Experimental Procedure: Laboratory Exercise: The Synthesis of KAl(SO 4 ) 2 . 12H

Question

Experimental Procedure:

        Laboratory Exercise: The Synthesis of KAl(SO4)2.12H2O (Alum)

Aluminum metal will reduce water, but the reaction is extremely slow under neutral conditions at room temperature.

                Al(s) + 3H2O(l) Al(OH)3(s) + H2(g)                             very, very slow

In the presence of aqueous base, however, Al will readily react to form the very soluble Al(OH)4–1(aq) complex anion which can be neutralized with H2SO4(aq) in two steps. The first produces the very insoluble Al(OH)3(s). If the Al(OH)3(s) slurry is then immediately boiled with additional H2SO4(aq) the insoluble hydroxide will react completely to give a clear, colorless solution. After excess water is boiled off, cooling produces KAl(SO4)2.12H2O(s), Alum, as fine, white needles.   

                Al(s)    +     3H2O     +   KOH               KAl(OH)4(aq) + H2(g)   fast and exothermic at room temperature

                KAl(OH)4(aq)   +   ½ H2SO4(aq)         Al(OH)3(s) + H2O(l) + ½ K2SO4(aq)

                Al(OH)3(s)     +     H2SO4(aq)         ½ Al2(SO4)3(aq) + 3 H2O   (requires heating)

                ½ Al2(SO4)3(aq)   +   ½ K2SO4(aq) KAl(SO4)2(s)

Combining these four equations gives the overall reaction.

                Al(s)   + KOH(aq) + 2 H2SO4(aq)   [KAl(SO4)2(aq)] + H2O(l) + H2(g) and

                [KAl(SO4)2(aq)] + 12 H2O(l)   KAl(SO4)2.12H2O(s) (Alum)

A student dissolved 0.946g Al in 35mL of 2.00M KOH and obtained 13.8g of dry alum by following the experimental procedure given in this exercise. Assume that the Alum was 97.7% pure. Calculate the volume of 6.00M H2SO4(aq) needed just to precipitate Al(OH)3(s).

Explanation / Answer

1 mol of Al --> 1 mol of alum

1 mol of KOH --> 1 mol of alum

so..

mol of Al -- > mass/MW = 0.946/26.981539 = 0.03506 mol of Al

mol of KOH --> MV = 2*35/1000 = 0.07 mol of K

ratio is 1:1 so..

aluminium is limiting reaction

13.8 g of alum

MW alum = 39.0983 +26.98+(32+16*4)*2 + 12*18 = 474.0783 g/mol

mol of alum = mass/MW = 13.8/474.0783= 0.02910

only 97.7% purity --> 0.977*0.02910= 0.0284307 mol of Aluminium reacts

then... mol of KOH used --> 0.0284307 mol

mol of KOH left = 0.07-0.0284307= 0.0415693 mol of KOH left...

mol of H2SO4 --> 0.0415693/2 = 0.02078465mol of H2SO4 required to neutralize

V = mol/M = 0.02078465/6 = 0.00346410833liters = 3.46 mL of acid required

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