Question

vinegar is an aqueous solution of acetic acid. By law, it must contain 4 g acetic acid (CH3CO2H 60.05 g/mol) per 100 mL solution although it is commonly up to 8% acetic acid. If a 15.00 mL aliquot of vinegar required 5.28mL of 0.10 M NaOH to reach the end point, is the sample legally vinegar?

Answer 1

. If a 15.00 mL aliquot of vinegar required 5.28mL of 0.10 M NaOH to reach the end point, is the sample legally vinegar?

4.0-g CH₃COOH x ( 1 mol CH₃COOH / 60.05-g CH₃COOH)/0.1-L = 0.666 M is concentration of standard solution.

Now for the titration:

1 mol acetic acid requires 1 mol NaOH. Since

moles = M x V, then

M x V (acid) = M x V(base)

M x 15-mL(acid) = 5.28-mL x 0.1 M(base)

M(acid) = 0.0352

=0.666/0.0352 = 18.9

'legal' vinegar is almost 19 times more concentrated