Benzoic acid (C₆H₅CO₂H = HBz)solutions are sometimes used in experiments to determine the molarity of a basic solution of unknown concentration.What is the pH of a 0.100 M solution of benzoic acid if K_a = 6.5 × 10^-5 and the equilibrium equation of interest is HBz(aq)+ H₂O(l)⇌ H₃O⁺ + Bz^-(aq)?

Question

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The correct answer is B because the pH of a 0.100 M solution of benzoic acid can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]), where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid. Substituting the given values into the equation, we get: pH = -log(6.5 × 10^-5) + log([Bz-]/[HBz]) = 4.19 + log([Bz-]/[HBz]). Since the solution is 0.100 M in benzoic acid, [HBz] = 0.100 M. At equilibrium, [Bz-] = [H3O+], so we can substitute [H3O+] for [Bz-] in the equation: pH = 4.19 + log([H3O+]/0.100 M). Solving for [H3O+], we get: [H3O+] = 10^(pH - 4.19) = 10^(2.59 - 4.19) = 10^-1.60 = 2.51 × 10^-2 M. Therefore, the pH of the solution is 2.59.

Benzoic Acid (C6H5CO2H = HBz)solutions Are Sometimes Used in Experiments

Benzoic Acid (C₆H₅CO₂H = HBz)solutions Are Sometimes Used in Experiments