A 10.00 mL sample of vinegar (an aqueous solution of acetic acid) is titrated with 0.5062 M NaOH(aq) and 16.58 mL is required to reach the equivalence point. If the density of vinegar is 1.006 g/mL, what is the percent by mass of acetic acid in vinegar

5.01%

Explanation:

Density of vinegar = mass/volume

Mass of 10.00 mL = density x volume

                           = 1.006 x 10 = 10.06 g

From the equation of reaction:

1 mole pf CH3COOH requires 1 mole of NaOH for neutralization.

mole of NaOH = molarity x volume

                      = 0.5062 x 0.01658

                       = 0.008392796‬ mole

0.008392796‬ mole of NaOH will therefore require 0.008392796‬ mole of CH3COOH.

mass of CH3COOH = mole x molar mass

                                     = 0.008392796‬ x 60.052

                                      = 0.504 g

Percentage by mass of acetic acid in the vinegar = 0.504/10.06 x 100%

    = 5.01%

The percent by mass of acetic acid in the vinegar is 5.01%

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