Acid-Base Titration and Volumetric Analysis
The purpose of this experiment is to determine the [NaOH] of a solution by titrating it with standard HCl solution, to neutralize a known mass of an unknown acid using the NaOH solution as a standard, to determine the moles of NaOH required to neutralize the unknown acid, and to calculate the molecular mass of the unknown acid. Procedure:
Standarized 0.10M HCl solution and unknown NaOH solution were poured into two beakers. The burets were then filled with the solution and put into the flask. Water was then added to the same flask. Then phenolphthalein indicator solution was added before we began titrating by adding more drops of the NaOH solution while swirling the flask until a pink color remained. The liquid levels were recorded on both burets. Part B:
Solid unknown acid was added to a flask with water and swirled until almost fully dissolved. Then phenolphthalein indicator was added. With standardized NaOH solution we titrated and recorded the initial and final volumes.
Part A Acid
Part A Base
Part B Base
11.64mL (2nd) Total:49.30mL
Part B Mass of Acid
1. The molarity of the base NaOH is 0.091M.
2. The number of moles of NaOH that reacted with the unknown acid is 0.0045mol.
3. There are 0.0045mol of hydronium ion in 1 gram of the unknown acid because the mole to mole ratio of NaOH to the unknown acid is 1:1. So if there was 0.0045mol of NaOH, because of this ratio of 1:1, there will also be 0.0045mol of H3O+.
4. The molar mass of the unknown acid is 220.
1. When the end point is reached in an acid-base titration, the relationship between the concentrations of OH- and H3O+ are that they are equal. This doesn’t mean that the pH will be neutral, but the concentrations of both will be the same in a titration.
2. The pH of the end point is determined by what kind of indicator is used to indicate when the end point I reached. In this experiment, phenolphthalein was used. It shows a color change in the presence of a base, which means that our solution had to be slightly basic for it to turn a pink-purple color. So, the type of indicator used for the experiment will overall determine the pH of the end point because some indicators turn a different color in the presence of an acid and others in the presence of a base.
3. Regardless of the amount of water that is used to dissolve the unknown acid, the amount of moles would not change. This is because when you are diluting a solution, you are affecting the volume of the solution, but not the number of moles present in the solution. So, in this experiment when 40cm3, 35 cm3 and 45 cm3 could have been used to dissolve the unknown acid and the number of moles would not be different.
4. If the unknown acid ad been diprotic, then the mole-to-mole ratio between the acid and NaOH would have been 2:1, the molarity and normality would have been 0.180, the number of equivalents would have been two and not one, the number of moles of the unknown acid would have been 0.0090mol instead of 0.0045mol, and the molar mass of the acid would have been 220. Therefore, if the unknown acid had been diprotic everything would have been doubled.
In this experiment, an acid-base titration was used to determine the molarity of a NaOH solution, the number of moles of NaOH that reacted with a different unknown acid, and the molar mass of this unknown acid. This was done by making the concentrations of 0.10M HCl and NaOH equal to determine the molarity of NaOH which is 0.091M. We then found that 0.0045mol of NaOH reacted with a different unknown acid by using the molarity of NaOH and the volume of NaOH that we used to titrate with the unknown acid. Since the mole-to-mole ratio of NaOH and the unknown acid were 1:1, we could use the same number of moles, 0.0045mol, for the acid to determine if molar mass. This was completed by using 0.0045mol and the mass in grams of the acid we used, which was 0.983g. By doing this we discovered that the molar mass of our acid was 220. By doing an acid-base titration