2+ Year Member
- Jul 26, 2016
- Dental Student
The solution in this question states that the Ksp = [Pb]^2. I am confused as to why it is not Ksp = [Pb][2I]^2?
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!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
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?Let me know if that's clear enough lol
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.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.
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.The solution in this question states that the Ksp = [Pb]^2. I am confused as to why it is not Ksp = [Pb][2I]^2?