Question

acetic acid is an important ingredient of vinegar. a sample of 50.0 ml of a commercial vinegar is titrated against a 1.00 M NaOH solution. What is the molarity of acetic acid present in the vinegar if 5.75 ml of the base are needed for the titration?

Answer 1

Consider reaction , CH₃COOH _(aq) + NaOH _(aq) CH₃COONa _(aq) + H₂O (l)

From reaction, stoichiometric ratio = No. of moles of acid / No. of moles of base = 1 /1 = 1

We have correlation, M _acid$\times$ V _acid = M _base$\times$ V _base$\times$ Stoichiometric ratio

i e M _CH3COOH$\times$ V _CH3COOH = M _NaOH$\times$ V _NaOH$\times$ Stoichiometric ratio

$\therefore$ M _CH3COOH$\times$ 50.0 ml = 1.00 M $\times$ 5.75 ml $\times$ 1

M _CH3COOH = 1.00 M $\times$ 5.75 ml $\times$ 1 / 50.0 ml = 0.115 M

ANSWER : Molarity of acetic acid present in the vinegar = 0.115 M