Question

A 15.3 mL sample of vinegar, containing acetic acid, was titrated using 0.809 M NaOH solution. The titration required 24.06 mL of the base. Assuming the density of the vinegar is 1.01 g/mL, what was the percent (by mass) of acetic acid in the vinegar?

Answer 1

Write down the balanced chemical equation for the reaction between acetic acid (CH₃COOH) and NaOH.

CH₃COOH + NaOH -------> CH₃COONa + H₂O

As per the stoichiometric equation,

1 mole CH₃COOH = 1 mole NaOH.

Moles of NaOH required for the titration = (24.06 mL)*(0.809 M) = 19.46454 mmole.

Therefore, millimoles of CH₃COOH titrated = 19.46454 mmole.

Molar mass of CH₃COOH (= C₂H₄O₂) = (2*12.01 + 4*1.008 + 2*15.9994) g/mol = 60.0508 g/mol.

Mass of CH₃COOH corresponding to 19.46454 mmole = (19.46454 mmole)*(1 mole/1000 mmole)*(60.0508 g/1 mole) = 1.16886 g.

Volume of vinegar taken = 15.3 mL; density of vinegar = 1.01 g/mL; therefore, mass of vinegar = (15.3 mL)*(1.01 g/mL) = 15.453 g.

Mass percent of acetic acid in vinegar = (1.16886 g)/(15.453 g)*100 = 7.5639% ≈ 7.56% (ans).