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 0.038 M in NH4Cl at 25 °C?

Answer 1

NH4Cl is the salt of strong acid and weak base

so cation involve in salt hydrolysis

NH4+ + H2O ---------------------> NH4OH + H+

0.038                                         0                  0 ------------------initial

0.038-x                                      x                 x -----------------at equilibrium

Ka = [NH4OH][H+]/[NH+]

Ka = x^2 / (0.038-x)

Kw / Kb = x^2 / (0.038-x)

1 x 10^-14 / 1.8 x 10^-5 = x^2 / (0.038-x)

5.56 x 10^-10 =x^2 / (0.038-x)

x^2 + 5.56 x 10^-10 x - 2.11 x10^-11 =0

x = 4.59 x 10^-6

x = [H+] = 4.59 x 10^-6 M

pH = -log[H+]

pH= -log (4.59 x 10^-6)

pH = 5.34