Synthesis of Alum
Experiment 4 Online Tutorial >> Synthesis of Alum >> Introduction
Experiment 4 - Synthesis of Alum
The goal of this experiment is to convert the Al in a beverage can into Alum KAl(SO4)2 . 12H2O(s). The reaction is done in two stages.
Stage 1: Addition of KOH to dissolve the aluminum sample
The KOH dissolves the beverage can by converting the aluminum from the metallic form, Al(s), to an ionic form, Al(OH)4-(aq), that is dissolved in the solution. Note that this is a redox reaction in which the Al loses three electrons to form an atom with a formal charge of 3: Al(OH)4-. This can be viewed as an Al3+ ion surrounded by 4 (OH-)'s, as in Al3+(OH-)4.
Stage 2: Addition of H2SO4.
Stage 1 converted the Al in the can from the metallic form to the oxidized Al3+ form. However, the Al3+ is complexed to 4 OH- ions to form Al(OH)4-(aq). We next want to strip away the OH- ions, which we can do by adding a strong acid (H2SO4). [The two H+'s in H2SO4 react with the OH- ions to form H2O.] An interesting thing happens here. The first OH- we strip away leads to Al(OH)3, which is an insoluble material and so forms a white sludge. If we continue to add H2SO4, and heat, we can dissolve this solid by stripping away the remaining OH- ions, giving us a solution with bare Al3+ ions. When we cool the solution, we get solid alum by the reaction:
Al3+(aq) + K+(aq) + 2SO42-(aq) + 12H2O(l) => KAl(SO4)2 • 12H2O(s)
Our goal was to convert the Al metal in the can to alum. The % yield is a measure of how well we succeeded at this goal: How much alum did we get compared to the amount we would get if the reaction had gone completely?
|Last Updated: Monday, April 6th, 2020 @ 10:13:15 am|