This experiment serves as an introduction to several concepts that will be central to our studies throughout the year. Over the next several days, you will prepare potassium aluminum sulfate, KAl(SO
4
)
2
 xH
2
O, commonly known as alum. This salt is interesting in that it is readily made from scrap aluminum (in this case a small amount of scrap foil, although an aluminum can could also be used) and thus demonstrates a way to recycle the metal remnants into useful compounds, as alum is a very important mordant for preparing fabric for dyeing. In addition, alum is interesting in that the salt contains two cations, one of plus one charge, while the other has plus three. Indeed, several of these types of ionic compounds exist, and are all called alums as a group.
The reaction to produce alum illustrates several concepts that we will study in much detail. Initially, you must oxidize the aluminum into its ions. This is done be immersing the foil in a concentrated potassium hydroxide solution. Interestingly, we would expect such a reaction to produce insoluble aluminum hydroxide. However, the KOH is in such excess that a complex ion , [Al(OH)
4
] is formed. We will apply some acid base chemistry, and acidify the solution with H
2
SO
4
. You will see the formation of aluminum hydroxide, followed by its disappearance in the highly acidic solution, so that the only ions present are Al 3+ , SO
4
2, and K + . By reducing the volume of the solution, we will be creating a situation of supersaturation , causing alum to precipitate out of solution. The actual (unbalanced) reactions of interest are:
Al(s) + OH
K + (aq) + [Al(H
2
(aq) + H
H + (aq) + [Al(OH)
4
Al(OH)
3
(s) + H +
O)
6
] 3+
]
2
-
O(l)  Al(OH)
(aq)  Al(OH)
(aq) + H
(aq) + SO
2
4
3
4
(aq) + H
2
(s) + H
O(l)  [Al(H
2
O)
2(aq)  KAl(SO
6
]
2
O(l)
3+
4
)
2
•
(g)
(aq)
12H
2
O(s)
1) Weigh out one gram of aluminum foil to the nearest 0.01 g. Tear the foil into small pieces and place in a 250 ml beaker.
2) Slowly add 25 ml of 3.0 M KOH solution. Put the beaker in the hood, and allow the reaction to proceed until the foil is dissolved.
3) Remove any undissolved solids such as carbon particles by filtration using a Buchner funnel while the solution is hot.
4) Cool the solution and acidify it slowly , with constant stirring, using 35 ml of 3.0 M
H
2
SO
4
. After acidifying and stirring, if a precipitate still remains, filter the solution and discard the solid. Heat the solution gently with a Bunsen burner until the volume is reduced to approximately 50 ml. Cover the beaker and allow it to sit over the weekend (or at least overnight).
5) Overnight, crystals will form. Collect the crystals via filtration with a Buchner funnel. Wash the crystals with 30 ml of a 50% by volume water and ethanol mixture
(use three successive 10ml washings). Allow the crystals to dry and weigh them .

Calculations:
1) The formula for the alum is KAl(SO
4
)
2
• 12H
2
O. The aluminum in your reaction was the limiting reactant, so use this information to calculate the percent yield of your crystals.
You should discuss two main concepts in your discussion section:
First, explain in your discussion section be how you were able to precipitate out the alum from aqueous solution, even though it is water-soluble . Be sure to include a discussion of the terms unsaturated, saturated, and supersaturated in your explanation. Explain. Finally, you want to discuss why your percent yield cannot be
100% (for experimental reasons). All of this should of course be in a well written paragraph or paragraphs, using transitions as the READER HAS NO IDEA WHAT
QUESTIONS YOU ARE ANSWERING!
One sentence stating what you have made and its percent yield.
Thus, to summarize, this write-up has no intro, just title, purpose, procedure, observations, data, calculations discussion and conclusion.