= Ion product
{title2=$Q_{\mathrm{sp}}$}
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.<Chemistry 2e by Paul Flowers and others/15.1 Precipitation and Dissolution>[Chemistry 2e, §15.1, Predicting Precipitation: comparison of reaction quotient and solubility product]{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 $\sqrt{K_{\mathrm{sp},c}}$.
Back to article page
