Vinegar is a solution of acid, HC_2 H_2 O_2 (aq) often made by fermenting grapes

Question

Vinegar is a solution of acid, HC_2 H_2 O_2 (aq) often made by fermenting grapes beyond the wine stage. The FDA require that an aqueous acetic acid solution must contain at least 4% acetic acid by mass to be legally called vinegar. In this experiment, you will titrate samples of distilled white vinegar to determine this percentage. 1. Set up a clean, rinsed buret (as you did on day 1). You must use the same NaOH solution from day 1. 2. Use the auto dispenser (as demonstrated by your instructor) to add 5.00 mL of vinegar into a clean Erlenmeyer flask. 3. As on day 1, add 2 drops of indicator and about 20 mL of deionized water, then titrate with your NaOH solution to the endpoint (= first drop of titrate that gives a permanent pink color). Record the data for 3 trials in Table 2. 1. Calculate the molarity of the acetic acid in vinegar for each trial using dimensional analysis Trial 1: ___ Trial 2: ____ Trial 3: ___ Avg: ___ 2. Use the average molarity above and a density of 1.00 g/mL for the vinegar to calculate the % acetic acid (w/w) using dimensional analysis 3. Another company sells vinegar containing 7.85% (w/w) acetic acid. What is the molarity of acid in this solution?

Explanation / Answer

1. Since molarity of base is not given.

We would assume molarity of base = 0.1 M

Trial 1 : molarity of acetic acid = 0.1 M x 20.10 ml/5 ml = 0.4 M

Trial 2 : molarity of acetic acid = 0.1 M x 19.70 ml/5 ml = 0.4 M

Trial 3 : molarity of acetic acid = 0.1 M x 20.39 ml/5 ml = 0.5 M

2. average molarity of acetic acid = 0.42 M

% acetic acid = 0.42 M x 0.005 L x 60 g/mol x 1000/5 ml x 1 g/ml = 2.52%

3. %(w/w) = 7.85%

molarity = 7.85 x 0.42/2.52 = 1.31 M

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