Question

A 50 ml sample of vinegar is titrated with 0.774 m naoh(ap). if the triation requires 41.6ml of naoh(ap), what is the concentration of acetic acid in the vinegar

Answer 1

CH3COOH + NaOH -----> CH3COONa + H2O

The number of mole CH3COOH is equal to the number of mole NaOH in this reaction, so we can write

M(CH3COOH)*V(CH3COOH) = M(NaOH)*V(NaOH)

M(CH3COOH)= M(NaOH)*V(NaOH)/V(CH3COOH)

M(CH3COOH)= 0.774 M*41.6 mL/50 mL ≈ 0.644 M

Answer is M(CH3COOH) ≈ 0.644 M.