A sparingly soluble ionic solid can establish a heterogeneous equilibrium with the ions it releases into solution. The equilibrium constant for this dissolution process is called the solubility product, $K_{sp}$.

For a general solid

$$\mathrm{M_pX_q(s)\rightleftharpoons p,M^{m+}(aq)+q,X^{n-}(aq)},$$

the concentration-based solubility-product expression is

$$K_{sp}=[\mathrm{M^{m+}}]^p[\mathrm{X^{n-}}]^q.$$

The pure solid does not appear in the expression because its contribution is constant while that phase is present.

For example,

$$\mathrm{CaF_2(s)\rightleftharpoons Ca^{2+}(aq)+2F^-(aq)}$$

gives

$$K_{sp}=[\mathrm{Ca^{2+}}][\mathrm{F^-}]^2.$$

The exponents come from the balanced dissolution equation. They are not ionic charges: the exponent 2 on $[\mathrm{F^-}]$ appears because two fluoride ions are produced per formula unit of $\mathrm{CaF_2}$.

A small $K_{sp}$ means that the equilibrium ion concentrations must satisfy a small product, but it should not be read as a direct numerical measure of molar solubility across salts with different dissolution stoichiometries. For example, a 1:1 salt and a 1:2 salt relate their ion concentrations to dissolved formula units differently.

$K_{sp}$ is an equilibrium constant. At a fixed temperature it characterizes the dissolution equilibrium, not how rapidly the solid dissolves. Grinding a solid may make equilibrium be reached faster by increasing surface area, but it does not by itself change $K_{sp}$.

The next step is to combine the $K_{sp}$ expression with dissolution stoichiometry to calculate how much solid can dissolve.