What is the pH of a 0.30 mol L^-1 pyridine solution that has K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is below: C₅H₅N(aq)+H₂O(l) ⇌ C₅H₅NH+(aq)+OH-(aq) A) 4. 62 B) 8.72 C) 9. 38 D) 10. 38

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To solve this problem, we need to use the equation for the dissociation of pyridine and the acid dissociation constant (Ka) which is related to the given base dissociation constant (Kb) by the following equation: Ka × Kb = Kw where Kw is the ion product constant of water, which is equal to 1.0 × 10^-14 at 25°C. From the given equation for the dissociation of pyridine, we can write the expression for Ka as follows: Ka = [H₅NH+][OH-]/[C₅H₅N] Since the solution is basic, we can assume that most of the pyridine is dissociated into H₅NH+ and OH- ions. Therefore, we can simplify the expression for Ka as follows: Ka ≈ [OH-]^2/[C₅] Substituting the given values, we get: Kb = 1.9 × 10^-9 [C₅] = 0.30 M Using the relationship between Ka and Kb, we get: Ka × Kb = Kw Ka = Kw/Kb Ka = 1.0 × 10^-14/1.9 × 10^-9 Ka = 5.26 × 10^-6 Finally, we can use the expression for the pH of a weak base solution: pH = pKb + log([H₅NH+]/[C₅]) At equilibrium, the concentration of H₅NH+ is equal to the concentration of OH-, which can be calculated from the expression for Ka: [OH-] = sqrt(Ka × [C₅]) ≈ 7.43 × 10^-3 M Substituting the values, we get: pH = 14.00 - pKb - log(Ka/[C₅]) pH = 14.00 - (-8.72) - log(5.26 × 10^-6/0.30) pH ≈ 9.38 Therefore, the answer is C.

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