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 0.084 M in NH4Cl at 25 °C?

Answer is: pH of solution is 5,17.

Kb (NH₃) = 1,8·10⁻⁵.

c (NH₄Cl) = 0,084 M = 0,084 mol/L.

Chemical reaction: NH₄⁺ + H₂O → NH₃ + H₃O⁺.

Ka · Kb = 10⁻¹⁴.

Ka (NH₄⁺) = 10⁻¹⁴ : 1,8·10⁻⁵.

Ka (NH₄⁺) = 5,55·10⁻¹⁰.

[H₃O⁺] = [NH₃] = x.

Ka (NH₄⁺) = [H₃O⁺] · [NH₃] : [NH₄⁺].

5,55·10⁻¹⁰ = x² : (0,084 M - x).

Solve quadratic equation: x = [H₃O⁺] = 6,8·10⁻⁶ M.

pH = - log[H₃O⁺].

pH = - log (6,8·10⁻⁶ M) = 5,17.