Question

4. In a typical titration a student needed 36.48 mL of 0.1067 M NaOH to reach the phenolphtha- lein endpoint. If the volume of the original vinegar sample was 5.00 mL, calculate the mass yo of acetic acid in the vinegar. Renart vour answer to the appropriate number or signincam figures.

Answer 1

2) Addition of water does not changes number of moles of acetic acid present in the sample.so that, no matter how much of water we add to the flask, it is just useful for changing titrating mixture volume.Hence titration can be done conveniently.

4)

CH3COOH + NaOH --> CH3COONa + H2O

no of mol of NaOH reacted = 36.48*0.1067 = 3.9 mmol

from equation, 1 mol CH3COOH = 1 mol NaOH

no of mol of Acetic acid(CH3COOH) reacted = 3.9 mmol

mass of Acetic acid(CH3COOH) reacted = n*M.wt = 3.9*10^-3*60 = 0.234 g

volume of vinegar solution = 5 ml

mass of vinegar solution = 5 g ( if density = 1 g/cc)

mass % of Acetic acid(CH3COOH) = 0.234/5*100 = 4.68%