= Saturation criterion for precipitation
{title2=$Q_{\mathrm{sp}}\gtrless K_{\mathrm{sp}}$}
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 (chemistry)>; equality marks saturation. These comparisons identify a driving direction favored by <thermodynamics>, not a <reaction rate>.<Chemistry 2e by Paul Flowers and others/15.1 Precipitation and Dissolution>[Chemistry 2e, §15.1, Predicting Precipitation: unsaturated, saturated and supersaturated cases]{ref}
A proof outline comes from the reaction <Gibbs free energy> for dissolution:
$$
\Delta_rG=RT\ln\!\left(\frac{Q_{\mathrm{sp}}}{K_{\mathrm{sp}}}\right).
$$
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>.
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