Question

Initial NaOH volume mL	Final NaOH volume mL	Total volume of NaOH used mL
9	0	9
9	0.2	8.8
9	0.3	8.7

Three trials, average volume of NaOH used mL = 8.8

Average volume of NaOH used mL	Concentration CH3COOH in vinegar mol/L	% CH3COOH in vinegar
8.8	72	83

1. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer statement? Show your work.

2. 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? Show your work.

Answer 1

1. From data in the first table,

average volume of NaOH used = 8.8 ml

moles of NaOH used = 1.5 M x 8.8 ml = 13.2 mmol

mass of acetic acid in vinegar = 13.2 mmol x 60.0 g/mol/1000 = 0.80 g

% CH3COOH in vinegar = 0.80 g x 100/volume of vinegar solution

volume of vinegar solution is not given above. Feed the value and determine the % final answer.

2. moles NaOH used = 1.5 M x 7.5 ml = 11.25 mmol = 0.01125 mol

mass percent CH3COOH in vinegar = (0.01125 mol x 60 g/mol x 100)/7.0 ml = 9.64%