Thanks for telling me what the problem is. That helps a lot.
This is the common ion effect.
HI is a strong acid. On ionization it give a (I^-) = 0.5 M for A and 0.2 M for B and 0.8 M for C. So here's the way it works, and I'll use the 0.8 M KI for the reasoning.
.......................PbI2(s) ==> Pb^2+ + 2I^-
I........................solid.............0...........0
C.......................solid.............x...........2x
E........................solid.............x...........2x
So, if I want to know the solubility of PbI2 in a solution, you see it is
Ksp = (Pb^2+)(I^-)^2 or (x)(2x) = 4x^3 = Ksp.
But, when you add a common ion, in this case I'm using KI, the KI (its true the other HI too) ionizes completely and it dissolves completely to give
..................KI(s) ==> K^+ + I^-
I..................0.8 M.......0........0
C.................-0.8 M.....0.8M.....0.8 M
E....................0.........0.8..........0.8 M
So now that Ksp expression looks this way:
Ksp = (Pb^2+)(I^-)^2
For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above. The bottom line is this.
.......................PbI2(s) ==> Pb^2+ + 2I^-
I........................solid.............0...........0
C.......................solid.............x...........2x
E........................solid.............x...........2x
Adding the I^- as a common ion, you INCREASE the 2I^- on the right which forces the equilibrium to the left to reduce the I^- anyway it can and that makes the PbI2 solid much less soluble. C is the answer because the I^- in KI is the largest of the three.
Explanation is long because of the typing but it's all there. Follow up if necessary. I'm here for another few hours.
In which aqueous system is PbI2 least soluble and why?
[A] 0.5 M HI [B] 0.2 M HI [C] 0.8 M KI
I understand that I- is the common ion. I don't understand why [C] 0.8 M KI is the correct answer.
2 answers
oops.
I didn't proof when I should have proofed and this
"For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above." should have been this.
"For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8)^2 so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/(0.8)^2 = much larger number than above when solving for x = solubility PbI2." Sorry about that.
I didn't proof when I should have proofed and this
"For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8) so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/0.8 = much larger number than above." should have been this.
"For Pb you substitute x.
For I^- you substitute x from PbI2 and 0.8 from the KI so the total is x+0.8
Ksp = (x)(x+0.8)^2 so now (neglecting the quadratic because x is small) you have x = solubility = Ksp/(0.8)^2 = much larger number than above when solving for x = solubility PbI2." Sorry about that.
2 answers