# Chemical Equilibrium, solubility

1. Mar 17, 2009

### AKilren

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. Mar 18, 2009

### Staff: Mentor

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

3. Mar 18, 2009

### 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

4. Mar 18, 2009

### Staff: Mentor

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

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

5. Mar 22, 2009

### 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

6. Mar 23, 2009

### Staff: Mentor

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