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 0.074 M in NH4Cl at 25 °C?

Answer 1

NH4Cl -----> NH4^+ + Cl^-1

Ka =Kw/Kb = 1*10^-14/1.8*10^-5 = 5.56*10^-10
NH4^+ + H2O -----> NH3 + H3O^+ (NH4^+ acts as a conjugate acid)
5.56*10^-10 = [NH3][H3O+]/[NH4+]

H3O+ is the conjugate acid and controls the pH
NH4^+ + H2O -----> NH3 + H3O^+

0.074                        0           0

0.074-x                      x          x

5.56*10^-10 = x^2/(0.074-x)

x = 6.4 * 10^-6

pH = -log(6.4*10^-6)

     = 5.19