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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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2 answers

  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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