Chapter 15, Problem 113MP

Aluminum ions react with the hydroxide ion to form the precipitate Al(OH)₃(s), but can also react to form the soluble complex ion Al(OH)₄⁻. In terms of solubility, Al(OH)₃(s) will be more soluble in very acidic solutions as well as more soluble in very basic solutions.

a. Write equations for the reactions that occur to increase the solubility of Al(OH)₃(s) in very acidic solutions and in very basic solutions.

b. Let’s study the pH dependence of the solubility of Al(OH)₃(s) in more detail. Show that the solubility of Al(OH)₃, as a function of [H⁺], obeys the equation

$S={[{\text{H}}^{\text{+}}]}^3{K}_{\text{sp}}/K_{\text{w}}{}^3+K{K}_{\text{w}}/[H^{+}]$

where S = solubility = [Al³⁺] + [Al(OH)₄⁻] and K is the equilibrium constant for

${\text{Al(OH)}}_{\text{3}}(s)+O{H}^{-}(aq)\rightleftharpoons Al{(OH)}_4{}^{-}(aq)$

c. The value of K is 40.0 and K_sp for Al(OH)₃ is 2 × 10⁻³². Plot the solubility of Al(OH)₃ in the pH range 4–12.