Q.$K_{sp}$ of $Al(OH)_3$ is $1 \times 10^{-15}$. At what pH does $1.0 \times 10^{-3}$ M $Al^{3+}$ precipitate on the addition of a buffer of $NH_4Cl$ and $NH_4OH$ solution?

For a sparingly soluble salt $X_mY_n(s) \rightleftharpoons mX^{n+}(aq)+nY^{m-}(aq)$, the solubility product is $K_{sp}=[X^{n+}]^m[Y^{m-}]^n$ -- the product of the constituent ions' molar concentrations, each raised to its stoichiometric coefficient, once the solid's own (constant) concentration is absorbed out of the equilibrium expression.

Predicting precipitation. Computing the same expression with the actual ion concentrations present (whether or not the solution is at equilibrium) gives the ionic product; comparing it to $K_{sp}$ predicts the outcome: ionic product $>K_{sp}$ means the solution is supersaturated and precipitation occurs; ionic product $<K_{sp}$ means the solution is unsaturated (no precipitate); ionic product $=K_{sp}$ means the solution is exactly saturated, at equilibrium.

Relating Ksp to molar solubility. If $s$ is the molar solubility of $X_mY_n$, then $[X^{n+}]=ms$ and $[Y^{m-}]=ns$, so $K_{sp}=(ms)^m(ns)^n=m^m n^n s^{m+n}$ -- e.g. for a 1:1 salt like $BaSO_4$, $K_{sp}=s^2$; for a 2:1 salt like $Ag_2CrO_4$, $K_{sp}=4s^3$. …