The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts wit

Question

The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion:
CH3COOH(aq)+NaOH(aq)?H2O(l)+NaC2H3O2(aq)

1.A titration of vinegar with a solution of NaOH was performed. If 3.10mL of vinegar needs 45.0mL of 0.130 M NaOH to reach the equivalence point in a titration, how many moles of NaOH were used in the titration?

2.How many moles of acetic acid were titrated?

3.Calculate the mass of acetic acid present in the vinegar sample.

4.How many grams of acetic acid are in each mL of vinegar?

5.How many grams of acetic acid would be in a 1.70qt sample of this vinegar?

Explanation / Answer

1. molarity = no. of moles / volume

so no.of moles of NaOH = 0.13M x 45mL = 5.85 mmol or 0.00585 mol

2. At equivalence point, same moles of acetic acid were titrated. So no. of moles acetic acid = 5.85 mmol or 0.00585 mol

3. no. of moles = mass taken/ molar mass = 0.00585 mol x 60.05 g/mol = 0.3513 g

4. because 0.3513 g is present in 3.1 ml, each mL has 0.3513/3.1 = 0.1133 g per mL

5. 1 quart = 32 fl oz
1 fl oz = 29.57 mL
1.70 quarts x 32 fl oz/qt x 29.57 mL/fl oz = 1608.6 mL
Molarity of acetic acid soln = 0.00585 moles / 3.1 mL = 1.88x10^-3 moles/mL
now, moles of new solution = 1.88x10^-3 mol/mL x 1.608.6 = 3.0356 mol
Mass of acetic acid present = 3.0356 mol x 60.05 g/mol = 182.29 g
so mass of acetic acid present in a 1.70qt of sample = 182.29 g

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