Question

The $K_{sp}$ for $Cr(OH)_3$ is $1.6 × 10^{-30}$. The molar solubiity of this compound in water is :

• A. $4\sqrt{1.6\times10^{-30}}$
• B. $4\sqrt{1.6\times10^{-30}/ 27}$
• C. $1.6\times10^{-30}/ 27$
• D. $2\sqrt{1.6\times10^{-30}}$

Answer 1

Answer: (B)

$Cr\left(OH\right)_{3}\left(s\right) {\rightleftharpoons} Cr^{3+}\left(aq.\right) + 3OH^{-}\left(aq.\right)$
$S\quad\quad3S
27\, S^{4 }= k_{sp}$
$S = \left(\frac{K_{SP}}{27}\right)^{1/ 4} = \left(\frac{1.6\times10^{-30}}{27}\right)^{1/ 4}$