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.<Chemistry 2e by Paul Flowers and others/15.1 Precipitation and Dissolution>[Chemistry 2e by Paul Flowers and others; §15.1, Common Ion Effect: adding an ion also produced by dissolution shifts the solubility equilibrium]{ref}

The preceding <solubility product and material balance> give a short proof without a qualitative slogan. Taking a <derivative> of $s_*^2+c_0s_*=K_{\mathrm{sp},c}$ at fixed $K_{\mathrm{sp},c}$ gives
$$
(2s_*+c_0)\frac{ds_*}{dc_0}+s_*=0,\qquad
\frac{ds_*}{dc_0}=-\frac{s_*}{2s_*+c_0}<0.
$$
Thus 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 $s_*$ beside $c_0$ replaces $s_*(c_0+s_*)=K_{\mathrm{sp},c}$ by $c_0s_{\mathrm{app}}=K_{\mathrm{sp},c}$. 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:
$$
s_{\mathrm{app}}=\frac{K_{\mathrm{sp},c}}{c_0}
=s_*+\frac{s_*^2}{c_0},\qquad
\frac{s_{\mathrm{app}}-s_*}{s_*}=\frac{s_*}{c_0}.
$$
At $c_0=0.030\,\mathrm M$, $s_{\mathrm{app}}=0.003333333\,\mathrm M$, approximately $10.1\%$ too high. At $c_0=0.100\,\mathrm M$, $s_*=0.000990195\,\mathrm M$ 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>.

\Image[/common-ion-solubility-exact-and-approximate.png]
{title=Common-ion suppression and the error from neglecting the <solid>'s own B <anions>}
{description=Original curves for the same stipulated AB model, with exact <molar solubility> and $K_{\mathrm{sp},c}/c_0$ approximation. The exact curve reaches $0.010\,\mathrm M$ at $c_0=0$; the approximation diverges there. The lower panel shows its exact <relative error>, with markers at $0.030$ and $0.100\,\mathrm M$. No curve represents experimental measurements of a named substance.}

For a desired fractional <relative error> at most $\varepsilon>0$, substituting $s_*/c_0\leq\varepsilon$ into the exact balance gives the equivalent criterion
$$
c_0\geq\sqrt{\frac{K_{\mathrm{sp},c}}{\varepsilon(1+\varepsilon)}}.
$$
For this model, a $5\%$ tolerance requires $c_0\geq0.0436436\,\mathrm M$. 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.


 Common-ion effect (source code)