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Chemical Equilibrium, solubility

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Hey all, I'm new here, so hello to all of you.
    Question: "Calculate the solubility of PbI2 in .300m AlI3
    (PbI2 - Ksp = 2.0x10-30)"
    So I'm not really sure how to do the symbols and such, but thats a try.
    Now I'm seeing it two ways. One: Right a disassociation equation as PbI2 + AlI3 <---> Al + I + Pb
    I'm too lazy to do the little numbers next to it right now, but you get the idea. Then I would write a Equilibrium expression as 2.0X10-30 = [Pb][Al]....
    In all honesty I'm really lost. I missed a couple days of this unit and I'm quite behind. Any help would be highly, highly appreciated.
    Thank you
    AK



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 18, 2009 #2

    Borek

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    Staff: Mentor

    You are dissolving PbI2 in th excess amount of I-.
     
  4. Mar 18, 2009 #3
    OK. So by doing that the molarity of I in the equation would be .300 because there is already so much of it, or its saturated with I already. So it seems to me that what I need to do is complete the Equilibrium Expression, but my other question is was my Dissassociation equation right?
    Thank you for the prompt response,
    AK
     
  5. Mar 18, 2009 #4

    Borek

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    Staff: Mentor

    No. How many moles of I- per mole of AlI3?

    No. You can't combine two separate reactions in one equation.
     
  6. Mar 22, 2009 #5
    OK so I think I understand this and have worked through it. My final answer was 6.67M PbI2
    Is that remotely correct?
    Thanks
    AK
     
  7. Mar 23, 2009 #6

    Borek

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    Staff: Mentor

    I am afraid it is incorrect at first sight, 6.67M is a highly concentrated solution, while PbI2 is very weakly soluble.

    Show your work.
     
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