Titration for Acetic Acid in Vinegar
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||10 mL||3.6 mL||6.4 mL|
|Trial 2||10 mL||3.2 mL||6.8 mL|
|Trial 3||10 mL||3.0 mL||7.0 mL|
|Average Volume of NaOH Used (mL) : 6.73 mL|
Data Table 2. Concentration of CH3COOH in Vinegar.
|Average volume of NaOH used (mL)||Concentration CH3COOH in vinegar (mol/L)||% CH3COOH in vinegar|
|6.73 mL||0.673 mol/L||4.0%|
A. The manufacturer of the vinegar used in the experiment stated that the vinegar contained 5.0% acetic acid. What is the percent error between your result and the manufacturer’s statement?
B. What challenges would you encounter with the titration if you had used apple cider vinegar or balsamic vinegar as the analyte instead of white vinegar?
We use white vinegar in this experiment so that we can see the color change. In addition, balsamic vinegar contains a few different elements.
C. 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.
You want to fill the tip to release any air bubbles and so you can have it be one drop at a time so you don’t let too much out at a time.
D. How would your results have differed if you had over-titrated (added drops of NaOH to the analyte beyond the stoichiometric equivalence point)?
This would most likely mess with our calculations and make our percentage wrong.
E. 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?
Macid = 1.6 M
1.6 M * 60g = 96g CH3COOH
96g/7.0mL = 13.71
Mass percent = 13.7%
F. Why is it important to do multiple trials of a titration, instead of only one trial?
It is important to do multiple trials of titration because you are dealing with such a precise amount such as a drop at a time. You could have messed up on one of your trials without realizing it. It is best to do more than one trial and take the average.
G. 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?
The student would need to start over. When doing the calculations at the end it is important to use the correct amount of NaOH. The ratio would be flipped and it most likely would turn pink a lot quicker.