Common ion effect
Any equilibrium involving ions (like solubility equilibria) can be shifted by dissolving into the solution any other compound that adds a common ion, or any compound that reacts with one of the ions already in solution. For example, if you add a few drops of HCl_{(aq)} to a saturated solution of NaCl_{(aq)}, some crystals of NaCl_{(s)} will precipitate because of the addition of chloride ions.
Usually, a common ion problem will give you two compounds. It will tell you the $\displaystyle K_{\text{sp}}$ value for one and the concentration of the other, and then it will ask you to find the solubility of the first in the second. To find the solution, you need an ICE table—this is the only time you’ll ever need one in a solubility-related problem.
Example
What is the molar solubility of lead(II) chromate in a 0.10 mol/L solution of sodium chromate? The solubility product constant for lead(II) chromate is 2.3 × 10^{−13}.
First, write the solubility equilibrium equation of the compound whose concentration is not given:
PbCrO_{4(s)} ⇌ Pb^{2+}_{(aq)} + CrO_{4}^{2−}_{(aq)}.
Let $\displaystyle x$ represent $\displaystyle \left \lvert \Delta{} \left [ \text{CrO}_{4}^{2 -} \right ] \right \rvert$. Now we can make our ICE table:
PbCrO_{4(s)} | Pb^{2+}_{(aq)} | CrO_{4}^{2−}_{(aq)} | |
I | — | $\displaystyle 0$ | $\displaystyle 0.10$ |
C | — | $\displaystyle + x$ | $\displaystyle + x$ |
E | — | $\displaystyle x$ | $\displaystyle x + 0.10$ |
Notice that we ignore lead chromate because it’s a solid, not an aqueous solution or gas. Unlike in a regular solubility problem, one of our ions has an initial concentration. The 0.10 mol/L of chromate ions comes the compound in the question whose concentration is given, sodium chromate.
Now we can write the solubility product constant expression:
$\displaystyle K_{\text{sp}} = \left [ \text{Pb}^{2 +} \right ] \left [ \text{CrO}_{4}^{2 -} \right ] = \left ( x \right ) \left ( x + 0.10 \right )$.
Since $\displaystyle \frac{0.10}{K_{\text{sp}}} > 100$, we can drop $\displaystyle x$ where it is added or subtracted:
$\displaystyle K_{\text{sp}} = \left ( x \right ) \left ( x + 0.10 \right ) \approx x \left ( 0.10 \right )$.
Now we can solve for $\displaystyle x$, which represents the molar solubility of lead(II) chromate in a 0.10 mol/L solution of sodium chromate:
$\displaystyle x = \frac{K_{\text{sp}}}{0.10} = \frac{2.3 \times 10^{- 13}}{0.10} = 2.3 \times 10^{- 12} \, \text{mol/L}$.