The pH of a 0.172 M solution of benzoic acid (K_a = 6.31 × 10^-5) is

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The correct answer is A because the pH of a 0.172 M solution of benzoic acid (Ka = 6.31 × 10-5) can be calculated using the Henderson-Hasselbalch equation: pH = pKa + log([A-]/[HA]). Plugging in the values, we get pH = 4.20 + log([A-]/[HA]). Since the solution is 0.172 M in benzoic acid, [HA] = 0.172 M. We can use the Ka expression to solve for [A-]: Ka = [A-][H+]/[HA]. Plugging in the values, we get 6.31 × 10-5 = [A-][H+]/0.172. Solving for [A-], we get [A-] = 1.08 × 10-5 M. Plugging this value into the Henderson-Hasselbalch equation, we get pH = 4.20 + log(1.08 × 10-5/0.172) = 2.48.

The PH of a 0