A common ion is an ion already present in solution that is also produced by the dissolution of a sparingly soluble solid. Its presence usually lowers the amount of that solid that can dissolve.

Consider

$$\mathrm{AgCl(s)\rightleftharpoons Ag^+(aq)+Cl^-(aq)}$$

with

$$K_{sp}=[\mathrm{Ag^+}][\mathrm{Cl^-}].$$

In pure water, dissolving silver chloride produces equal concentrations of $\mathrm{Ag^+}$ and $\mathrm{Cl^-}$. If instead the solution already contains chloride from a soluble salt such as $\mathrm{NaCl}$, the initial chloride concentration is not zero.

Suppose

$$K_{sp}=1.8\times10^{-10}$$

and the solution initially contains

$$[\mathrm{Cl^-}]_0=0.010,\mathrm M.$$

Let $s$ be the additional molar concentration of $\mathrm{AgCl}$ that dissolves. At equilibrium,

$$[\mathrm{Ag^+}]=s,$$

$$[\mathrm{Cl^-}]=0.010+s.$$

Therefore

$$1.8\times10^{-10}=s(0.010+s).$$

Because the dissolved amount is expected to be tiny compared with $0.010,\mathrm M$, first approximate

$$0.010+s\approx0.010.$$

Then

$$s\approx\frac{1.8\times10^{-10}}{0.010} =1.8\times10^{-8},\mathrm M.$$

The result confirms the approximation:

$$\frac{s}{0.010}=1.8\times10^{-6},$$

so the added chloride from dissolving $\mathrm{AgCl}$ is negligible compared with the chloride already present.

This solubility is far smaller than the pure-water value,

$$\sqrt{1.8\times10^{-10}}\approx1.3\times10^{-5},\mathrm M.$$

The reason is not a separate empirical rule. Adding chloride makes the ion product larger than it would be at the same silver concentration. To satisfy the same fixed $K_{sp}$, the equilibrium silver concentration must therefore be smaller. Equivalently, adding a dissolution product drives the equilibrium toward the solid.

The same logic applies when the common ion is the cation rather than the anion and to salts with other stoichiometries.

The common-ion effect is a special case of ordinary equilibrium response. It does not mean every added electrolyte reduces solubility in the same way, and other coupled reactions can reverse the simple prediction. For example, protonation or complex formation can remove one dissolved ion and increase total solubility; those effects require additional equilibrium chemistry.