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Question

Calculate the solubility of lead chloride in water, if its solubility product is 1.7×105.
(Pb = 206, Cl = 35.5)

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Solution

Given : Solubility product of PbCl2=1.7×105
Now, PbCl2 dissociates in water as under
PbCl2Pb2++2Cl
Ksp=[S][2S]2
1.7×105=4S3
S=31.7×1054
or S=0.0162moles/l=0.0162×277=4.487g/l

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