# Chemistry

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.

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1. 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.

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DrBob222
2. 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.

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DrBob222

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