Lab 14: Titration Lab

Last lab!

In this lab we determined the percent ionization of acetic acid in vinegar.  We filled a burette with 50 mL of NaOH (the base) that had a molarity of 0.25 M and slowly added drops to a flask with 7.6 mL vinegar (the acid) and water while stirring.  We added 5 drops of phenolphthalein to the solution as an indicator that would turn from clear to pink when the solution changed from an acidic form to a basic form.  The drops of NaOH created a momentary pink color in the solution, and once the solution reached a very light pink, we knew that it was the equivalence point between H+ ions and OH- ions.  We then recorded the volume of NaOH used by looking at the volume on the burette.

Titration setup

Analyte at equivalence point

Trial 1:

Molarity of acid: 0.86 M
Volume of acid: 7.60 mL

Molarity of base: 0.25 M
Volume of base: 26.0 mL

Trial 2:

Molarity of acid: 0.87 M
Volume of acid: 7.40 mL

Molarity of base: 0.25 mL
Volume of base: 25.7 mL

Average concentration of acetic acid: 0.87 M
[H3O+] = 10^(-pH) = 10^(-2.4) = 0.0040
Percent ionization: [H3O+] / molarity of acetic acid x 100 = 0.0040 / 0.87 = 0.46%

The percent ionization of acetic acid in vinegar is 0.46%.  This is such a low number because acetic acid is weak, which means that not all of the ions break apart and ionize, so there are not many H+ and OH- ions in the solution.