Saturation criterion for precipitation ($Q_{\mathrm{sp}}\gtrless K_{\mathrm{sp}}$)

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.