At low Iodide (I-) concentrations, the solubility of lead is governed by the reaction: PbI2(s) ⇄ Pb2+ + 2I- Ksp = [Pb2+][I-]2 = 7.9 x 10-9 However at high iodide concentrations, these species of lead are formed, PbI+, PbI2(aq) , PbI3-, and PbI42-. Write one balanced chemical reaction to explain this observation.
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At low Iodide (I-) concentrations, the solubility of lead is governed by the reaction:
PbI2(s) ⇄ Pb2+ + 2I-
Ksp = [Pb2+][I-]2 = 7.9 x 10-9
However at high iodide concentrations, these species of lead are formed, PbI+, PbI2(aq) , PbI3-, and PbI42-. Write one balanced
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