millimoles Pb(NO3)2 = 50.0 x 0.17 = 8.5
mmoles KCl = 50.0 x 1.6 = 80.0
...Pb(NO3)2 + 2KCl ==> PbCl2(s) + 2KNO3
I....8.50.......80.0.......0.......0
C...-8.50......-17.0.....8.50....17.0
E......0.......63.0......8.50.....17.0
(KCl) remaining = 63.0 mmoles/100 mL = 0.63M
Let x = solubility PbCl2 (to find the Pb^2+ in the final solution and note it has a common ion of Cl^- from the excess KCl.)
PbCl2 ==> Pb^2+ + 2Cl^-
x........x.......2x
Ksp = (Pb^2+)(Cl^-)^2 = 1.6E-5
Substitute x for Pb^2+.
Substitute 2x+0.63 for Cl and don't forget to square it. The concn Cl is 0.63 from the excess KCl and 2x from the PbCl2.
Then solve for x to obtain the concn Pb2+ in the final solution.
A solution is prepared by mixing 50.0 mL of 0.17 M Pb(NO3)2 with 50.0 mL of 1.6 M KCl. Calculate the concentrations of Pb2+ and Cl - at equilibrium. Ksp for PbCl2(s) is 1.6X 10-5.
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