Q. $K_{sp}$ for $\text{Cr(OH)}_3$ is $1.6\times10^{-30}$. What is the molar solubility of this salt in water?
A) $\frac{1.8\times10^{-30}}{27}$
B) $\sqrt[5]{1.8\times10^{-30}}$
C) $\sqrt[4]{\frac{1.6\times10^{-30}}{27}}$
D) $\sqrt[2]{1.6\times10^{-30}}$
Explanation
[Solution by Dr. Abubacker] $\text{Cr(OH)}_3\rightleftharpoons\text{Cr}^{3+}+3\text{OH}^-$. $K_{sp}=s\,(3s)^3=27s^4$, so $s=\left(\frac{K_{sp}}{27}\right)^{1/4}=\sqrt[4]{\frac{1.6\times10^{-30}}{27}}$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.