Titration of Vinegar




Introduction:

Vinegar is one of a standard household substance that you have no doubt used. We add vinegar to our salad dressing and other dishes. Sometimes we take advantage of its acidic properties and use it for a cleanser. Household vinegar contains acetic acid in water. The amount of acetic acid is generally 5 % by mass acetic acid in the vinegar solution. You will determine the amount of acetic acid in vinegar by titration.

Discussion:

Analysis by titration is a very useful and commonly used technique in chemistry. The sample to be analyzed is usually a solution in an Erlenmeyer flask. A reagent is carefully added to the sample using a buret. A buret is like an upside down graduated cylinder. The reagent is added until the entire sample has reacted. Knowledge of the volume of reagent added and the concentration of the reagent allows us to calculate the amount of the substance in the sample that we are interested in. How will you know when to stop adding the reagent in the buret? An indicator is used to determine the equivalence point or endpoint of the reaction. The equivalence point is where the number of equivalents or in the sample is equal to the number of equivalents added with the buret. The equivalence point of the reaction is determined by the use of a visual indicator. For this titration the indicator is phenolphthalein, which changes from colorless at pH = 8 to red at pH = 10. Our sample, vinegar, is an acid, with pH less than 7. Our reagent will be sodium hydroxide, a base. It is fairly easy to follow the color change from colorless to red, and so the vinegar is placed in an Erlenmeyer flask with a few drops of phenolphthalein. The base is added in small portions so that the addition can be stopped when a permanent pinkish color appears.

You can determine the % by mass of acetic acid in vinegar from the following information:

  1. The volume of NaOH(aq) used to titrate the sample.
  2. The concentration of NaOH(aq) in moles/L.
  3. The stoichiometry of the reaction;

    4.  NaOH(aq) + HC2H3O2(aq) ® H2O(l) + NaC2H3O2(aq)

  4. Volume of vinegar and density of the vinegar.
  5. Molecular Weight of acetic acid.

Materials:

2 125 mL Erlenmeyer flasks, vinegar, 1 buret, 1 buret stand and holder, funnel, phenolphthalein, deionized water, 0.2 Molar NaOH(aq).

Procedure:

1. Rinse a 50 mL buret with water. Be sure to let the water run out of the tip. Rinse the buret twice with approximately 5 mL portions of 0.2 M NaOH(aq). Rotate the buret to make sure the sides are coated with the NaOH(aq). Let some of the NaOH(aq) out of the tip and pour the rest down the sink.
2. Obtain and clean a funnel and fill the buret with the NaOH solution until the volume reads about 10.00 mL. Let some the NaOH(aq) solution out of the tip into beaker. Be sure that the tip of the buret is filled and that there are no air bubbles in it. Now fill the buret to the 0.00 line.
3. Obtain two Erlenmeyer flasks and clean them well. Add 5.0 mL of vinegar from the dispenser to each flask. Add 25 mL of deionized water and two drops of phenolphthalein to each flask as well.
4. Titrate the vinegar solution by carefully adding NaOH(aq) from the buret into the Erlenmeyer flask containing the vinegar. Stop the titration when a faint pink color appears. This is the equivalence point. If the color is hot pink, redo the titration. (Run 1)
5. Refill the buret.
6. Repeat this titration for the second sample. (Run 2)

Disposal

Pour the unused NaOH(aq) from the buret back into the bottle. Pour the remaining solutions down the drain. Clean and return the buret.

Calculations:

The percent by mass of acetic acid present in a sample of vinegar is determined by first finding the amount of acetic acid present in the sample titrated. The moles or equivalents of acetic acid are equal to the moles or equivalents of NaOH used in the titration.

(1) Moles of NaOH = vol. of NaOH solution in liters x Molarity of NaOH.
Moles of NaOH = moles of acetic acid.
Grams of acetic acid = moles acetic acid/molar mass acetic acid.
The density of vinegar is 1.0 g/mL, so 5.0 mL = 5.0 grams vinegar.
% By mass acetic acid in vinegar is grams of acetic acid/5.0 grams vinegar.

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