The solubility of lead (II) chloride (PbCl2) is 1.6 x 10-2 M. What is the Ksp of PbCl2?
2 Answers
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The equilibrium is
PbCl2 (s) <=> Pb2+ (aq) + 2 Cl- (aq)
the requirement for equilibrium is given by
Ksp = [Pb2+][Cl-]^2
we are told that the concentrations are
[Pb2+] = 1.6 x 10^-2 M
conforming to the chemical equation [Cl-]= 2 x 1.6 x 10^-2 = 3.2 x 10^-2
Ksp = ( 1.6 x 10^-2)(3.2 x 10^-2)^2 =1.6 x 10^-5
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Lead 2 Chloride
