Determine the lactic acid content (in M) of the 10.-mL fermented milk sample.

Question

Mola and Rity were tasked to analyze the acidity of a 10.-mL fermented milk sample diluted with water in a 250-mL volumetric flask. Mola obtained a 20.-mL aliquot of this solution, diluted it to 50. mL, and then titrated this diluted sample. Titration data were given to Rity and she found out that the 50.-mL sample contains 0.75 mmol of lactic acid.

Answer 1

0..75 mmol lactic acid in 50 ml = 0.75 mmol lactic acid in the 20 ml aliquot.

The 20 ml aliquot came from 250 ml so amount in the 250 ml = 12.5 x 0.75 mmol = 9.375 mmoles lactic acid in the 250 ml = 9.375 mmol lactic acid in the original 10 ml.

To convert this to M (molar = moles/L) we do the following:

9.375 mmoles / 10 ml x 1000 ml / L x 1 mol / 1000 mmoles = moles / L = 0.9375 M = 0.94 M (2 sig. figs.)