Question

A titration of 25.00 mL of an acetic acid solution with 0.1220 M NaOH solution starts at a burette reading for NaOH of 0.17 mL. The phenolphthalein indicator turns light pink in the acid solution for over 30 seconds at a burette reading of 36.32 mL.

The volume of NaOH used to neutralize the acid is _______ mL.

Answer 1

The reaction between CH3COOH and NaOH is as follows:

CH3COOH + NaOH = CH3COONa + H2O

Given that

25.00 mL of an acetic acid

Molarity of NaOH = 0.1220 M

The volume of NaOH used to neutralize the acid is = final reading – initially reading in mL.

= 36.32 ml -0.17 ml

= 36.15 ml

Moles of NaOH = moalrity * volume in L

= 0.1220 * 36.15/1000

= 0.0044 Moles NaOH

Mole sof CH3COOH = 0.0044 Moles NaOH *1/1

= 0.0044 Moles CH3COOH

Moalrity = number of moles / volume in L

= 0.0044 Moles CH3COOH / 0.025

= 0.176 M