Name _____________________________________________________ Date ________________________
NORTH ALLEGHENY SENIOR HIGH SCHOOL
AP Chemistry
SYNTHESIS OF ALUM
INTRODUCTION:
In this experiment the ionic compound, potassium aluminum sulfate (KAl(SO4)2.12H2O, alum), will be
prepared from a water solution that contains K+, A13+ and SO42- (potassium, aluminum, and sulfate ions,
respectively). The aluminum ions will be formed by oxidizing aluminum from aluminum foil. The "double
salt" potassium aluminum sulfate dodecahydrate is commonly referred to as alum. Many combinations of
mono- and tri-positive cations yield crystals of the same stoichiometry and structure, and alum is a general
name for this type of compound. For example, there is chrome alum, KCr(SO4)2.12H2O, which is a deep
purple color, as well as alums where either sodium or ammonium ions are present instead of the potassium ion.
The crystals are usually in the form of octahedra.
If an aqueous solution which contains [Al(H2O)6]3+ ions, K+ ions and SO42- ions is allowed to evaporate,
the compound KAl(SO4)2.12H2O will crystallize. Within the alum crystal, six waters of hydration are bonded
directly to the aluminum ion to give [Al(H2O)6]3+ ions, while the other six surround the K+ ion.
Alum crystals of great purity are easily prepared. Because of this purity, alum is useful in the dyeing of
cloth, where the alum acts as a source of Al3+ ions that are not contaminated with Fe3+. The Al3+ is precipitated
on the cloth as aluminum hydroxide which acts as a binding agent for dyes. It is necessary that no Fe3+ be
present in order to produce clear colors.
Aluminum is considered a reactive metal, but because its surface is usually protected by a thin film of
aluminum oxide, it reacts only slowly with acids. It does, however, dissolve quickly in basic solutions. Excess
hydroxide ion converts the aluminum to the tetrahydroxoaluminate(III) ion, [Al(OH)4]-. When acid is slowly
added to this ion, white, gelatinous aluminum hydroxide (AI(OH)3) precipitates. Continued addition of acid
causes the hydroxide ions to be completely neutralized, and the aluminum exists in solution as the hydrated ion
[Al(H2O)6]3+. Aluminum hydroxide is considered to be an "amphoteric" hydroxide because it dissolves in both
acids and bases.
MATERIALS OF NOTE: Scrap Aluminum foil, 1.4 M KOH, 9.0 M H2SO4, Buchner funnel and filter flask
PROCEDURE:
Day 1:
1. Cut from an aluminum pie pan (or another source of scrap aluminum) approximately 1 gram of
aluminum. Record the mass of aluminum, then cut the aluminum into small pieces.
2. Place the aluminum pieces in a 250 mL beaker and add 50 mL of 1.4 M KOH. The reaction that occurs
is below
2Al(s) + 2KOH(aq) + 6H2O(l) → 2K[Al(OH)4](aq) + 3H2(g)
3. Heat the beaker gently on a hot plate to speed up the reaction. Do not use a Bunsen burner. The
hydrogen generated is flammable. When you are finished heating the solution, the volume should be
about 30 mL. If the volume falls below this level while heating, add some distilled water to replenish it.
The reaction is not complete until the bubbling stops.
4. Using a Buchner funnel, filter the hot solution using an aspirator vacuum. The filtrate should be clear,
with any dark residue left on the filter paper.
5. Rinse the beaker twice with 5 mL portions of distilled water, pouring each rinse through the filter
residue.
6. Transfer the clear residue into a clean 250 mL beaker. Rinse the filtering flask with 5 mL of distilled
water and pour this into the beaker. Allow the solution to cool.
7. When cool, slowly and carefully, while stirring add 20 mL of 9 M H2SO4 to the solution.
8. As you are adding the H2SO4, a white precipitate should form. However, when all of the sulfuric acid
has been added, the precipitate should have been redissolved. If necessary, warm the solution gently on
a hot plate to insure that all of the precipitate dissolves.
9. If any solid residue remains and will not dissolve, filter and work with the clear filtrate.
10. STOP! Allow beaker to sit overnight. Alum crystals should precipitate.
Day 2:
Alum crystals should now have precipitated. The reaction from part (7) and (8) can be summarized as:
K[Al(OH)4](aq) + 2H2SO4(aq) + 8H2O(l) → KAl(SO4)2.12H2O(s)
alum
1. Measure out 20 mL of the chilled ethanol/water solution that is on the central table.
2. Clean and reassemble the suction filtration equipment.
3. Transfer the alum crystals to the filter paper in the Buchner funnel. Use your stirring rod to spread the
crystals uniformly on the filter paper.
4. Wash your crystals using the 20 mL of ethanol/water obtained in (1).
5. Allow the aspirator to pull air through the crystals for about 3-4 minutes.
6. Determine the mass of the dry crystals. Record the mass in your data table.
CALCULATIONS:
1. Calculate the theoretical yield of alum based on the equations given and your starting mass of
aluminum.
2. Calculate the percent yield of alum based on the mass of crystals you recovered and the theoretical yield
calculated in question 1.
QUESTIONS:
1.
2.
3.
4.
5.
Why did the solution become hot when the sulfuric acid was added to the solution?
Why should you NOT expect a 100% yield of alum crystals?
How does the solubility of alum in water change with temperature?
Why should you NOT wash the crystals with pure water?
What do your crystals look like? Describe them.