What is the concentration of acetic acid, given it takes 22 mL of 0.2 M NaOH to neutralize a 52 mL sample of the acids?

Answer 1

Acetic acid = CH3COOH

Reaction: CH3COOH + NaOH ==> CH3COONa + H2O

moles of NaOH used = 22 ml x 1 L / 1000 ml x 0.2 mol / L = 0.0044 mols NaOH

moles CH3COOH present = 0.0044 mols NaOH x 1 mol CH3COOH / mol NaOH = 0.0044 mols CH3COOH

Concentration of CH3COOH = 0.0044 mols / 52 ml x 1000 mls / L = 0.0846 mol / L = 0.08 M (1 sig fig)