Aluminum is hardly a precious metal, on par with
gold, silver, or platinum. In fact, it is so common today that we
use it for our cheap aluminum foil and soda cans. In the 19th century
however, it was a luxury. Aluminum was made into expensive bracelets
alongside gold, and Napoleon III was presented with a set of aluminum
forks and spoons and a valuable centerpiece crafted from aluminum,
gold, and bronze. This stark contrast in availability stems from
the Hall-Heroult process developed in 1886, allowing the economically
viable, commercial production of aluminum. Society was thrilled
with this practical new metal, and many household products were
made from it-from teakettles to playing cards. Less than a hundred
years after the design of this process, the United States experienced
an abundance of aluminum, at which time people even made rugs out
of it!
You will have the opportunity to work with this unique
metal today. From aluminum foil, you will make a useful compound
commonly called alum (specifically, aluminum potassium sulfate,
KAl(SO4)2). There are many practical uses
of alum, the product you will be isolating today. It is used industrially
in manufacturing dyes and dyeing fabric. Alum is also used in making
paper, cement and explosives. Interestingly enough, this chemical,
with all of its diverse uses, is also employed in pickling cucumbers!
The conversion of aluminum foil to alum involves
several chemical steps. The procedures for each step of today's
experiment are more complicated in comparison to the type of lab
work you have done so far. It is a good idea to see what is happening
qualitatively before you begin.
In the first chemical step of the experiment, you
will add a strong base, potassium hydroxide (KOH), to your aluminum
foil. You will heat this mixture on a hot plate. The aluminum and
hydroxide will combine to form Al(OH)4- and
H2(g). The chemical equation for this reaction is shown
below:
2Al(s) + 2KOH(aq) + 6H20(l)
2K+(aq)
+2Al(OH)4-(aq) + 3H2(g)
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The next step in the procedure is the addition of
sulfuric acid (H2SO4), to your sample. Adding
acid causes two chemical reactions to occur. First, the Al(OH)4-(aq)
species is converted into Al(OH)3 (s), which is a white
precipitate. The chemical equation for this conversion is:
2Al(OH)4-(aq)
+ H2SO4(aq) 2Al(OH)3(s)
+ 2H2O(l) + SO42-(aq)
This white precipitate will disappear as the acid
is stirred into the solution, however, as Al3+(aq) and
water form from the addition of acid to Al(OH)3 (s).
The second chemical change caused by the addition of acid is:
2Al(OH)4-(aq)
+ 3H2SO4(aq)
2Al3+(aq) + 3SO42-(aq)
+ 6H2O(l)
Finally, we see the formation of hydrated alum crystals (KAl(SO4)2·12H2O(s))
as the solution is cooled.
K+(aq) + Al3+(aq)
+2SO42-(aq) +12H2O(l)
KAl(SO4)2·12H2O(s)
Note that the alum formed is Al(SO4)2·12H2O(s)
The ·12H2O means that 12 molecules of water are
associated with each molecule of the potassium aluminum sulfate.
These molecules are the water of hydration, and in many compounds
are associated with a color change. You will measure this product
after it "air dries" but still has its twelve associated
water molecules- the water of hydration. .
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