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

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Discussion Overview

The discussion revolves around calculating the solubility of lead iodide (PbI2) in a solution of aluminum iodide (AlI3) with a concentration of 0.300m. Participants explore the dissociation equations, equilibrium expressions, and the implications of the presence of excess iodide ions on solubility.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant presents a dissociation equation for PbI2 and suggests writing an equilibrium expression based on the solubility product constant (Ksp).
  • Another participant notes that dissolving PbI2 in the presence of excess I- ions affects the molarity of I in the equilibrium expression.
  • A participant questions the correctness of the dissociation equation proposed by the original poster, stating that combining two separate reactions into one equation is not valid.
  • The original poster later claims to have calculated a final solubility of 6.67M for PbI2, seeking validation of this result.
  • Another participant challenges this result, asserting that 6.67M is too high for a substance that is known to be very weakly soluble.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the dissociation equation and the calculated solubility of PbI2. There is no consensus on the final answer, and the discussion remains unresolved regarding the accuracy of the calculations and the approach taken.

Contextual Notes

Participants have not fully clarified the assumptions regarding the dissociation of AlI3 and its impact on the concentration of iodide ions. The calculations presented are based on incomplete information and may depend on additional context regarding the solubility of PbI2.

AKilren
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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

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