The solubility product of $\mathrm{Cr}(\mathrm{OH}{)}_3$ at $298 \mathrm{K}$ is $6.0\times {10}^{-31}$. The concentration of hydroxide ions in a saturated solution of $\mathrm{Cr}(\mathrm{OH}{)}_3$ will be

 The solubility product of $\mathrm{Cr}(\mathrm{OH}{)}_3$ at $298 \mathrm{K}$ is $6.0\times {10}^{-31}$.

$\mathrm{Cr}(\mathrm{OH}{)}_3\rightleftharpoons {\mathrm{Cr}}^{+3}+3{\mathrm{OH}}^{-}$

s                       o         o

•                s         3s

${\mathrm{K}}_{\mathrm{sp}}=\left(\mathrm{S}\right)(3\mathrm{S}{)}^3$

$6\times {10}^{-31}=27{ \mathrm{S}}^4$

${ \mathrm{S}}^4=\frac{6\times {10}^{-31}}{27}$

$\mathrm{S}={\left(\frac{6\times {10}^{-31}}{27}\right)}^{1/4}$

$\left[{\mathrm{OH}}^{-}\right]=3 \mathrm{S}=3\times {\left(\frac{6\times {10}^{-31}}{27}\right)}^{1/4}$

$=(81{)}^{1/4}\times {\left(\frac{6\times {10}^{-31}}{27}\right)}^{1/4}$

$\left[{\mathrm{OH}}^{-}\right]={\left(\frac{81\times 6}{27}\times {10}^{-31}\right)}^{1/4}={\left(18\times {10}^{-31}\right)}^{1/4}$

Hence the correct option is (D) ${\left(18\times {10}^{-31}\right)}^{1/4}$