Question

if 14.00 ml of 0.4 M NaOH is used to titrate the acid in an aliquot of acetic acid-butanol mixture, how many moles of acetic acid is present

Answer 1

Reaction between NaOH and acetic acid

NaOH (aq) + CH₃ COOH (aq) --- > CH₃COONa(aq) + H₂O (l)

Mol ratio:

1 mol NaOH : 1 mol CH₃ COOH

Calculation of moles of acetic acid

n CH₃ COOH = mol NaOH x 1 mol CH₃ COOH / 1 mol NaOH

mol NaOH = Volume of NaOH in L x molarity

Volume = 0.014 L

[NaOH] = 0.4 M

Lets plug these value to to get mol of acetic acid

n CH₃ COOH = ( 0.014 L x 0.4 M ) x 1 mol CH₃ COOH / 1 mol NaOH

                       = 5.60E-03 mol CH₃COOH

Ans : Moles of acetic acid would be 5.60E-03