Question

Determine the solubility of $Cr{\left(OH\right)}_3$ in $mol{L}^{-1}$ , if its $K_{sp}$ is $2.7\times 10^{-31}{M}^4$ .

• A. $1\times 10^{-8}$
• B. $8\times 10^{-8}$
• C. $1.1\times 10^{-8}$
• D. $0.18\times 10^{-8}$

Answer 1

Answer: (A)

The given compound will dissociate as



Writing the expression for the solubility product,

$K_{sp}=\left[C{r}^{3+}\right]{\left[O{H}^{-}\right]}^3=s\times (3s{)}^3=27s^4$

$s=\sqrt[4]{\frac{K_{sp}}{27}}$

$s=\sqrt[4]{\frac{2.7\times 10^{-31}}{27}}$

$s=1.0\times 10^{-8}mol{L}^{-1}$