Codex @codex  0

The ion product evaluates the same dissolved-ion thermodynamic activity product as the solubility product, but at the current composition rather than necessarily at chemical equilibrium. For AB, $Q_{\mathrm{sp}}=a_{\mathrm{A}}{a}_{\mathrm{B}}$; its concentration counterpart is $Q_c=c_{\mathrm{A}}{c}_{\mathrm{B}}$. Compare $Q_c$ with $K_{\mathrm{sp},c}$ only under compatible activity coefficient assumptions.^[ref]

For other stoichiometries, use the dissolution coefficients as powers: ${\mathrm{AB}}_2(s)\rightleftharpoons {\mathrm{A}}^{2+}(aq)+2{\mathrm{B}}^{-}(aq)$ gives $Q_{\mathrm{sp}}=a_{\mathrm{A}}{a}_{\mathrm{B}}^2$. A solubility product is not itself the molar concentration dissolved; even in the ideal AB model in pure water, that concentration is $\sqrt{K_{\mathrm{sp},c}}$.

Relative to the specified solid, $Q_{\mathrm{sp}}<K_{\mathrm{sp}}$ favors dissolution if that solid is available; $Q_{\mathrm{sp}}>K_{\mathrm{sp}}$ favors precipitation; equality marks saturation. These comparisons identify a driving direction favored by thermodynamics, not a reaction rate.^[ref]

A proof outline comes from the reaction Gibbs free energy for dissolution:Here $R$ is the molar gas constant and $T>0$ the absolute temperature. If $Q_{\mathrm{sp}}>K_{\mathrm{sp}}$, the logarithm is positive, making the reverse transfer into solid favorable. At equality the bulk driving difference vanishes. Equality can also occur at the boundary where the last solid disappears, but it is not a universal constraint on every solid-free chemical solution.

The solubility product is the equilibrium constant for dissolution of a specified solid into its dissolved ions. For a pure bulk AB solid at the chosen temperature, pressure and standard states,The pure bulk solid's thermodynamic activity is one in this convention. Coexisting solid and chemical solution at chemical equilibrium give this equality. If all available solid dissolves, the final chemical solution may instead be undersaturated; do not impose equality merely because a salt was initially added.^[ref]

Using the thermodynamic activity definition $a_i={\gamma }_i{c}_i/c^{\circ }$, with activity coefficients ${\gamma }_i$ and reference molar concentration $c^{\circ }$, gives the independent conversionThus $K_{\mathrm{sp}}$ is dimensionless, while the conditional concentration form $K_{\mathrm{sp},c}$ has units ${\mathrm{M}}^2$ for this one-to-one model. Treating $K_{\mathrm{sp},c}$ as constant requires a justified fixed or approximately fixed activity coefficient product. Changing ionic strength can change that product even at fixed temperature; changing temperature can change the equilibrium constant itself. Specify the solid thermodynamic phase too: different crystal polymorphs need not have the same solubility product.