Ksp

[LTami]

(1) The Ksp of PbCl2 is 1.6x10E-5 at 25C. what is the solubility of PbCl2 in 0.01 M KCl?
PbCl2--> Pb2+ +2Cl-
Ksp=(Pb2+)(Cl-)^2 -->1.6x10^-5=x(2x+.01)^2-->1.6x10^-5=x(0.01)^2
-->x=.16M

I know that the addition of an common ion will push the equilibrium in a direction which tends to reduce the concentration of that ion. That means the equilibrium will move to the left, and the solubility of the compound will reduce. But please check my calculation below. I will just calculate the solubility of PbCl2 without a common ion.

(2) Ksp=(Pb2+)(Cl-)^2-->1.6x10^-5=x(2x)^2-->1.6x10^-5=4X^3
-->x=0.016M

when I compared the 2 solubilities, the one with an common ion is more soluble, did I do anything wrong? Shouldn't (2) be more soluble than (1)?

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[Ryltar]

Looks like the common ion, KCl, is too small in relation to Ksp, thus the estimation method cannot be used. 0.016 is greater than 0.01. For example, here is the solution using 0.1M KCl instead of 0.01M.

Ksp = x(0.1)^2
x(0.01) = 1.6E-5
x= 1.6E-3 = 0.0016

Ksp = 4X^3
4X^3 = 1.6E-5
X^3 = 0.4E-5
X = 0.16

 

[LTami]

Thanks for replying. It is clear now. Thank you.