How Do You Calculate the Solubility of PbI2 in .300m AlI3?

[AKilren]

Homework Statement

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 that's 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

Homework Equations

The Attempt at a Solution

 

[Borek]

You are dissolving PbI₂ in th excess amount of I^-.

 

[AKilren]

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

 

[Borek]

molarity of I in the equation would be .300

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

my other question is was my Dissassociation equation right?

No. You can't combine two separate reactions in one equation.

 

[AKilren]

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

 

[Borek]

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.