Question

Exercise 1: Determining the Concentration of Acetic Acid

Data Table 1. NaOH Titration Volume

	Initial NaOH Volume (mL)	Final NaOH Volume (mL)	Total volume of NaOH used (mL)
Trial 1	7ml	0	7
Trial 2	7ml	0.4	6.7
Trial 3	7ml	0.6	6.5

Data Table 2. Concentration of CH₃COOH in Vinegar

Average volume of NaOH used (mL)	Concentration CH3COOH in vinegar (mol/L)	% CH3COOH in vinegar
6.7	65mol/L	77%

Questions:

If the manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid, what would the percent error between your result and the manufacturer’s statement be? Show your calculations.

What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte instead of distilled white vinegar?

How would your results have differed if the tip of the titrator was not filled with NaOH before the initial volume reading was recorded? Explain your answer.

How would your results have differed if you had over-titrated (added drops of NaOH to the analyte beyond the stoichiometric equivalence point)? Explain your answer.

If a 7.0 mL sample of vinegar was titrated to the stoichiometric equivalence point with 7.5 mL of 1.5M NaOH, what is the mass percent of CH₃COOH in the vinegar sample? Show your calculations.

Why is it important to do multiple trials of a titration, instead of only one trial?

A student did not read the directions to the experiment properly and mixed up where to place the NaOH solution and the vinegar. He put the vinegar in the titrator and the measured amount of NaOH in the beaker. He then added a drop of the phenolphthalein to the solution in the beaker. Does the student need to empty out all of the solutions and start over again or can he go ahead and run the titration? If he runs the titration using the solutions as given above, what should he expect to see happen for results?

Answer 1

1) Volume of vinegar = 7.0 mL

     Volume NaOH = (7+6.6+6.4)/3

                               = 6.7 mL

Concentration of NaOH = 1.5M

Concentration acetic acid = (concentration of NaOH*volume of NaOH)/volume of vinegar

                                             = (6.7mL*1.5M)/7.0M

                                             = 1.44M

Let us take 100g (100mL) of vinegar

Moles of acetic acid in vinegar = 1.44M*0.10 L

                                                      = 0.144 mol

Mass of acetic acid in 100g of vinegar = 0.144 mol *60.0g/mol

                                                                   = 8.64 g

perccentage of acetic acid in vinegar = 8.64 %

Percentage error in percentage of acetic acid = {(8.64%-5.0%)/5.0} *100

                                                                                = 72.8 %

2) If we use apple cider or balsamic vinegar instead of distilled white vinegar then there may be other acids and they consume some amount of NaOH. Then we will get greater amount of acetic acid than of original amount.

3) If tip of titrator was not filled then we record more volume of NaOH than amount of NaOH original used for titration. So we will get more amount of acetic acid than original amount.

4) If we add drops of NaOH beyong stoichiometric point, it also lead to note more volume of NaOH than original amount needed.

5) Volume of vinegar = 7.0 mL

     Volume NaOH = 7.5 mL

Concentration of NaOH = 1.5M

Concentration acetic acid = (concentration of NaOH*volume of NaOH)/volume of vinegar

                                             = (7.5mL*1.5M)/7.0M

                                             = 1.61 M

Moles of acetic acid in vinegar in 7.0 mL = 1.61M*0.007 L

                                                                       = 0.01127 mol

Mass of acetic acid in 100g of vinegar = 0.01127 *60.0g/mol

                                                                   = 0.676 g

Mass of acetic acid (CH3COOH) in vinegar sample = 0.676 g