Problem: Enter the solubility-product expression for Al(OH)3(s).

Recall that K_sp is an equilibrium constant related to slightly soluble ionic compounds. 

The dissociation of slightly soluble Al(OH)₃ in solution is as follows:

Al(OH)₃(s) ⇌  Al³⁺(aq) + 3 OH^–(aq)

We can construct an ICE table for this dissociation. 

where x = molar solubility of the compound.

The K_sp expression is:

$\boxed{{\mathbf{K}}_{\mathbf{sp}}\mathbf{=}\frac{\mathbf{products}}{\cancel{\mathbf{reactants}}}\mathbf{=}\mathbf{\left[}{\mathbf{Al}}^{\mathbf{3}\mathbf{+}}\mathbf{\right]}\mathbf{\left[}{\mathbf{OH}}^{\mathbf{-}}\mathbf{\right]}}$

Remember that solids are ignored in the ICE table and K_sp expression.