Finding the Concentrations - Common Ion

Let’s solve this problem:

The \(K_{sp}\) of \(PbCl_2\) is \(1.7 \cdot 10^{-5}\). What is the concentration of chloride ion 

in a saturated solution that already has \([Pb^{+2}]\ = 0.25~M\)?

There are several things that are implied, but not stated, that we need to recognize:

The reaction involves \(PbCl_2\) breaking up into ions (this comes from the definition of \(K_{sp}\)).

The reaction will run forward until equilibrium is reached. (Again, this is the definition of saturated.)

So, our reaction is:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

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

We can now do the following math:

\(K_{sp}=[Pb^{+2}] \cdot [Cl^{-1}]^2~~~~~1.7 \cdot 10^{-5} = [0.25 + x] \cdot [2x]^2\)

This will be a “little x” problem, so we can assume that 0.25 + x = 0.25. That simplifies the math to:

\(1.7 \cdot 10^{-5} = (0.25)\cdot 4x^2\)

which gives us x=0.0041 M

Since \([Cl^{-1} = 2x\) we know that \([Cl^{-1}]=0.0082~M\).

This is an example of the common ion effect. Simply put, the presence of lead ions in the solution did not allow as much of the lead II chloride to dissolve as would have in pure water.

Another, perhaps more useful, way to think about the common ion effect is to relate it to Le Chatelier’s Principle. For the reaction:

\(PbCl_{2~(s)} \rightleftarrows Pb^{+2}_{(aq)} + 2~Cl^{-1}_{(aq)}\)

additional amounts of \(Pb^{+2}\) ions will drive the reaction to the left, reducing the amount of chloride ion in solution.