Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: pear on February 25, 2010, 01:09:50 AM
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Hi,
So I'm trying to understand what happens when excess I- is added to the equilibrium PbI2(s) ::equil:: Pb2+(aq) + 2I-(aq)
What I'm thinking is that first the reaction is governed by solubility product. And eventually, [I-] is so high that Q >>>> Ksp and the lead re-dissolves to instantly form covalent bonds with the two outer electrons && the two electrons from the unfilled d shell, thus removing twice the [I-] as the PbI2 molecule.
Am I thinking logically/correctly?
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So here's the answer, for anyone else wondering..
Eventually the amount of energy possessed by a solution with excess negatively charged ions will overcome the high energy input required for the 2 d-shell electrons of Pb to create covalent bonds alongside the 2 outer p-shell electrons. So instead of the solid PbI2, which only takes care of 2 I- ions, the covalent, aqueous molecule PbI42- forms and lessens the amount of excess energy in the solution.
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Isn't this an application of Le Chatelier's principle where if you add excess product, it drives the reaction to favour the reverse reaction and you will precipitate the reactant?
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PbI42- formation is a separate reaction, just applying LeChetelier's principle to half of the system won't get you a correct result.
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PbI42- formation is a separate reaction, just applying LeChetelier's principle to half of the system won't get you a correct result.
But the explanation I posted is correct, right?