A50 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

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.

the reaction mixture contains some ammonia, plus a lot of unreacted nitrogen and hydrogen. the mixture is cooled and compressed, causing the ammonia gas to condense into a liquid. the liquefied ammonia is separated and removed. the unreacted nitrogen and hydrogen are then recycled back into the reactor .