Question

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

• A. $\left(18\times10^{-31}\right)^{1/ 2}$
• B. $\left(2.22\times10^{-31}\right)^{1/ 4}$
• C. $\left(18\times10^{-31}\right)^{1/ 4}$
• D. $\left(4.86\times10^{-29}\right)^{1/ 4}$

Answer 1

Answer: (C)

$K_{sp}=27\left(s\right)^{4}=6\times10^{-31}$
$\Rightarrow \left[3\left(s\right)\right]^{4}=18\times10^{-31}$
$\left[OH-\right]=3\left(s\right)=\left[18\times10^{-31}\right]^{1 /4}$