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 0.084 M in NH4Cl at 25 °C?

Answer 1

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.