Calculate the pH of a solution prepared by dissolving 0.350 mol of benzoic acid (C₇H₅O₂H)and 0.250 mol of sodium benzoate (NaC₇H₅O₂)in water sufficient to yield 1.00 L of solution. The K_a of benzoic acid is 6.50 × 10^-5. A)4.333 B)4.041 C)9.959 D)9.667 E)5.190

Question

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First, we need to calculate the concentration of benzoic acid and sodium benzoate in the solution: [C₇H₅O₂H]=0.350 mol/1.00 L=0.350 M [NaC₇H₅O₂]=0.250 mol/1.00 L=0.250 M Next, we need to use the Henderson-Hasselbalch equation: pH = pKa + log([base]/[acid]) where pKa is the dissociation constant of benzoic acid, which is given as 6.50 × 10^-5, [base] is the concentration of sodium benzoate, and [acid] is the concentration of benzoic acid. Plugging in the values, we get: pH = 4.82 + log(0.250/0.350) pH = 4.041 Therefore, the pH of the solution is 4.041. The correct answer is B.

Calculate the PH of a Solution Prepared by Dissolving 0