What is the pH of a 2.4 M pyridine solution that has K_b = 1.9 × 10^-9? The equation for the dissociation of pyridine is C₅H₅N(aq)+ H₂O(l)⇌ C₅H₅NH⁺(aq)+ OH^-(aq).

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We can start by writing the equilibrium constant expression for the dissociation of pyridine: K_b = [C₅H₅NH⁺][OH^-]/[H₅N] From the equation, we can see that pyridine acts as a base and accepts protons (H+) to form its conjugate acid, which is the species present in the numerator of the equilibrium constant expression. Therefore, we can write: Kw = [H+][OH-] = 1.0 × 10^-14 We can also write the equilibrium constant expression for the autoionization of water: Kw = [H+][OH-]/[H2O] = 1.0 × 10^-14 Assuming that the concentration of water remains essentially constant, we can use the value of Kw to calculate [H+]: [H+][OH-] = 1.0 × 10^-14 [OH-] = Kw/[H+] = 1.0 × 10^-14/[H+] Now, we need to consider the effect of pyridine on the acidity of the solution. Since pyridine is a weak base, it will react with H+ to form the pyridinium ion, which is a weak acid: H₅N + H+ ⇌ H₅NH+ The equilibrium constant expression for this reaction is: Ka = [H₅NH+]/[H₅N][H+] We can write a mass balance equation for pyridine: [H₅N] + [H₅NH+] = [H₅N]T = 2.4 M At equilibrium, we can assume that [H₅NH+] << [H₅N], so we can simplify the equilibrium constant expression to: Ka ≈ [H₅NH+]/[H₅N] We can use the value of Ka and the mass balance equation to solve for [H+]: Ka = [H₅NH+]/[H₅N] = [H+] 1.9 × 10^-9 = [H+] pH = -log[H+] = -log(1.9 × 10^-9) ≈ 9.83 Therefore, the pH of the 2.4 M pyridine solution is approximately 9.83. The correct answer is choice C.

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