[Olympiad] Calculate the pH of a 0.05 M solution of Ammonia (NH₃). (Kb = 1.8 × 10⁻⁵)

[OH⁻] = √(Kb * C) = √(1.8 × 10⁻⁵ * 0.05) = √(9 × 10⁻⁷) = 9.49 × 10⁻⁴ M. pOH = -log(9.49 × 10⁻⁴) ≈ 3.02. pH = 14 - 3.02 = 10.98.

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