[trenda]

Stuck on this assignment question:
The Ksp for PbI2 (s) is 1.4 x 10^-8. Calculate solubility of PbI2 (s) in 0.41 M Pb(NO3)2.

I know that the balanced chemical equation for the dissolution of PbI2 is:
PbI2 (s) Pb2+ (aq) + 2I- (aq)

And I know that the Ksp expression is:
Ksp = [Pb2+][I-]^2

I did:
[Pb2+] = 2 x 0.41 M = 0.82 M

[Pb2+] = 0.82 + x M
[I-] = 2 x M

Ksp = (0.82 + x)(2x)^2

I am stuck on this step above and I'm not sure what I'm supposed to do here.

[Hunter2]

Why times 2 for lead.
You have Ksp = [Pb2+][I-]^2
KSP and the lead concentration is given. Solve for I- . From this you get the amount of PbI2 what can be dissolved.

[Hunter2]

. Deleted

[Borek]

More accurate would be with cPb2+(add) = 0.5cI-

Which is basically what OP did.

Ksp = cPb2(total)/(2cI-)^2

Huh?

[Borek]

Ksp = (0.82 + x)(2x)^2

Apart from the unnecessary factor of 2 already mentioned, you are on the right track. Just solve for x.

Note: to make things easier you can assume solubility will be so low that the concentration of [Pb²⁺] doesn't change during dissolution. That will simplify the equation to solve (cubic -> quadratic) but after finding the solution you need to check if the assumption was correct (typically we assume if the concentration change is below 5% things are OK).

[Hunter2]

We have still no Response. So I try to give the solution.

Please confirm it's correct.

Simple way
Ksp =cPb2+ *c(I-)^2
1.4*10^(-8) =0.41 *c(I-)^2
M (PbI2) = 461 g/mol
cI- = 0.00018 mol/l = 85,2 mg  PbI2

More accurate way.
Ksp =cPb2+ *c(I-)^2

cPb2+ = 0.41 + cPb2+ (add)
cPb2+ add = 2* c(I-)

1,4*10^(-8) =(0.41 + 2*c(I-)* c(I-)^2

c(I-) = x

1.4*10^(-8) = 0.41* x^2 + 2 x^3

2x^3 +0.41*x^2 -1.4*10^(-8) = 0

Solving of cubic equation gives for x = 0.0002 mol/l means 92,2 mg/l PbI2

It's a deviation  of 8%

[mjc123]

I think there are a couple of errors:
c(PbI2) = 0.5c(I-) = 0.000092 M = 42.6 mg/L
In part 2, c(Pb2+) = 0.41 + 0.5c(I-)   
I haven't tried solving the amended cubic.

[Borek]

2x^3 +0.41*x^2 -1.4*10^(-8) = 0

Solving of cubic equation gives for x = 0.0002 mol/l means 92,2 mg/l PbI2

It's a deviation  of 8%

This is wrong, and it is obvious even without checking the exact solution of the cubic equation. Solubility calculated assuming concentration of Pb²⁺ didn't change is about four orders of magnitude lower than the concentration of Pb²⁺ already present, so the assumption is perfectly valid. No way to get substantially different result with exact calculations.

And then, Wolfram Alpha to the rescue. 0.00018 again.

(Then, there is the mistake pointed out by mjc123, but it doesn't change the fact both solutions have to be almost identical.)

[Hunter2]

Thanks for the checking and calculation.
My Grapfer Free gave me  for x = -0.205 , - 0.0002 and 0.0002 as solutions, probably it's not accurate enough.
Yes the iodide concentration hast to be transformed to lead iodide, That was a mistake in deed.