If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass percent of CH3COOH in the vinegar sample?

Answer 1

Hi Alina,

In order to solve this problem, we need to assume that the density of vinegar is 1.0 g/mL. We will need this information to calculate the mass% of acetic acid (HAc) in the vinegar sample.

1) Find the moles of NaOH required to reach the equivalence point.

1.5 mol/L NaOH (7.5x10^-3 L) = 0.1125 mol NaOH

2) The stoichiometry of the reaction is one-to-one, so there are also 0.1125 moles of HAc in the 7.0 mL sample.

3) Convert moles of HAc to grams of HAc

0.1125 mol HAc (60.05 g/mol) = 0.6756 g HAc in the sample

4) Assume the density of the 7.0-mL vinegar sample to be 1.0 g/mL = 7.0 g of vinegar

5) Calculate the mass/mass% of the HAc in the sample.

0.6756 g HAc (100)/7.0 g sample = 9.7 mass/mass% => 2 sig. fig.'s

Best,

Randall