If 29.7 ML of 0.205 M KOH is required to completely neutralize 25.0 ML of a CH3COOH Sol what is the molarity?

Question

What is the molarity of the acetic acid solution
CH3COOH(aq)+KOH(aq)=CH3COOK(aq)+H2O (l)

Answer 1

One way to do this is to use M1V1 = M2V2, since n1 = n2.

Solve for M2 = M1V1/V2

M2 = 0.205 M x 29.7 mL/25.0 mL = 0.244 M CH3COOH

Answer 2

With monobasic acids and bases, [Vol * molarity ]acid = [Vol * molarity.] base. You have 3 terms known.