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The common-ion effect is the change in a chemical equilibrium caused by adding an ion already participating in it. In the fixed-activity coefficient AB model, adding the common B anion lowers the molar solubility. The shared ion participates regardless of which salt supplied it.^[ref]

The preceding solubility product and material balance give a short proof without a qualitative slogan. Taking a derivative of $s_{\ast }^2+c_0{s}_{\ast }=K_{\mathrm{sp},c}$ at fixed $K_{\mathrm{sp},c}$ givesThus the molar solubility decreases continuously as $c_0$ increases. At $c_0=0$, the derivative is $-1/2$; for a very large common-ion background, the rationalized expression approaches $K_{\mathrm{sp},c}/c_0$. The solubility product is fixed within this model even while the molar solubility changes.

Neglecting $s_{\ast }$ beside $c_0$ replaces $s_{\ast }(c_0+s_{\ast })=K_{\mathrm{sp},c}$ by $c_0{s}_{\mathrm{app}}=K_{\mathrm{sp},c}$. This approximation always overestimates the exact molar solubility, since it neglects the B anions supplied by AB itself. For $c_0>0$, its relative error has an exact simple form:At $c_0=0.030\mathrm{M}$, $s_{\mathrm{app}}=0.003333333\mathrm{M}$, approximately $10.1\%$ too high. At $c_0=0.100\mathrm{M}$, $s_{\ast }=0.000990195\mathrm{M}$ and $s_{\mathrm{app}}=0.001000000\mathrm{M}$, approximately $0.990\%$ too high. A small solubility product by itself does not justify the approximation: the common-ion background must be large compared with the resulting molar solubility.

For a desired fractional relative error at most $\epsilon >0$, substituting $s_{\ast }/c_0\le \epsilon$ into the exact balance gives the equivalent criterionFor this model, a $5\%$ tolerance requires $c_0\ge 0.0436436\mathrm{M}$. This bound concerns only the algebraic approximation. In a real chemical solution, changes in ionic strength and activity coefficients, or competing chemical reactions, can invalidate a constant $K_{\mathrm{sp},c}$ even when this bound is met.

A solubility product constrains a product of thermodynamic activities; the available inventories determine which combinations of concentrations are possible. Their intersection determines the molar solubility. Consider an original, abstract AB solid, with the preceding one-to-one dissolution chemical equation. Stipulate $c^{\circ }=1\mathrm{M}$, ${\gamma }_{\mathrm{A}}={\gamma }_{\mathrm{B}}=1$ and $K_{\mathrm{sp},c}=10^{-4}{\mathrm{M}}^2$ at fixed temperature and pressure. Thus $K_{\mathrm{sp}}=10^{-4}$ is dimensionless. These are chosen model parameters, not data for an identified material.

Initially, a completely dissociated soluble XB salt supplies B anions at molar concentration $c_0\ge 0$ and X cations at the same molar concentration; A is absent. All three ions have unit electric charge magnitude. X remains a spectator ion, and only AB can form a solid. Assume constant volume, no additional chemical reactions, and negligible contributions from self-ionization of water. Dissolving AB raises both reacting-ion concentrations by the same $s$:Charge neutrality is satisfied: $c_{\mathrm{A}}+c_{\mathrm{X}}-c_{\mathrm{B}}=s+c_0-(s+c_0)=0$. Adding B alone without its accompanying cation would omit the initial electric charge balance.

In the plane of the two reacting-ion concentrations, this inventory constraint is the line $c_{\mathrm{B}}=c_{\mathrm{A}}+c_0$. Saturation supplies the hyperbola $c_{\mathrm{B}}=K_{\mathrm{sp},c}/c_{\mathrm{A}}$ for $c_{\mathrm{A}}>0$. Their positive intersection solves a quadratic equation:The quadratic formula supplies two roots of a polynomial; only the positive one is a physical concentration. Uniqueness also follows because $s(s+c_0)$ strictly increases for $s\ge 0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.

For pure water, $c_0=0$ gives $s_{\ast }=\sqrt{K_{\mathrm{sp},c}}=0.010\mathrm{M}$. For $c_0=0.030\mathrm{M}$,Their product is $10^{-4}{\mathrm{M}}^2$, although neither concentration equals the pure-water value. This is the useful geometrical distinction: an inventory line and an equilibrium constant specify different constraints.

Check the solid inventory before imposing saturation. If the initially added AB has amount of substance $n_0$ in volume $V$, the saturated branch requires $n_0>V{s}_{\ast }$ for excess solid to remain. For example, choose $n_0=0.0020\mathrm{mol}$ and $V=0.100\mathrm{L}$. In pure water, $0.0010\mathrm{mol}$ dissolves and the same amount of substance remains as solid. At $c_0=0.030\mathrm{M}$, approximately $0.0003027756\mathrm{mol}$ dissolves, leaving $0.0016972244\mathrm{mol}$ as solid. In this model, an insufficient supply instead givesIf $n_0/V<s_{\ast }$, all AB dissolves and the ion product is below the solubility product. At $n_0/V=s_{\ast }$, the last solid disappears exactly at saturation.