Finding Molar Solublity

Let’s solve this problem:

The $K_{sp}$ of $PbC{l}_2$ is $1.7\cdot {10}^{-5}$, what is the molar solubility of the compound?

This problem is the SAME as finding the concentrations in a saturated solution, with the exception that the answer is ALWAYS “x”. Be sure to read all the way to the bottom of this page to make sure that you understand the implications of that statement!

Our reaction is:

$PbC{l}_{2(s)}\rightleftarrows P{b}_{(aq)}^{+2}+2C{l}_{(aq)}^{-1}$

We can create an ICE table for this reaction, which would look like this:

We can now do the following math:

$K_{sp}=[P{b}^{+2}]\cdot [C{l}^{-1}{]}^{2}1.7\cdot {10}^{-5}=[x]\cdot [2x{]}^2$

Simplified, that is:

$1.7\cdot {10}^{-5}=4x^3$

which gives us x=0.016 M

Molar solubility is “x” therefore the answer is 0.016M.

What's interesting here is that it does NOT matter what the formula for the ionic compound is, the answer is always "x". 

A simple ionic compound like $AgCl$ would simplify to:

$K_{sp}=x^2$

while a compound like $P{b}_3(P{O}_4{)}_2$ would work out to be:

$K_{sp}=108x^5$

However, in both cases, the left column of the ICE table would be "some - x" and therefore the molar solubility is just "x".