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 excess KOH(aq) and the volume of 6.00M H2SO4(aq) needed to neutralize it.

Explanation / Answer

In the reaction,

1 mole of Al requires 4 moles of KOH

moles of Al = 0.946/27 = 0.035 mols

moles of KOH = 2 M x 0.035 L = 0.07 mols

alum = 13.8 x 0.977 = 13.4826 g

moles of alum = 13.4826/474.4 = 0.028 mols

moles of H2SO4 required = 0.028 x 2 = 0.057 mols

ml of H2SO4 required = 0.057/6 = 9.47 ml H2SO4

moles of KOH needed = 0.035 x 4 = 0.14 mols KOH

ml of KOH needed = 0.14/2 = 70 ml

balanced KOH needed to add = 70 - 35 = 35 ml

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