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ACS Analytical Chemistry Practice Exam

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What is the value of the equilibrium constant for the reverse reaction of a base reacting with water (i.e., base on reactants, conjugate acid and hydroxide on products)?
Correct Answer:
1/Kb
Explanation:
The key idea is that the equilibrium constant for a reaction and for its reverse are reciprocals. When you reverse a reaction, you swap products and reactants, so the expression inverts. For base reacting with water, the forward hydrolysis is B + H2O ⇌ BH+ + OH−, with Kb = [BH+][OH−]/[B]. If you look at the reverse, BH+ + OH− ⇌ B + H2O, the equilibrium constant is K_reverse = [B][H2O]/([BH+][OH−]). Water is the solvent, so its activity is effectively 1, giving K_reverse ≈ [B]/([BH+][OH−]) = 1/Kb. So the equilibrium constant for the reverse reaction is 1/Kb. (Remark: Ka of the conjugate acid BH+ relates via Ka(BH+) × Kb(B) = Kw, but that detail isn’t needed to answer this question.)

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