Calculate the solubility of lead chloride in water, if its solubility product is 1.7×10−5. (Pb = 206, Cl = 35.5)
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Solution
Given : Solubility product of PbCl2=1.7×10−5 Now, PbCl2 dissociates in water as under PbCl2⇌Pb2++2Cl− ∴Ksp=[S][2S]2 1.7×10−5=4S3 S=3√1.7×10−54 or S=0.0162moles/l=0.0162×277=4.487g/l