[Olympiad] Calculate the pH of a 0.02 M solution of Pyridine (C₅H₅N). (Kb = 1.7 × 10⁻⁹)
-
5.23
-
8.77
-
9.12
-
10.45
📝 Detailed explanation
[OH⁻] = √(Kb * C) = √(1.7 × 10⁻⁹ * 0.02) = √(3.4 × 10⁻¹¹) = 5.83 × 10⁻⁶ M.
pOH = -log(5.83 × 10⁻⁶) = 5.23.
pH = 14 - 5.23 = 8.77.
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