= Solubility product and material balance
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> $c_0\geq0$ 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 (thermodynamics)>, no additional <chemical reactions>, and negligible contributions from <self-ionization of water>. Dissolving AB raises both reacting-ion <concentrations> by the same $s$:
$$
c_{\mathrm A}=s,\qquad c_{\mathrm B}=c_0+s,\qquad c_{\mathrm X}=c_0.
$$
<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>:
$$
s(s+c_0)=K_{\mathrm{sp},c},\qquad
s_* =\frac{\sqrt{c_0^2+4K_{\mathrm{sp},c}}-c_0}{2}
=\frac{2K_{\mathrm{sp},c}}{\sqrt{c_0^2+4K_{\mathrm{sp},c}}+c_0}.
$$
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 $s\geq0$, starting at zero and growing without bound. The second expression, obtained by multiplying by the conjugate, avoids subtraction of nearly equal numbers at large $c_0$.
\Image[/solubility-product-balance-intersection.png]
{title=<Solubility product and material balance> and the conditional <solubility product> select the dissolved <concentrations>}
{description=Original stipulated AB model with $K_{\mathrm{sp},c}=10^{-4}\,\mathrm M^2$ and unit <activity coefficients>. Intersections of the saturation <hyperbola> with inventory <lines> give the pure-<water> and common-ion <chemical equilibrium> compositions. The dashed projections read the $c_0=0.03\,\mathrm M$ intersection; these curves are calculated constraints, not measured data or a time evolution.}
For pure <water>, $c_0=0$ gives $s_*=\sqrt{K_{\mathrm{sp},c}}=0.010\,\mathrm M$. For $c_0=0.030\,\mathrm M$,
$$
s_*\approx0.003027756\,\mathrm M,\qquad
c_{\mathrm B}\approx0.033027756\,\mathrm M.
$$
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 (thermodynamics)> $V$, the saturated branch requires $n_0>Vs_*$ 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 gives
$$
s=\min\{n_0/V,s_*\}.
$$
If $n_0/V<s_*$, all AB dissolves and the <ion product> is below the <solubility product>. At $n_0/V=s_*$, the last <solid> disappears exactly at saturation.
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