Question

a 2.00 mL sample of vinegar is titrated with 15.86 mL of 0.350 M NaOH to a phenolphthalein end point

A) calculate the molar its of acetic acid in the vinegar solution

B) calculate the % of acetic acid in the vinegar

Answer 1

The reaction between Acetic acid and sodium hydroxide is given by

CH₃COOH + NaOH ⟶ CH₃COONa + H₂O

       1mole of CH₃COOH = 1mole of NaOH

CH₃COOH solution                                                                NaOH solution

M₁ = molarity of CH₃COOH ?                            M₂ = Molarity of NaOH=0.35M

V₁ =Volume of CH₃COOH =2 mL             V₂ = Volume of NaOH used=15.86mL

n₁ = 1                                                                           n₂ = 1 (From equation)

                                                   

                                                        M₁ =

                                                M₁ =

Molarity of Aceticacid in vinegar, M₁ = 2.775 M

Strength of CH₃COOH solution = Molarity x Molecular mass

                                              = 2.775 X 60.06= 166.66 g/lit.

% of CH₃COOH in 2.0 mL of Vinegar solution = Strength x

                                                                  = 166.66 X 0.002 X 100

% of CH₃COOH in 2.0 mL of Vinegar solution = 33.33%