Question

A food chemist determines the concentration of acetic acid in a sample of apple cider vinegar by acid-base titration. What is the concentration of acetic acid in the vinegar if the density of the sample is 1.01 g/ml, the titrant is 1.002M NaOH and the average volume of titrant required to titrate 25.00 mL subsamples of the vinegar is 20.78mL? Express your answer the way a food chemist probably would; *as percent by mass*

Answer 1

Values:

Titrant NaOH, Concentration of titrant 1.002M, Volume to titrant is 20.78 mL

Sample Vinegar (CH₃-COOH), concentratio of sample ¿X?, Volume of sample 25 mL, sample density 1.01 g/ml

Stoichiometric Reaction:

1 mol CH₃-COOH + 1 mol NaOH --------> 1 mol CH₃-COONa + 1 mol H₂O

1 mol_x60 g/mol CH₃-COOH + 1 mol_x40 g/mol NaOH --------> 1 mol_x82 g/mol CH₃-COONa + 1 mol_x18 g/mol H₂O

60 g CH₃-COOH + 40 g NaOH --------> 82 g CH₃-COONa + 18 g H₂O

If:

[NaOH]=1.002 M = 1.002 mol/L

V_NaOH= 20.78 mL

Mass_NaOH= ¿?

And If:

Mass of NaOH used for titrant is 0.8329g, and stoichiometric 40g of NaOH can be neutralized by 60g of acetic acid CH₃-COOH. The amount of acetic acid in the sample neutralized by 1.002M NaOH is:

Then:

The acetic acid concentration by mass / mass percentage in the sample of vinegar is:

• Amount of sample volume expressed as mass, helping with the value of the density of the sample.

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• Expression of the mass concentration of vinegar percentage mass (%m/m)

or