial)
1.2 -X                    X         X (at equilibrium)

Ka of NH4+ = 10^-14 / Kb
     = 10^-14 / (1.8*10^-5)
     = 5.56*10^-10

Ka = [NH3+] [H+] / [NH4+]
5.56*10^-10 = X*X / (1.2-X)

since Ka is very small, X will be small and it can be ignored as compared to 1.2
So, above expression becomes,
5.56*10^-10 = X*X / (1.2)
X = 2.58*10^-5 M

So,
[H+]= X = 2.58*10^-5 M