2. Chem Lab Days Experiments
  3. Recycling Waste Aluminum: Synthesis of Potassium Alum

Results Scenario

Scenario

You are a student attending the Chem Lab Days. You will find below the data obtained in the lab so you can complete the calculations just as if you were visiting the University of Waterloo!

Results

Use the given to state the results of qualitative tests:

Results of Qualitative Tests:

The following videos depict each qualitative test performed on the crude alum. Answer whether the test is positive (ion present) or negative (ion absent).

  1. Aluminum Test:

     

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    To test for aluminum ions in the crude potassium alum, we take some solid alum and place it into a glass tube. Add some deionized water to dissolve the solid, agitating the tube gently. The quantities of alum and water are only approximate as this is a qualitative test. Based on the results, we will know whether aluminum ions are present but not in quantity.

     

    Add two drops of the aluminon indicator from the dropper bottle, then add drops of ammonia solution until you see a pink precipitate. This precipitate is quite fine so it can be difficult to see.  By comparing with the result for a sample without aluminum ions, it is far easier to see the difference.

    In a second tube, add deionized water water only. Then add two drops of the aluminon solution followed by a few drops of ammonia. The solution is more see-through and the hue is more orange-red without  aluminum ions than pinkish red with aluminum ions.

     

     

     

  2. Potassium Test:

     

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    To test for potassium ions in the crude alum, we will use the flame test. Take a metal wire loop and wet it by dipping in the 3 M HCl solution, then dip the wire loop in your potassium alum to get some solid.

     

    The bunsen burner will be lit for you. Place the wire loop in the flame and wait for the solid to heat up until the red-hot stage. At this point, you should see a lavender colour coming out from the wire loop.

    Heat from the flame gives energy to the potassium ion and cause it to emit a characteristic colour. Potassium is pretty faint with the lavender color but it should still be noticeable when compared to the regular blue gas flame.

     

     

  3. Sulfate Test:

     

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    To test for the presence of sulfate ions in the crude potassium alum, take a small amount of solid and put it in a glass tube. Add a few drops of water and flick the glass tube to dissolve. You do not need to measure exact quantities of alum and water as this test is qualitative, meaning it will only indicate whether the ion is present but not in what quantity.

     

    Take the dropper bottle containing the barium chloride solution and add two drops into the glass tube. If the sulfate ions are present, you will see a white milky precipitate form, which is barium sulfate, a compound insoluble in water.

    For comparison, let's see what a negative test looks like. In a second glass tube, add only deionized water. Then add two drops of barium chloride solution. No precipitate form meaning there is no sulfate in the sample.

     

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Balancing the Overall Reaction

Use the equations previously balanced chemical equations to write the balanced overall reactions of \(\ce{Al(s)}\) to make potassium alum. Note that stoichiometric equivalents will be different as you need to 'cancel out' species who do not appear in the overall balanced equation.

  1.  \(\ce{Al(s)}\:+\:\)  \(\ce{KOH(aq)}\:+\:\)  \(\ce{H2O(l)}\:\rightarrow\:\)  \(\ce{Al(OH)4^{-}(aq)}\:+\:\)  \(\ce{K+(aq)}\:+\:\)  \(\ce{H2(g)}\)
  2.  \(\ce{Al(OH)4^{-}(aq)}\:+\:\)  \(\ce{H2SO4(aq)}\:\rightarrow\:\)   \(\ce{Al(OH)3(s)}\:+\:\)  \(\ce{H2O(l)}\:+\:\)  \(\ce{SO4^{2-}(aq)}\)
  3.  \(\ce{Al(OH)3(s)}\:+\:\)  \(\ce{H2SO4(aq)}\:\rightarrow\:\)  \(\ce{Al^{3+}(aq)}\:+\:\)  \(\ce{SO4^{2-}(aq)}\:+\:\)  \(\ce{H2O(l)}\)
  4.  \(\ce{Al^{3+}(aq)}\:+\:\)  \(\ce{K+(aq)}\:+\:\)  \(\ce{SO4^{2-}(aq)}\:+\:\)  \(\ce{H2O(l)}\:\rightarrow\:\)  \(\ce{KAl(SO4)2.12H2O(s)}\)
  1. Overall Balanced Equation:
     \(\ce{Al(s)}\:+\:\)  \(\ce{KOH(aq)}\:+\:\)  \(\ce{H2SO4(aq)}\:+\:\)  \(\ce{H2O(l)}\:\rightarrow\:\)  \(\ce{KAl(SO4)2.12H2O(s)}\;+\)  \(\ce{H2(g)}\)
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Calculations

Useful Data

  • Aluminum: \(\ce{AW}=26.98\:\ce{g/mol}\)
  • \(\ce{KOH}\): \(\ce{C\:=\:}1.5\:\ce{M}\) ; \(\ce{V\:=\:40\:mL}\)
  • \(\ce{H2SO4}\): \(\ce{C\:=\:}9\:\ce{M}\) ; \(\ce{V\:=\:20\:mL}\)
  • Potassium Alum (\(\ce{KAl(SO4)2.12H2O}\) ): \(\ce{MW}=474.37\:\ce{g/mol}\)

Calculate the Mass of Alum

Use the given to state the mass of the crude product:

Mass of Crude Product:

  1. Mass of aluminum can pieces: 0.95 g
  2. Mass of large petri dish: 16.00 g
  3. Mass of large petri dish + crude product: 27.19 g
     
    • Calculate the mass of crude product:    g

Determining the % Yield

To determine the % yield we must first determine the theoretical yield (maximum mass) that could have been obtained, below are the steps needed to do this.

  1. Determine the limiting reagent: 
    (Hint: Calculate the number of moles of reagent used!)
    1. Calculate the moles of aluminum used:    mol
    2. Calculate the moles of \(\ce{KOH}\) used:    mol
    3. Calculate the moles of \(\ce{H2SO4}\) used:    mol
    4. Therefore, the limiting reagent is: 
  2. Determine the theoretical yield in grams: 
    (Hint: Your limiting reagent determines how many moles of product you will produce)

    Theoretical yield (use two decimal places):    g

 

  1. Determine the % yield of crude product: 
    Hint: \(\ce{\%\:Yield\:=\:\dfrac{Experimental\;Mass\;(mass\;crude\;alum)}{Theoretical\;Mass}\times100\%}\)

    % yield (no decimal places):    %
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