What are the moles of acetic acid?

Question

A student used 27.48 mL of solution of NaOH 0.31 M to titrate 10.00 mL of vinegar.

a.  Calculate the moles of acetic acid in the sample

b.  Calculate the grams of acetic acid (HC₂H₃O₂) in the vinegar sample (MM acetic acid: 60.05 g/mol)

c.  Calculate the percent of acetic acid in the vinegar sample.

Answer 1

CH3COOH + NaOH ==> CH3COONa + H2O ... balanced equation for acetic acid + sodium hydroxide

a). moles NaOH used = 27.48 ml x 1 L / 1000 ml x 0.31 mol / L = 0.008519 moles NaOH

moles acetic acid = 0.008519 mols NaOH x 1 mol CH3COOH / mol NaOH = 0.0085 mols CH3COOH

b). grams CH3COOH = 0.0085 mols x 60.05 g/mol = 0.51 g CH3COOH

c). Cannot answer this part without knowing the mass of the original sample, or the density of the vinegar solution, or some other data allowing us to find the mass of the original 10 ml sample.