Asked by Alex
What is the maximum mass(g) of KCL that can be added to 1.00L of a 0.0100 M lead(ii) chloride solution without causing any precipitation of PbCl2? Assume the additional KCl does not affect the volume of the solution. For, PbCl2, Ksp=1.6x10^-5
Personally I'm at a loss here. I cant figure out where to start since Cl is in both compounds. I imagine the K ends up on it's own and that it doesn't really effect the equation but how can the Qsp of PbCl2 be raised by adding more Cl? Is there an excess of Pb that I am not aware of? Or will it force more Cl to bind with the Pb? I might be over thinking this.....
Personally I'm at a loss here. I cant figure out where to start since Cl is in both compounds. I imagine the K ends up on it's own and that it doesn't really effect the equation but how can the Qsp of PbCl2 be raised by adding more Cl? Is there an excess of Pb that I am not aware of? Or will it force more Cl to bind with the Pb? I might be over thinking this.....
Answers
Answered by
DrBob222
I think you are trying to make this too hard. To start the 0.01 PbCl2 is in solution.
......PbCl2 ==> Pb^2+ + 2Cl^-
I.....0.01......0.01....0.02
You can show this is in solution.
Qsp = (0.01)(0.02)^2 = 4E-6 which is < Ksp so it hasn't pptd and all is in solution.
What must (Cl^-) be to cause pptn?
Ksp = (Pb^2+)(Cl^-)^2
1.6E-5 = (0.01)(Cl^-)^2
(Cl^-) = 0.04 but you should confirm that. You already have 0.02 there. So you must add 0.02 M KCl. Convert that to grams in the 1L. Right?
......PbCl2 ==> Pb^2+ + 2Cl^-
I.....0.01......0.01....0.02
You can show this is in solution.
Qsp = (0.01)(0.02)^2 = 4E-6 which is < Ksp so it hasn't pptd and all is in solution.
What must (Cl^-) be to cause pptn?
Ksp = (Pb^2+)(Cl^-)^2
1.6E-5 = (0.01)(Cl^-)^2
(Cl^-) = 0.04 but you should confirm that. You already have 0.02 there. So you must add 0.02 M KCl. Convert that to grams in the 1L. Right?