Q. The molar solubility(s) of zirconium phosphate with molecular formula $(\mathrm{Zr^{4+}})_3(\mathrm{PO_4^{3-}})_4$ is given by relation :
A) $\left(\frac{K_{sp}}{9612}\right)^{1/3}$
B) $\left(\frac{K_{sp}}{6912}\right)^{1/7}$
C) $\left(\frac{K_{sp}}{8435}\right)^{1/7}$
D) $\left(\frac{K_{sp}}{5348}\right)^{1/6}$
Explanation
[Solution by Dr. Abubacker] $\mathrm{Zr_3(PO_4)_4}\rightleftharpoons3\mathrm{Zr^{4+}}+4\mathrm{PO_4^{3-}}$, so $K_{sp}=(3s)^3(4s)^4=27\cdot256\,s^7=6912\,s^7$ and $s=\left(\frac{K_{sp}}{6912}\right)^{1/7}$. Answer (2).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.