Question

The solubility product of $Cr(OH{)}_3$ at $298\text{K}$ is $6\times {10}^{-31}$ . The concentration of hydroxide ions in a saturated solution of $Cr(OH{)}_3$ will be:

• A. $(18\times {10}^{-31}{)}^{1/4}$
• B. $(18\times {10}^{-31}{)}^{1/2}$
• C. $(2.22\times {10}^{-31}{)}^{1/4}$
• D. $(4.86\times {10}^{-29}{)}^{1/4}$

Answer 1

The correct option is A $(18\times {10}^{-31}{)}^{1/4}$

$\begin{array}{ccccc}Cr(OH{)}_{3\left(s\right)}& \to & C{r}_{(aq.)}^{3+}& +& 3O{H}_{(aq.)}^{-}\\ 1-S& & S& & 3S\end{array}$
$K_{sp}=27S^4$
$6\times {10}^{-31}=27S^4$
$S={[\frac{6}{27}\times {10}^{-31}]}^{1/4}$
$\left[O{H}^{-}\right]=3S=3\times {[\frac{6}{27}\times {10}^{-31}]}^{1/4}=(18\times {10}^{-31}{)}^{1/4}\text{M}$