HC2H3O2(aq) + KOH(aq) → KC2H3O2(aq) + H2O(l)

Question

If 32.7 mL of 0.255 M KOH is required to completely neutralize 21.0 mL of a HC2H3O2 solution, what is the molarity of the acetic acid solution?

Answer 1

Neutralization reaction between an acid (HC₂H₃O₂) and a base (KOH)

HC₂H₃O₂(aq) + KOH(aq) ==> KC₂H₃O₂(aq) + H₂O(l) ... note the 1:1 mole ratio between acid and base

moles of KOH used = 32.7 ml x 1 L/1000 ml x 0.255 moles/L = 0.008339 moles KOH

moles of HC₂H₃O₂ present = 0.008339 moles KOH x 1 mole acid/mole KOH = 0.008339 moles HC₂H₃O₂

Molarity of HC₂H₃O₂ = 0.008339 moles/0.0210 L = 0.3971 M = 0.397 M (to 3 significant figures)