what mass of Pb^2+ ion is left in solution when 50 ml of 0.2M PbCl2 is added to 50ml of 1.5 NaCl?[Ksp for PbCl2= 1.7* 10^(-4)]

Answer 1

50 mL x 0.2M = 10 millimols Pb^2+
50 mL x 1.5M = 75 mmols Cl^-
........Pb^2+ + 2Cl^- ==> PbCl2
I.......10......75.........0
C......-10.....-20........10
what's left.0....55.......10

So we have left a PbCl2 solid in a solution with an excess of 55 mmoles Cl^- in a volume of 100 mL.
(Cl^-) = 55/100 = 0.55M
Ksp = (Pb^2+)(Cl^-)^2
Ksp = 1.7E-4 = (Pb^2+)(0.55)^2
Solve for (Pb^2+).