Analysis: In order to test the null hypothesis of this lab, it was first necessary to calculate the theoretical concentration of acetic acid in a bottle of vinegar because it can be used to compare with the results of the titration experiments in order to see whether or not the vinegar contains 5% acetic acid by mass. Therefore, the number of moles of acetic acid was calculated through the use of its density, its molar mass, and the total volume of the vinegar. Then its molarity was calculated using its number of moles and the volume of vinegar used in each trial. In order to obtain the actual molarity of the acetic acid in the samples, the vinegar must be titrated with solution of NaOH. So the solution of NaOH was prepared with the molarity of 0.285 M. Based on the differences between the molarities of the acid and NaOH, the ratio between them is about 1 and 3, which means that for every 75 mL of acetic acid, …show more content…
The stopcock was opened for the NaOH solution to drip down and after every few drops, the solution was swirled to check if the two substances had completely reacted with each other and reached the equivalence point. When the reaction has reached that point, the solution changed color to pink for 1 minute or more and the volume of NaOH titrated was recorded into the table. Next, the different trial results were used to calculate the molarity of the acetic acid using the formula (MAcid)(VAcid) = (MBase)(VBase), which the results are averaged and would be about 0.84284 M CH3COOH. Using the standard deviation with the 95% confidence intervals, the acceptable ranges for the acetic acid molarity would be between 0.78184-0.90384 M. Based on the calculation and experiment results, the null hypothesis is correct since the molarities for all 3 trials were in the calculated range. Therefore, 5% of a vinegar bottle contains acetic acid by