A slightly soluble ionic solid exists in a balance between undissolved solid and dissolved ions. Adding an ion already present disturbs this balance, reducing additional dissolution and often promoting precipitation.

Core idea: the common-ion effect

What it is and when it matters

This effect is most noticeable for slightly soluble salts (those that establish a dissolution equilibrium rather than dissolving completely). The key is that the solution already has a common ion (an ion that appears on the product side of the dissolution equilibrium).

Dissolution is a reversible equilibrium

For a salt $M_aX_b(s)$, dissolution and precipitation occur simultaneously. At equilibrium, the rate of dissolution equals the rate of precipitation, and ion concentrations stay constant.

The equilibrium constant expression is useful here because it shows that the ion concentrations are linked at equilibrium; if one ion concentration is forced higher, the other cannot remain as high without violating the equilibrium condition.

Qualitative explanation using Le Châtelier’s principle

Applying Le Châtelier correctly

Le Châtelier’s principle states that a system at equilibrium responds to a stress by shifting in the direction that reduces that stress.

This diagram shows how adding a strong acid increases $H_3O^+$, a product in the weak-acid equilibrium, causing the equilibrium to shift left. It visually reinforces the Le Châtelier idea that increasing a product concentration drives the reaction toward reactants, reducing ion formation at the new equilibrium. Source

For the dissolution equilibrium, a “stress” can be the addition of an ion already present among the dissolved products.

What happens when you add a common ion

If you add a soluble compound that provides $M^{n+}$ (or $X^{m-}$), you immediately increase the concentration of a product of the dissolution reaction.

The system responds by shifting left, toward the solid, which reduces the added ion’s effect by removing ions from solution.

Qualitatively, this means:

• More solid remains undissolved (or forms if ions were already present at high enough levels).

• The salt’s molar solubility decreases in the presence of the common ion.

• The solution contains fewer ions from further dissolution of that salt than it would in pure water.

Why the shift reduces solubility (molecular-level picture)

Adding a common ion increases the frequency of ion-ion encounters near the solid surface. This makes it more likely for ions to recombine into the ionic lattice (precipitation) rather than escape into solution (dissolution). The dynamic equilibrium still exists, but the new balance favours the solid phase.

“Less soluble” does not mean “insoluble”

The salt does not necessarily stop dissolving entirely; rather, the equilibrium is re-established at a lower concentration of the ions that come from the slightly soluble salt. Some dissolution still occurs, but to a smaller extent.

Recognising common-ion situations (no calculations)

Typical ways a common ion is introduced

• Adding a soluble salt that contains one of the ions of the slightly soluble salt

• Mixing two solutions where one already contains a relevant ion

• Using an electrolyte “background” solution that supplies a shared ion

Observable consequences (qualitative)

Depending on concentrations, you may observe:

• Suppressed dissolving of a solid added to a solution containing a common ion

• Increased cloudiness or formation of a precipitate if the ion product becomes large enough that precipitation is favoured

• A mixture that looks unchanged but has a different equilibrium composition (equilibrium changes can be invisible)

Common pitfalls in reasoning

• Do not say equilibrium “stops”; it shifts to a new position while remaining dynamic.

• The stress is adding a product ion, so the shift is toward reactants (solid).

• The qualitative prediction assumes temperature constant, since equilibrium constants depend on temperature.