Question

The${\mathit{K}}_{\mathbf{s}\mathbf{p}}$ of$\mathit{A}\mathit{l}{\left(OH\right)}_{\mathbf{3}}$ is$\mathbf{2}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{32}}$ . At what pH will a $\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{}\mathit{M}\mathbf{}\mathit{A}{\mathit{l}}^{\mathbf{3}\mathbf{+}}$solution begin to show precipitation of$\mathit{A}\mathit{l}{\left(OH\right)}_{\mathbf{3}}$ ?

Short answer

At a pH of3.19, the solution of 0.2 M $A{l}^{3+}$begins to show precipitation of$Al{\left(OH\right)}_3$ .

Step-by-step solution

The concentration of hydrogen ions

Given,

$$\begin{gathered} K_{sp}\left(Al(OH{)}_3\right)=2\times {10}^{-32} \\ \left[A{l}^{3+}\right]=0.2M \end{gathered}$$

The solubility equation is denoted asfollows:

$Al(OH{)}_3=A{l}^{3+}+3O{H}^{-}$

On substituting the values,

$$\begin{gathered} K_{sp}=\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^3 \\ 2\times {10}^{-32}=0.2M\times {\left[O{H}^{-}\right]}^3 \end{gathered}$$

Therefore,

$\left[O{H}^{-}\right]=1.55\times {10}^{-11}$

The pH at which the solution of 0.2 M Al3+begins to show precipitation

The pOH is calculated asfollows:

$$\begin{gathered} pOH=-\log \left[O{H}^{-}\right] \\ pOH=-\log (1.55\times {10}^{-11}) \\ pOH=10.81 \end{gathered}$$

Also,

$pH+pOH=14$

Therefore,

$$\begin{gathered} pH=14-10.81 \\ pH=3.19 \end{gathered}$$