Question

Find [H+] in 0.40 M NH₄NO₃ at 25 degrees Celsius given that K_b = 1.8 x 10^-5 for NH₃ (aq).

Answer 1

Lets make an ICE table for NH4NO3 in water,

                        NH4NO3 <==> NH4+ + NO3^-

initial                  0.40                  0            0

change                 -x                   +x          +x

Equilibrium       0.40-x                  x            x

So we can write,

Kc = [NH4+][NO3^-]/[NH4NO3]

     = x^2/(0.40-x)

NH4+ <===> NH3 + H+ .......Kb

[NH4+] = [H+]

H2O <==> H+ + OH- ........... Kw

So, Kc = Kw/Kb = 1 x 10^-14/1.8 x 10^-5

                        = 5.56 x 10^-10

5.56 x 10^-10 = x^2/(0.40-x)

x^2 + 5.56 x 10^-10 - 2.22 x 10^-10 = 0

x = 1.49 x 10^-5 M

[H+] = 1.49 x 10^-5 M