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Synthesis of Alum


  -Tutor 1

  -Tutor 2

  -Tutor 3

Evaluation Question

Experiment 4 Online Tutorial >> Synthesis of Alum >> Introduction

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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

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?

    - Tutor 3 helps you determine the % yield, based on

    • The amount of Al you started with (the Al in the alum product must come from the Al in the beverage can you started with)
    • The amount of K+ you added in stage 1 (the K+ in the alum product must come from the KOH you added in stage 1)

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Last Updated: Monday, April 6th, 2020 @ 10:13:15 am