Destroyer GC 264

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Itsnotludwigs

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The solution in this question states that the Ksp = [Pb]^2. I am confused as to why it is not Ksp = [Pb][2I]^2?

Thanks again!

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Since there's already Pb ions in the solution it's less soluble so you get Ksp= [Pb]^2. Feel free to pm me if you wanna ask about gchem/math. My test is tomorrow but helping others out and talking might calm me down a bit
 
Since there's already Pb ions in the solution it's less soluble so you get Ksp= [Pb]^2. Feel free to pm me if you wanna ask about gchem/math. My test is tomorrow but helping others out and talking might calm me down a bit
Ah, I see now. I misread the question. Didn't realize that there were already Pb ions in the solution of PbI2 itself. Whoops. Thanks!
 
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On second thought, I still don't exactly understand it. I ran it by a few other kids from my review class, and they seem to be stumped as well. @LuckBloodandSweat, do you think you'd mind elaborating further? (i'm doing my best to help ease your nerves here!)
 
Let me know if that's clear enough lol
 

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Let me know if that's clear enough lol
ha yes, I was able to see it. No worries, and thank for the awesome response! But I unfortunately am still unclear as to why it is X instead of 2X exactly. I understand that there's a common ion effect taking place here, but how exactly did you calculate/determine that it is X instead of 2X for I- ions? Did you divide by two?
 
I think the way any common ion effect works is that you just take away the coefficient you'd normally have but you would still square/cube/whatever function it would normally have. Sorry if I didn't make that clear.
 
I think the way any common ion effect works is that you just take away the coefficient you'd normally have but you would still square/cube/whatever function it would normally have. Sorry if I didn't make that clear.
Hmm ok. Yeah, that does seem to be the case, I'm just still a bit curious for even more elaboration as to exactly what is going on here and how the common ion effect affects the coefficients.

Thank you very much for the help!

I wonder, @orgoman22, would you possibly be able to elaborate on this further?
 
Maybe it has something to do with when you throw I ions in the solution it's not like you add them in a +2X fashion. But when I combined with Pb it's gonna be 2I's for every 1 Pb. I think it's pretty common (haha) for common ion effect problems. Basically you don't have the coefficients when you do the ice table because the ions aren't dissociating in a set fashion. You're just throwing ions in the solution which then they combine in an orderly fashion (for example 2 I's for every Pb) to form the precipitate. maybe someone else can add something. Orgoman22 would definitely be your best bet haha
 
The solution in this question states that the Ksp = [Pb]^2. I am confused as to why it is not Ksp = [Pb][2I]^2?

Thanks again!
Recall Ksp is the Products divided by the Reactants.......like any equilibrium constant, the solid is not used. PbCl2 dissociates into Pb+2 ions and two Chloride ions.......thus the Ksp is the product of The Pb+2 and Cl- squared. I square the Cl- since two chloride ions are dissociated .This is a very very important problem. Please redo the problem. If you need more work on this, I like the Raymond Chang text book as well as Brown and LeMay.

Hope this helps.

Dr. Romano
 
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