Q. $K_{sp}\ for\ Cr(OH)_{3}\ is\ 2.7 \times 10^{- 31}$ . What is its solubility in mol/L?
A) $1 \times 10^{- 8}$
B) $8 \times 10^{- 8}$
C) $1.1 \times 10^{- 8}$
D) $0.18 \times 10^{- 8}$
Explanation
(a)
$Cr(OH)_{3} \rightleftharpoons {Cr}^{3 +} + 3{OH}^{-}$
$(s)$ ($3s)$
$K_{sp} = s \times (3s)^{3}$
$= 27s^{4}$
$2.7 \times 10^{- 31} = 27s^{4}$
$\therefore$ $s = \sqrt[4]{\frac{K_{sp}}{27}} = \sqrt[4]{\frac{2.7 \times 10^{- 31}}{27}}$
$= \sqrt[4]{10^{- 32}}$
$\ = 10^{- 8}$ mol/L
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.