Solubility HELP!!!!!!!?

Question

What is the solubility (in mol/L) of PbCl2 in a 0.15 M solution of CaCl2? [Ksp (PbCl2)=1.6 x 10-5]
what am i doing wrong here?
1.6*10^-5= x(.15)^2
x=7.11*10^-4

and that should be [Ca]=x=[PbCl2] right?

Answer 1

You have miscalculated [Cl-]

We assume that dissolving CaCl2 is 100%.
Then [Cl-]o= 2*[CaCl2]= 2*0.15=0.3

.. .. .. .. .. PbCl2 <=> Pb+2 + 2Cl-
Initial .. .. .. .. .. .. .. .. .. .. .. .. .. 0.3
Dissolve .. x
produce .. .. .. .. .. .. .. .. x .. .. .2x
At Equil .. .. .. .. .. .. .. .. .x .. .. 0.3+2x

Ksp = [Pb+2][Cl-]^2= x(0.3+2x)^2= 1.6*10^-5
If you don't want to solve the polynomial then assume 0.3 >>x and 0.3+2x =0.3, thus
x*0.3^2 =1.6*10^-5 =>
x= (1.6*10^-5)/0.09 =1.78 *10^-4 M which is <<0.3 so our assumption is valid.

Don't mix Ca+2 into this. That would be wrong.

Answer 2

You are not interpreting your result correctly.
You have determined the conc of Pb+2 ion in the solution. The [Cl-] associated with this is much less than that available from CaCl2, which is presumed to have a much higher solubility. There is no reason to expect that the calcium ion = x, since it equals 0.15. If you had started with PbCl2 and no background CaCl2, the conc of PbCl2 would be about 2.3x10-2 M. The large amount of background Cl- depresses the Pb+2 content.