The purpose behind (the first step in) this experiment is to show that similarly to week 1, the molarity of an acid or base in solution can be determined (so long as one value’s is known) using titration. In this case though, finding the molarity of the acid used in the reaction is then used to determine the percent of that acid in a vinegar solution and compared to the standard value for % acid present in vinegar. The second part of the experiment was to see if by titrating a solution of NaOH and an unnamed mystery acid, you could find the molar mass of the unknown acid (solving the mystery). It must be understood that the number of moles of the reacting NaOH and the number of moles of the product NaX acid, must both equal (in this case 1:1) in order for the calculation to find the molar mass to work.
Begin the procedure by first making sure all glassware has been cleaned. Next set up the buret for the titration using the same method as week 1. Similar to week 1 the titration is performed using NaOH as a base in the solution so the buret should filled with the NaOH and the initial volume may be recorded. The acid (acetic acid) should be mixed in solution with 5.00 mL of acid and 50.0 mL of water. The water can be added in excess because the hydrogen that bonds with the OH to make water as a product is already present in vinegar and so it doesn’t bond with other water molecules. Once the solution is mixed the indicator is added and the titration process may be initiated. As the base is added to the solution a faint purple color appears and disappears. The color fades much slower as the titration is almost complete. After completion the final volume of the NaOH in the buret was recorded and used in calculation. This entire process was repeated for a second titration. The average molarity of NaOH was found and used to determine the percent acetic acid present in vinegar. After calculating the percent acetic acid in vinegar the second...
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