Experiment 7 Conclusion: Antacid Analysis and Percent Acetic Acid in Vinegar

In order to determine the percent acetic acid in vinegar, we used a simple titration method. We titrated NaOH into a 20 mL flask of vinegar until the color changed into a faint pink. From there we could calculate the number of moles of NaOH we needed by taking the volume used and multiplying it by the molarity of NaOH that was given to us. Knowing that the ratio of CH3COOH to NaOH is 1:1 we could then determine the moles of CH3COOH in the vinegar. From there determining the molarity of the acetic acid could be determined by dividing the moles found by the 20 mL of original acetic acid. Lastly to determine the percent acetic acid in vinegar we divided the found molarity of by 17.4M which is the glacial CH3COOH and then multiplied that by a 100. The net ionic equation for the reaction is.

CH3COOH(aq) + OH-(aq) ---> CH3COO-(aq)+ H2O(l)

Our average percent that we found for the percent acetic acid in vinegar was found to be 2.74% which was short of the range that we were given (2.85% - 3.15%). Our average deviation that was calculated was determined to be 0.057. Which is extremely low. A plausible reason why our percent was systematically lower than the given 3% was perhaps due to the fact that we needed to titrate more NaOH into the solution, maybe we stopped to soon all three times. Perhaps our measurements of the amount of NaOH titrated was off. Both of these reasons could explain why our percent was systematically