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An impure sample of benzoic acid (HC7H5O2) is titrated with 0.1278M NaOH. A 0.5038 gram sample...

An impure sample of benzoic acid (HC7H5O2) is titrated with 0.1278M NaOH. A 0.5038 gram sample requires 23.81mL of sodium hydroxide to reach the endpoint. What is the percent by mass of benzoic acid in the sample?

HC7H5O2(aq) + NaOH(aq) = NaC7H5O2 + H2O (l)

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Answer #1

benzoic acid + NaOH -----------> Sodium Benzoate + H2O

Molarity of NaOH = 0.1278 M

Volume of NaOH = 23.81 mL

No. of moles of NaOH consumed by benzoic acid = 3.043 mmol.

3.043 mmoles of benzoic acid is present in the impure sample.

Molecular weight of benzoic acid = 144.11 g/mol

so 3.043 moles of benzoic acid corresponds to : 3.043 mole x 144.11 g/mole = 438.52 mg = 0.4385 g

Amount of impure sample = 0.5038 g

So the percentage by mass of benzoic acid in the sample = (0.4385/0.5038) x 100 = 87.04%

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