Question

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

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

Answer 1

The correct option is A

${10}^{-3}M$

$pH=4\Rightarrow \left[H^{+}\right]={10}^{-4}M\Rightarrow \left[O{H}^{-}\right]={10}^{-10}M$

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

$K_{sp}\left(Al\right(OH{)}_3)=[A{l}^{3+}\left]\right[O{H}^{-}{]}^3$

$\left[A{l}^{3+}\right]{\left[O{H}^{-}\right]}^3=1\times {10}^{-33}$

$\left[A{l}^{3+}\right]({10}^{-10}{)}^3=1\times {10}^{-33}\Rightarrow \left[A{l}^{3+}\right]={10}^{-3}M.$