Difficult Volumetric calculation?

Question

100cm^3 of vinegar at 15 degrees centigrade was made upto 250cm^3 of solution with distilled water. 25.0cm^3 of this solution required 16.9cm^3 of 0.5M NaOH for neutralisation. Assuming that all the acidity of vinegar is caused by ethanoic acid, CH3COOH, calculate the percentage by mass of the acid in the original vinegar. The density of vinegar at 15 degree centigrade is 1.02g cm^-3.

Answer 1

Acetic acid is monoprotic, so the neutralization is 1:1. The concentration of the acid in the 25 ml (and also in the 250 ml) is:

Ma = Mb * Vb / Va = 0.5 * 16.9 / 25 = 0.338 M

The dilution factor used in the first step is 2.5, so the concentration of the acid in the original 100 ml is 0.338 * 2.5 or 0.845 M. In 100 ml, that is 0.0845 moles, which given acetic acid's molecular weight of 60 g/mole, is 5.07 g of acetic acid.

With the density of vinegar at 1.02 g/ml, 100 ml would weigh 102 g. The mass percent of the acid is then 5.07 / 102 * 100, or 4.97%.