Solubility Equilibria

How Insoluble Is Insoluble?

No salt is completely insoluble. A sparingly soluble solid in contact with its saturated solution is an equilibrium, described by the solubility product. For $\mathrm{A_xB_y(s) \rightleftharpoons xA^{y+} + yB^{x-}}$:

$$K_{sp} = [\mathrm{A^{y+}}]^x[\mathrm{B^{x-}}]^y$$

The solid itself does not appear, being a pure phase. Converting $K_{sp}$ into molar solubility $s$ requires care with the stoichiometry:

The factor of 4 arises because two ions form per formula unit, so that ion's concentration is $2s$ and it is squared. Comparing $K_{sp}$ values across different salt types is therefore meaningless, silver chromate has a smaller $K_{sp}$ than silver chloride yet is more soluble.

The common ion effect follows from Le Chatelier: adding an ion the solid already contains pushes the equilibrium back toward the solid, so solubility falls. Silver chloride is markedly less soluble in sodium chloride solution than in pure water.

Whether a precipitate forms is decided by comparing the ion product $Q$ with $K_{sp}$: if $Q > K_{sp}$ the solution is supersaturated and solid appears; if $Q < K_{sp}$ any solid present dissolves. Selective precipitation exploits the gap between two salts' $K_{sp}$ values, adding a reagent slowly so the less soluble one drops out first.

Common pitfall: comparing $K_{sp}$ values of salts with different formulas. $K_{sp}$ has different units and different exponents for AB and $\mathrm{AB_2}$ salts, so only solubilities computed from them are comparable, a smaller $K_{sp}$ does not reliably mean a less soluble salt.