Question

A food chemist determines the concentration of acetic acid in a sample of apple cider vinegar by acid–base titration.
The density of the sample is 1.01 g/mL. The titrant is 1.002 M NaOH.
The average volume of titrant required to titrate 25.00 mL subsamples of the vinegar is 20.78 mL.

What is the concentration of acetic acid in the vinegar?

Express your answer the way a food chemist probably would: as percent by mass.
CH3COOH + NaOH → CH3COONa + H2O

Answer 1

4.95 % is the percent by mass concentration of acetic acid in vinegar

Explanation:

We have the following chemical reaction:

CH₃COOH + NaOH → CH₃COONa + H₂O

First we determine the number of moles of NaOH.

molar concentration = number of moles / volume of solution (L)

number of moles = molar concentration × volume of solution (L)

number of moles of NaOH = 1.002 × 0.02078 = 0.02082 moles

Knowing the chemical equation, we devise the following reasoning:

if        1 mole of acetic acid (CH₃COOH) reacts with 1 mole of sodium hydroxide (NaOH)

then  X moles of acetic acid (CH₃COOH) reacts with 0.02082 moles of sodium hydroxide (NaOH)

X = (1 × 0.02082) / 1 = 0.02082 moles of acetic acid

mass = number of moles × molecular weight

mass of acetic acid = 0.02082 × 60 = 1.249 g

Knowing that the density of the vinegar sample is 1.01 g/mL and the volume of 25 mL, we calculate the mass of the sample solution.

density = mass / volume

mass = density × volume

mass of  vinegar solution = 1.01 g/mL × 25 mL = 25.25 g

Now to calculate the concentration of acetic acid in the vinegar as percent by mass, we use the following formula.

concentration = (mass of solute / mass of solution) × 100

concentration = (1.249 / 25.25) × 100 = 4.95 %

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