Question

NH3 is a weak base (Kb = 1.8 × 10–5) and so the salt NH4Cl acts as a weak acid. What is the pH of a solution that is 0.066 M in NH4Cl at 25 °C?

Answer 1

NH4+ + H2O <---> NH3 + H3O+
so
Kh = [NH3] [H3O+] / [NH4+]
and
Kh = Kw / Kb = 1x 10^-14 / 1.8x 10^-5 = 5.556x 10^-10
if we let the amount of NH4+ hydrolyzed be X
then
[NH4+] = 0.066 - X
[NH3] = X
[H3O+] = X
so
5.556x10^-10 = X² / (0.066 - X)
X² = 5.556 x 10^-10 (0.066 - X)
since Kh is very small, X*Kh will be likewise very small
so we can eliminate this
giving
X² = 5.556x 10^-10 (0.066) = 36.67 x 10^-12
X = 6.055 x 10^-6
then
pH = -log[H3O+] = -logX = -log(6.055 x 10^-6) = 5.2218