Question

What is the solubility of $Al\left(O{H}_3\right),K_{sp}=1\times {10}^{-33}$, in a solution having $pH=4$?

• A. $6\times {10}^{-3}$M
• B. ${10}^{-6}$M
• C. $1.5\times {10}^{-4}$M
• D. $2.47\times {10}^{-9}$M

Answer 1

Answer: (D)

$2.47\times {10}^{-9}$M

The solubility equilibrium of $Al(OH{)}_3$ is

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

Therefore $x=1,y=2$ and $K_{sp}$=$\left[A{l}^{+3}\right]\times \left[O{H}^{-}\right]$ = $27S^4$

$S$ = $(\frac{{10}^{-33}}{27}{)}^{1/4}$

Therefore $S$= $2.47\times {10}^{-9}M$ in $mold{m}^{-3}$