The concentration of ethanoic acid in diluted vinegar is determined by titrating with a standard NaOH solution and using stoichiometry to calculate moles and concentration.
The question deals with calculating the concentration of ethanoic acid in a diluted vinegar solution and subsequently in the original vinegar sample by performing a titration with a standard NaOH solution.
The neutralization reaction that takes place during the titration is: CH3COOH + NaOH → CH3COONa + H2O.
The student performed titrations using 25 cm3 portions of 0.12 M NaOH, and the mean titration volume was 20.5 cm3.
To find the concentration of the diluted ethanoic acid, the number of moles of NaOH used is calculated by multiplying the volume (in liters) by the molarity (mol/L).
Moles of NaOH used = 0.0205 L * 0.12 mol/L. Since the reaction ratio is 1:1, the moles of ethanoic acid is the same as the moles of NaOH.
To find the concentration of ethanoic acid in mol/L, divide the moles of ethanoic acid by the volume of the diluted acid that was titrated (expressed in liters).