An impure sample of benzoic acid (C₆H₅COOH,122.12 g/mol) is titrated with 0.9855 M NaOH.A 3.412-g sample requires 24.43 mL of titrant to reach the endpoint.What is the percent by mass of benzoic acid in the sample?
C₆H₅COOH(aq) + NaOH(aq) → NaC₆H₅COO(aq) + H₂O(l)

Question

Quizplus · Accepted Answer

To find the percent by mass of benzoic acid, calculate the moles of NaOH used: 0.02443 L * 0.9855 mol/L = 0.02408 mol. Since the reaction is 1:1, moles of benzoic acid = moles of NaOH = 0.02408 mol. Mass of benzoic acid = 0.02408 mol * 122.12 g/mol = 2.940 g. Percent by mass = (2.940 g / 3.412 g) * 100% = 86.18%.

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