Common Ion Effect in Solubility — Definition & Overview

The presence of a common ion in solution can significantly reduce the solubility of a substance.

For example, consider calcium sulfate ($CaSO_4$), $K_{sp} = 9.1 \times 10^{-6}$. It dissolves according to the following equation:

$$CaSO_4 \rightleftharpoons Ca^{2+} + SO_4^{2-}$$

If we complete an ICE table for the dissolution of solid calcium sulfate in water, we find that the final concentrations of the $Ca^{2+}$ and $SO_4^{2-}$ ions are equal to $3.02 \times 10^{-3} M$.

Now, consider solid calcium sulfate dissolved in a solution that is already $0.100 M$ in $Ca^{2+}$ ions. The change in concentration of the calcium and sulfate ions is now only about $9.1 \times 10^{-5} M$. This is because the presence of calcium ions in the solution causes the equilibrium position to shift to the left by Le Chatelier's Principle. As a result, less of the solid $CaSO_4$ dissolves into calcium and sulfate ions.

(Video) 17.5 Common Ion Effect and Precipitation | General Chemistry

by Chad's Prep

In this section of the video, Chad explains the common ion effect. You'll learn how a common ion changes molar solubility. Chad will show you two example problems. This is an excellent video for learning how to do common ion problems.