Question

What is the pH of a solution that is 1.3 M NH3? ( Ka=5.70 x 10^-10 for NH4+).

Answer 1

Since K_a*K_b = 10^-14

Thus, K_b = 10^-14/(5.7*10^-10) = 1.75*10^-5

The reaction taking place is :

                      NH₃ + H₂O ----> NH₄⁺ + OH^-

Initial                1.3                   0           0

Eqb                1.3-x                  x           x

Thus, Kb = [NH₄⁺]*[OH^-]/[NH₃] = x*x/(1.3-x) = 1.75*10^-5

Solving we get :

x = [OH] = 4.76*10^-3 M

Thus, pOH = -log([OH-]) = 2.32

Thus, pH = 14- pOH = 11.68