Chapter 16, Problem 121MP

Aluminium ions react with the hydroxide ion to form the precipitate ${\text{Al(OH)}}_{\text{3}}\text{(s)}$, but can also react to form the soluble complex ion ${\text{Al(OH)}}_{\text{4}}{}^{-}$. In terms of solubility, ${\text{Al(OH)}}_{\text{3}}\text{(s)}$will be more soluble in very acidic solutions as well as more soluble in very basic solutions.

1. a. Write equations for the reactions that occur to increase the solubility of ${\text{Al(OH)}}_{\text{3}}\text{(s)}$in very acidic solutions and in very basic solutions.
2. b. Let’s study the pH dependence of the solubility of ${\text{Al(OH)}}_{\text{3}}\text{(s)}$in more detail.  Show that the solubility of ${\text{Al(OH)}}_{\text{3}}$, as a function of ${\text{[H}}^{+}]$, obeys the equation

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

  Where $\text{S}={\text{ solubility = [Al}}^{3+}]+[{\text{Al(OH)}}_{\text{4}}^{\text{-}}]$and $K$ is the equilibrium constant for

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

1. c. The value of $K$ is $40.0$ and $K_{sp}$ for ${\text{Al(OH)}}_{\text{3}}$ is $2\times {10}^{-32}$.  Plot the solubility of ${\text{Al(OH)}}_{\text{3}}$ in the pH range $4-12$.