# A 10.7 mL sample of vinegar, containing acetic acid, was titrated using 0.439 M NaOH solution. The titration required 22.83 mL of the base.?

A 10.7 mL sample of vinegar, containing acetic acid, was titrated using 0.439 M NaOH solution. The titration required 22.83 mL of the base. What was the molar concentration of acetic acid in the vinegar? Assuming the density of the vinegar is 1.01 g/mL, what was the percent (by mass) of acetic acid in the vinegar?

Whats the chemical formula?

Relevance

The formula for acetic acid is HC2H3O2 or CH3COOH. The neutralization reaction is:

HC2H3O2 + NaOH --> NaC2H3O2 + H2O

Moles NaOH = 0.439 mol/L X 0.02283 L = 0.01002 mol NaOH

Moles NaOH = moles acetic acid

Molarity acetic acid = 0.01002 mol / 0.0107 L = 0.937 mol/L acetic acid in vinegar

To calculate the % by mass, assume you have 1.00 L of vinegar. That volume has a mass of:

1000 mL X 1.01 g/mL = 1010 g

The mass of acetic acid in that volume = 0.937 mol X 60.05 g/mol = 56.2 g acetic acid

%acetic acid = (56.2 g / 1010 g) X 100 = 5.56%