Question

A food chemist determines the concentration of acetic acid in a sample of apple vinegar by acid-base titration. The density of the sample is 1.01 g/mL. The titrant is 1.022 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. ____ %

Answer 1

First find moles
0.025 Litres of NaOH @ 1.022 mol/litre = 0.025 moles of NaOH

by the eqaution we have
1 NaOH & 1 H C2H3O2 --> 1 H2O & 1 NaC2H3O2
0.025 moles of NaOH reacts with an equal number of moles of acetic acid = 0.025 mol acid

now find mass of acid in vinegar, using molar mass of acetic acid
0.02078 mol acid @ 60.05 g / mol HC2H3O2 = 1.249 grams of acid


now find the mass of the sample of vivegar
25.00 g @ 1.01 g/ml = 25.25 g of vinegar

now the acetic acid's concentration in the vinegar as percent by mass:
1.249 grams of acid / 25.25 g of vinegar =
4.947 % acid in the vinegar

4.95 % acid