Synthesis of Alum from Scrap Aluminum

Objective: In this experiment, you will be converting the aluminum metal from a
beverage can into the chemical compound potassium aluminum sulfate, KAl(SO4)2•12 H2O, commonly referred to as alum.

Introduction
It had been a good year for the O’Keefe farm with just the right amount of sun and rain. In fact, the cucumbers had grown so well that Richard was afraid he wouldn’t be able to sell them fast enough. Faced with the prospect of a warehouse full of rotten cukes, he contemplated the idea of preserving them and selling them as “O’Keefe’s Farm Fresh Pickles”. He discussed the plan with his wife, Diane, who agreed it was a good idea and said she had a delicious pickling recipe that had been handed down to her from her mother. “I recall one of the ingredients was alum. I think it helps to keep the pickles firm and crisp,“ Diane said. “It just so happens that I know of a recipe for making alum from aluminum cans,” Richard said. “We can use some of those soda pop cans that have been piling up, and that should save us some money.” “How can you convert aluminum metal into an aluminum salt?” Diane asked. Richard explained the chemistry behind the process: One of the interesting properties of aluminum is that it is amphoteric, meaning it will dissolve in both strong, aqueous acids and strong, aqueous bases In both cases, the formation of hydrogen gas is observed:
 

2 Al (s) + 6 H+ (aq) = 2 Al3+ (aq) + 3 H2 (g)

2 Al (s) + 6 H2O (l) + 2 OH– (aq) = 2 Al(OH)4– (aq) + 3 H2 (g)

The first reaction that needs to be carried out is that of aluminum and potassium
hydroxide, KOH:

2 Al (s) + 2 KOH (aq) + 6 H2O (l) = 2 Al(OH)4– (aq) + 2 K+ (aq) + 3 H2 (g)

Adding sulfuric acid, H2SO4, to the resulting solution will cause the compound Al(OH)3
to precipitate; however, Al(OH)3 is also amphoteric and will re-dissolve when more acid
is added:

2 K+ (aq) + 2 Al(OH)4– (aq) + H2SO4 (aq) =2 K+ (aq) + 2 Al(OH)3 (s) + 2 H2O(l) + SO42–(aq)

2 Al(OH)3(s) + 3 H2SO4 (aq) = 2 Al3+ (aq) + 3 SO42–(aq) + 6 H2O(l)

Crystals of the double salt KAl(SO4)2⋅12 H2O (s), or alum, will form upon cooling this
final solution since the solubility of alum in water decreases as the temperature is
lowered:

K+ (aq) + Al3+ (aq) + 2 SO4 2– (aq) + 12 H2O(l) = KAl(SO4)2•12 H2O (s)

“But how will we know that everything really worked ok and that it’s really alum that you produced? I think we need to implement some quality control measures,” Diane observed. “In order to confirm that your synthesis of alum resulted in the desired product, we need to perform a qualitative analysis of the compound." Qualitative analysis is the name given to the process whereby the identities of the elements or substances present in a sample are determined through the use of simple chemical tests. In this case, chemical reactions will be performed with an alum sample that will confirm the presence of K+, Al3+ and SO42–.
In the first test, alum will be reacted with barium chloride in an aqueous solution,
which will result in the following chemical reaction:

KAl(SO4)2 (aq) + 2 BaCl2 (aq) = 2 BaSO4 (s) + KAlCl4 (aq)

The precipitation of solid material, BaSO4, from a solution of alum upon reaction with
barium chloride is a positive test for the presence of sulfate ion, SO42–. The second test is performed to confirm the presence of potassium and is a flame test. Potassium is volatilized at the very high temperature of a flame (about 1000°C) at which point it imparts a bluish-purple color to the flame. After a few seconds in the flame, the sulfur dioxide will be driven off of your alum sample, and the solid material remaining will consist of aluminum oxides. The third test confirms the presence of aluminum ion and involves its reaction with potassium hydroxide. A wispy, gelatinous precipitate of Al(OH)3 will form upon addition of a small amount of KOH to the aqueous alum solution. Further addition of KOH will cause the precipitate to re-dissolve.
“We should also make sure that we’re really saving some money. We’ll need to
calculate the percent yield of the process once we’ve made the alum,” Diane said.
When carrying out chemical reactions, it is of interest to know what percentage of
the starting material is recovered in the form of the desired product. It is possible to
calculate this percent yield in the following manner:

1. Calculate the number of moles of reactant
2. From the stoichiometry of the reaction, determine the expected number of moles of product. A review of the above reactions reveals that there is a 1:1 relationship between the aluminum containing reactant and the aluminum containing product in all cases. Therefore, one mole of aluminum metal should produce one mole of alum.
3. From the expected number of moles of product, calculate the expected mass of product or the theoretical yield.
4. Calculate the percent yield by dividing your actual yield by the theoretical yield and multiplying by 100.