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,

$K _{ sp }=\left[ Cr ^{3+}\right]\left[ OH ^{-}\right]^{3}= s \times(3 s )^{3}=27 s ^{4}$

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

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

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