Vinegar Molarity

Determining the Molarity of Acetic Acid in Vinegar

Abstract
This experiment was done to determine the molarity of acetic acid in vinegar. The acetic acid in vinegar was titrated with a strong base sodium hydroxide to determine the equivalence point of this chemical reaction. The indicator phenolphthalein was used because the moment it changes color is on the basic side of the pH scale. To insure the experiment was done with the most accuracy, the two conductors of this experiment took three different trials to attempt to reach the best endpoint possible between the two. After punching in numbers and calculating results through a stoichiometry equation, the resultant molarity was .960M, but the actual molarity was .845M. We found that it took 9.60 mL of NaOH to
neutralize 10.00 mL of acetic acid. These results left the individuals experiment with a percent error of around 13.6%. There were many places where human error could have affected the results of this experiment. If burettes are used properly, the issues should become less prominent.

Introduction and Background Information
Within vinegar, acetic acid is the second most prominent component that can be found. This experiment explores acids and bases in common everyday forms. In this conducted the individuals running the experiment will determine the concentration of acetic acid after taking a sample of vinegar and titrating the acetic acid with a strong base (sodium hydroxide). The chemical equation for the chemical reaction observed in this particular lab is shown below.
HC2H2O2 (aq) +NaOH (aq) NaC2H3O2 (aq) +H2O (l)
When the starting amount of moles of H+ from the acetic acid (HC2H2O2) is completely neutralized by the exact number of OH- ions from the sodium hydroxide (NaOH), the equivalence point of the titration has been met. The equivalence point of weak-acid and strong-base titrations are for the most part basic and because of this, an indicator must be chosen that changes color in the basic region of the pH scale. The indicator used for this particular experiment was phenolphthalein which is colorless in acid and red/magenta in base.
Purpose
In this experiment, the conductors of the experiment are going to determine the concentration of acetic acid in a vinegar sample by titrating the acetic acid (HC2H2O2) with the strong base that is sodium hydroxide (NaOH).
Materials List Pipets | Vinegar | Indicator (phenolphthalein) | Sodium Hydroxide (1.00M) | 50 mL burette for NaOH | One 150 mL Erlenmeyer flask |

Procedures amd Description of Experimental Set-up 1. Obtain 10.00 mL of vinegar into the Erlenmeyer flask. 2. Record the brand of vinegar used. 3. Add 2-3 drops of indicator solution (phenolphthalein). Swirl around your flask to make sure the solution is mixed well. 4. Record Molarity of base (NaOH). 5. Drain enough sodium hydroxide into the waste bucket so the tip of the burette is filled without air bubbles. Make sure to go slowly or the amount of base may go below the meniscus. 6. Record the initial volume reading of the NaOH to the nearest .01mL. 7. Titrate first sample with good mixing to indicator endpoint. The solution should be a pale pink that is neither too light nor too dark. 8. Record final burette reading to the nearest .01mL. 9. As refilling the burette, use no more solution hydroxide than needed. 10. When completed, clean up glassware and the entire station used. Trial Number | Initial Volume | Ending Volume | Total Volume Used | 1 | 50.00 mL | 40.25 mL | 9.75 mL | 2 | 40.25 mL | 30.50 mL | 9.75 mL | 3 | 30.50 mL | 20.90 mL | 9.60 mL |
Data
Total Volume of Vinegar | 10.00mL | Molarity of Vinegar Sample | .960M | Average Molarity | .875M | Relative Standard Deviation of Class Data | .0430M | Number of Grams contained in 10.00mL of Original Vinegar ( % concentration of vinegar) | .576g HC2H3O2 | Correct Value for Molarity of Original Vinegar | .845M | Percent Error in Result | 13.6% |

Analysis of Data
Molarity of Acetic Acid:
HC2H2O2(aq) +NaOH(aq) NaC2H3O2(aq) +H2O(l) L x M =moleL *L L x M =moleL
.01000L .00960L 1.00M NaOH* .00960 L NaOH =.00960 mole NaOH
? M 1.00M .00960 mole NaOH*1 moleHC2H2O21 mole NaOH =.00960 mole HC2H2O2
? mole ? mole .00960 mole HC2H2O2.01000 L HC2H2O2 = .960 M HC2H2O2
Grams/Percent Concentration:
Molar Mass of HC2H2O2 = 60.5 grams/mole HC2H2O2
.960 M HC2H2O2*60.5 grams HC2H2O21 mole HC2H2O2 = .576 grams HC2H2O2
Standard Deviation:
Standard Deviation = .0430M
Barred X = .875M
N = 38 Results or Values
List of X is below (ALL M) .830 | .940 | .840 | .881 | .830 | .850 | .840 | .870 | .840 | .960 | .860 | .860 | .860 | 875 | .875 | .910 | .860 | .830 | .910 | .910 | .770 | .850 | .910 | .895 | .955 | .850 | .900 | 1.00 | .900 | .870 | .880 | .870 | .865 | .800 | .874 | |
Error Analysis:
.960M- .845M.845M*100=13.6% Error
Conclusion and Discussion of Results
In this experiment, we attempted to determine the molarity of acetic acid in vinegar. My partner and I used 10.00mL of vinegar to conduct a titration with sodium hydroxide, a base of high strength. The acetic acid in vinegar was titrated with a strong base sodium hydroxide to determine the equivalence point (endpoint) of this chemical reaction. The indicator phenolphthalein was used because the moment the solution changes color it is on the basic side of the pH scale. When the acetic acid was entirely neutralized with the sodium hydroxide, all of the solution turned to a pale pink (due to the indicator) if the trial was done correctly/successfully. To insure the experiment was done with the most accuracy, the two conductors of this experiment took three different trials to attempt to reach the best endpoint possible between the two. In the first trial, we used too much NaOH (around 9.75mL of base) and made the solution a deep pink or magenta. In the second trial, we accidently used too much NaOH (ironically the same amount as the first trial) and the solution was too dark once again. Lastly, in our final trial we were able to successfully titrate acetic acid with 9.60 mL of NaOH. After punching in numbers and calculating results through a stoichiometry equation, the resultant molarity was .960M, but the actual molarity was .845M. The standard deviation of our entire class data was calculated to three sig figs, ending in .0430M. We showed that for our results, 9.60 mL of 1.00 M NaOH was needed to titrate 10.00 mL of vinegar. These results left the individuals experiment with a percent error of around 13.6%. There were many places where human error could have affected the results of this experiment. NaOH could have spilled out of the burette; making for an inaccurate measurement of how much NaOH was really used. An incorrect measurement of the vinegar could have been used because a person could have blown out the last drop of vinegar in the pipet accidently. If burettes and pipets are used properly, the issues should become less prominent and less significant on the final results ending in a lower percent error.
Questions - If Asked 1. If the tip of the burette was not filled with sodium hydroxide before the initial volume reading was recorded, explain how the resultant molarity of acetic acid would be affected.
The volume of the NaOH would be greater than the actual recorded, making the molarity of the acetic acid increase as well.

2. If a few drops splashed out of the Erlenmeyer flask during the titration, explain how the resultant molarity of acetic acid would be affected.
The resultant molarity of the acetic acid would be greater than what it should have been in reality.

3. If the last drop of vinegar solution were blown out of the pipet into the Erlenmeyer flask, explain how the resultant molarity of acetic acid would be affected.
The resultant molarity of the acetic acid would be lower than expected.

4.
If the wet burette was not rinsed with sodium hydroxide before filling, explain how the resultant molarity of acetic acid would be affected.
Water doesn’t affect the Molarity of the concentration of acetic acid because the 1:1 ratio of H+ and OH- ions does not acid the acidity or basicness of the reaction.

5. If a volume of water was added to the Erlenmeyer flask, explain how the resultant molarity of acetic acid would be affected.
See Number 4 for answer because response is almost identical

6. If methyl orange was used as the indicator, explain how the resultant molarity of acetic acid would be affected.
Phenolphthalein is more basic than methyl when comparing one another on the pH scale. If a different indicator such as methyl was used, a different amount of base would have been required to change the color of the solution, offsetting the molarity of the acetic acid with phenolphthalein.

7. Explain why the color of the endpoint fades upon standing.
The color of the endpoint fades upon standing because the solution will have all H+ and OH- ions neutralized eventually and change the pH to where the indicator is. NaOH eventually ionizes and the color will change
afterward.