Question

What is the solubility of $Al(OH{)}_3$
$(K_{sp}=1\times {10}^{-33})$ in a buffer solution of $pH=4$?

• A. ${10}^{-3}M$
• B. ${10}^{-6}M$
• C. ${10}^{-4}M$
• D. ${10}^{-10}M$

Answer 1

The correct option is A ${10}^{-3}M$

Given, $pH=4$
$\therefore pOH=14-pH=10\Rightarrow -log\left[O{H}^{-}\right]=10\Rightarrow \left[O{H}^{-}\right]={10}^{-10}$

Again,
$\begin{array}{cccc}Al(OH{)}_3& \leftrightharpoons & A{l}^{3+}+& 3O{H}^{-}\\ & & S& {10}^{-10}\end{array}$

$\therefore K_{sp}=S\times ({10}^{-10}{)}^3\Rightarrow {10}^{-33}=S\times {10}^{-30}\Rightarrow [A{l}^{3+}]=S={10}^{-3}M$