[Lt.Kill]

Consider this reaction:
PbCl2 (s) <--> Pb^2+ (aq) + 2Cl^-

a. Write down the equilibrium constant K for the above reaction.

b. Suppose you are performing an experiment where you are dissolving PbCl2 (s) in water at 25 degrees C. You added 1.11 g of solid PbCl2 to 100 mL of water. Calculate the reaction quotient Q for the above reaction assuming all the salt dissolves.

c. The equilibrium expression you wrote in part a is called the solubility product, describing the solubility equilibrium between an ionic solid and the dissolved ions. For this particular reaction:
K=1.6 * 10^-5 at 25 degrees C.
Can the solution you prepared above dissolve additional CaF2 without adding water?

d. If excess PbCl2 (s) was dissolved in water which initially contains zero ions, what are the concentrations of Pb^2+ and Cl^- ions at the equilibrium condition? Use K given in part C

At part A, I know the answer is Kc=(Pb^2+)(Cl^-)^2 because I don't factor in the solid into the equation.

Part B, what I did was this:
1.11g PbCl2 / 278.2 g PbCl2 = 0.00398994 mol PbCl2 (I divided grams by Molec Weight)
0.00398994 mol Pb2+ / 0.100L = 0.0398994 M Pb2+ (I was hoping that by splitting it, I'd use mols from the original equation and insert them for the mols for the individual atoms. So because Pb is only 1, I'd just insert the mol from PbCl2 into Pb and do the same for 2Cl but multiplied by two because of the coefficient.
0.00797988 mol Cl- / 0.100L = 0.0797988 M Cl-
Qsp = [Pb2+] [Cl-]² = 0.0399 x (0.0798)² = 2.54x10^-4
Even though I got that answer, I feel like there's a better explanation than just "splitting it".

Part C, I was completely lost because there was no mention of CaF2 so I wasn't sure what dissolving the CaF2 would do.

Part D, I set 1.6 x 10^-5= x(2x)^2 (I used the Ice chart off hand and again, didn't calculate the solid.)
1.6 x 10^-5=4x^3
x^3=4 x 10^-6
x=.01587
so plugging in
Pb^2+=.01587 mol
Cl^-=.0317 mol

I feel like A and D are solid. Part B has me questioning weather I followed the correct procedure and C just lost me completely.

[Borek]

You have to show your attempts at solving the question to receive help. This is a forum policy.

[Lt.Kill]

Thanks. I fixed it and as I said, the only questions I have concern with are B and C.

[Borek]

No idea why CaF₂, IMHO it is just an error in the problem (perhaps they meant additional PbCl₂).

Your logic looks good.

[Lt.Kill]

Well I checked with someone else for part C and here is what they said:

"Qsp = 2.54 x 10-4;        Ksp = 1.6 x 10-5
Ksp represents the ion concentrations in a saturated solution. Since Qsp > Ksp, the concentration of ions is greater than in a saturated solution. The solution is supersaturated, so it cannot add more PbCl2 without adding more water. However, it can dissolve additional CaF2, because these ions are not common to the equilibrium."

Assuming that CaF2 isn't an error, would this answer still be correct?

[Borek]

Ah, obvious. Yes, that's a correct answer, somehow the wording confused me.

Note that instead of getting supersaturated solution will most likely never dissolve all PbCl₂.

[Lt.Kill]

Thank you very much for your time! Just wanted to make sure I understand everything before finals.