1).An aluminium oxalate complex was prepared by dissolving aluminium metal in po
ID: 981676 • Letter: 1
Question
1).An aluminium oxalate complex was prepared by dissolving aluminium metal in potassium hydroxide, followed by reaction with oxalic acid.The resulting complex was analysed for oxalate content by titration against potassium permanganate:
A 0.1105 g sample of the aluminium oxalate complex was dissolved in deionized water (30 cm3) and bench dilute (2 M) sulphuric acid (10cm3)added. The resulting solution was heated to about 60 ºC and titrated against standardized potassium manganate (VII) solution.14.35 cm3 of a 0.02 mol dm-3 KMn04 solution was required.
(i) Write out a balanced equation for the overall redox process.( 3 marks
(ii) Use the analytical data to deduce if the prepared complex was K3Al(C2O4)3.3H2O or KAl(C2O4)2(H2O)2.2H2O. (7 marks).
Molar masses /g mol-1
K, 39.00; Al, 26.98; C, 12.01; O, 16.00; H, 1.0
Explanation / Answer
potassium tris(oxalato)aluminate(III)
Aluminium will react with KOH to give aluminium hydroxide
When it reacts with oxalic acid it will form the aluminium oxalate complex
(i) the overall redox reaction will be
oxalate ion get oxidized to CO2
And KMnO4 gets reduced to Mn+2 ( from +7 to +2)
2MnO4- + 5H2C2O4 + 6H+ =>10 CO2 + 2 Mn2+ + 8H2O
(ii) A 0.1105 g sample of the aluminium oxalate complex was dissolved in deionized water (30 cm3) and bench dilute (2 M) sulphuric acid (10cm3)added. The resulting solution was heated to about 60 ºC and titrated against standardized potassium manganate (VII) solution.14.35 cm3 of a 0.02 mol dm-3 KMn04 solution was required.
The amount of complex dissolved = 0.1105 grams
Volume = 30mL
The Moles of KMnO4 used = Molarity X volume = 0.02 X 14.35 = 0.287 millimoles
Millimoles of oxalate reacted = 5 X 0.287 / 2 = 0.717 millimoles
So mass of oxalate reacted = Moles X molecular weight
Mass = 0.717 X 88 = 0.063 grams
these grams = x% of 0.1105 grams
so x % = 57%
In complex
K3Al(C2O4)3.3H2O
462 = molecular weight
oxalate in this is = 264 grams
Which is 57% of the given complex
So the complex should be K3Al(C2O4)3.3H2O
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