Q. The $K_{sp}\ for\ Cr(OH)_{3}\ is\ 1.6\ \ 10^{30}.$ The molar solubility of this compound in water is
A) $\sqrt[2]{1.6\ \ 10^{30}}$
B) $\sqrt[4]{1.6\ \ \ 10^{30}}$
C) $\sqrt[4]{1.6\ \ \frac{10^{30}}{27}}$
D) $\frac{1.6\ \ \ 10^{30}}{27}$
Explanation
(c)
Let molar solubility of $Cr(OH)_{3} = s\ mol\ L^{- 1}$
$Cr(OH)_{3}(s) \rightleftharpoons Cr^{3 +}(aq) + 3OH^{-}(aq)$
$s$ $3s$
$K_{sp} = 1.6 \times 10^{- 30} = \left\lbrack Cr^{3 +} \right\rbrack\left\lbrack OH^{-} \right\rbrack^{3}$
$= (s)(3s)^{3} = 27s^{4}$
$\therefore\ \ \ \ \ \ \ \ s^{4} = \frac{1.6 \times 10^{- 30}}{27}$
$\therefore\ \ \ \ \ \ \ \ \ s = \sqrt[4]{\frac{1.6 \times 10^{- 30}}{27}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.