What concentration of chloride ion is required to precipitate 99.9% of the lead ion from a solution with [Pb^2+]= 0.045 mol/L ?
Start with 0.045M. If we ppt 99.9% it means we leave 0.1% which in decimal form is 1E-3M for Pb^2+. Therefore (Pb^+) = 1E-3*0.045 = 4.5E-5.
PbCl2 ==> Pb^2+ + 2Cl^-
Ksp = (Pb^2+)(Cl^-)^2
You know Ksp and you know the Pb you want, solve for Cl^-.