Chemistry Lab: Titration of Vinegar

In class we were looking for the Molar Concentration of Acetic Acid in house hold vinegar.

Balanced equation: NaOH + CH3COOH ----> NaCH3COOH + H2O

Lab:

Trial                                        1                                2                                  3
Initial Volume                       33.2                           41.5                             49.9
Final Volume                        24.7                           33.2                             41.5
NaOH added                        8.5                             8.3                               8.4

Determine the average volume of NaOH added:

8.5 + 8.4 + 8.3 = 8.4mL -----> .0084L
           3

Determine the concentration of Acetic Acid.

1.00 mol x 0.0084L x 1 x 1 mol = .84 mol/L
      L                             1    .010 L

The actual value was .85 M.
My percent error was 1.2%

-Thomas H