Question

What is the solubility of $Al(OH{)}_3$, $(K_{sp}=1\times {10}^{-33})$ in a buffer solution having 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, the pH of buffer solution is 4.
So, $\left[H^{+}\right]={10}^{-4}M$
$\Rightarrow \left[O{H}^{-}\right]={10}^{-10}M$
Again, $Al(OH{)}_3\rightleftharpoons A{l}^{3+}+3O{H}^{-}$
So, $K_{sp}=\left[A{l}^{3+}\right][O{H}^{-}{]}^3$
$\Rightarrow 1\times {10}^{-33}=\left[A{l}^{3+}\right][O{H}^{-}{]}^3$
$\Rightarrow 1\times {10}^{-33}=\left[A{l}^{3+}\right]({10}^{-10}{)}^3$
$\Rightarrow \left[A{l}^{3+}\right]={10}^{-3}M$
So the solubility of $Al(OH{)}_3$ in a buffer solution of pH=4 is ${10}^{-3}M$.