TVS AP Chemistry Lab The Synthesis of Alum

TVS AP Chemistry LabThe Synthesis of AlumIntroductionThis experiment serves as an introduction to several concepts that will be central toour studies throughout the year. Over the next several days, you will prepare potassiumaluminum sulfate, KAl(SO 4 ) 2 xH 2 O, commonly known as alum. This salt is interestingin 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 themetal remnants into useful compounds, as alum is a very important mordant forpreparing fabric for dyeing. In addition, alum is interesting in that the salt contains twocations, one of plus one charge, while the other has plus three. Indeed, several of thesetypes 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 muchdetail. Initially, you must oxidize the aluminum into its ions. This is done be immersingthe foil in a concentrated potassium hydroxide solution. Interestingly, we would expectsuch a reaction to produce insoluble aluminum hydroxide. However, the KOH is in suchexcess that a complex ion, [Al(OH) 4 ] - is formed. We will apply some acid basechemistry, and acidify the solution with H 2 SO 4 . You will see the formation of aluminumhydroxide, followed by its disappearance in the highly acidic solution, so that the onlyions present are Al 3+ , SO 42-, and K + . By reducing the volume of the solution, we will becreating a situation of supersaturation, causing alum to precipitate out of solution. Theactual (unbalanced) reactions of interest are:Al(s) + OH - (aq) + H 2 O(l) Al(OH) 4 - (aq) + H 2 (g)H + (aq) + [Al(OH) 4 ] - (aq) Al(OH) 3 (s) + H 2 O(l)Al(OH) 3 (s) + H + (aq) + H 2 O(l) [Al(H 2 O) 6 ] 3+ (aq)K + (aq) + [Al(H 2 O) 6 ] 3+ (aq) + SO 42-(aq) KAl(SO 4 ) 2 •12H 2 O(s)Procedure: Synthesis of the crystal1) Weigh out one gram of aluminum foil to the nearest 0.01 g. Tear the foil into smallpieces 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 allowthe reaction to proceed until the foil is dissolved.3) Remove any undissolved solids such as carbon particles by filtration using a Buchnerfunnel while the solution is hot.4) Cool the solution and acidify it slowly, with constant stirring, using 35 ml of 3.0 MH 2 SO 4 . After acidifying and stirring, if a precipitate still remains, filter the solutionand discard the solid. Heat the solution gently with a Bunsen burner until the volumeis reduced to approximately 50 ml. Cover the beaker and allow it to sit over theweekend (or at least overnight).5) Overnight, crystals will form. Collect the crystals via filtration with a Buchnerfunnel. 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 wasthe limiting reactant, so use this information to calculate the percent yield of yourcrystals.Discussion:You should discuss two main concepts in your discussion section:First, explain in your discussion section be how you were able to precipitate out thealum from aqueous solution, even though it is water-soluble. Be sure to include adiscussion of the terms unsaturated, saturated, and supersaturated in yourexplanation. Explain. Finally, you want to discuss why your percent yield cannot be100% (for experimental reasons). All of this should of course be in a well writtenparagraph or paragraphs, using transitions as the READER HAS NO IDEA WHATQUESTIONS YOU ARE ANSWERING!Conclusion: 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.