Chet, that's true for a constant mass of solute, but it can be shown that for salts like PbCl2, the solubility in water at 25oC is proportional to its Ksp value. For a 1:2 ionization ratio at 25oC, the Solubility = (Ksp/4)1/3. The published Ksp-value for PbCl2 at saturation from standard Ksp tabels = 1.6 x 10-5. This gives Solubility PbCl2 = (1.6 x 10-5/4)1/3 = 0.016M = 2.00 gms/L (max solubility at saturation at 25oC in DI water ).
Now, if I understand the above problem data of adding 200-mg PbCl2 into 15-ml water and recovering 80-mg by filtration gives a mass of 120-mg of the salt that remains dissolved in solution. Based on this and assuming 120-mg did dissolve, then the concentration of PbCl2 would be [(0.120-gm/15-ml) = [(0.120/277)mole/(0.015L)] = 0.029M in PbCl2 x (277-gm PbCl2/mole) would equal 8.00 g/L. This suggests that Lead(II) Chloride in the problem is 4 times more soluble than what would be calculated from published standard Ksp tables. This might be achievable by increasing the pH with OH- to generate Pb-Hydroxide complex that would shift the PbCl2 <=> Pb+2 + 2Cl- to the right, but unless something else is going on this problem, the most PbCl2 will dissolve is 2.00-g/L = 2.00-mg/ml = 30.0-mg/15-ml max.; not 120-mg/15-ml.
Also, for salts having published Ksp values, S = solubility in moles/L:
Salts with 1:1 ionization ratios (AgCl) => S = (Ksp)1/2
Salts with 1:2 or 2:1 ionization ratios (PbCl2) => S = (Ksp/4)1/3
Salts with 1:3 or 3:1 ionization ratios (Al(OH)3) => S = (Ksp/27)1/4
Salts with 2:3 or 3:2 ionization ratios (Pb(AsO4)2) => S = (Ksp/105)1/5
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*Conditions are deionized water, 25oC, 1.00 Atm. No common ion effect or complex-ion effect.
Caution: Do not use Ksp values to compare solubility unless all being considered have the same ionization ratio. Otherwise, use the appropriate solubility equation associated with the ionization ratio to compare solubility.