The molar solubility $(s)$ of zirconium phosphate with molecular formula $\left(\mathrm{Zr}^{4+}\rig

Question

The molar solubility $(s)$ of zirconium phosphate with molecular formula $\left(\mathrm{Zr}^{4+}ight)_{3}\left(\mathrm{PO}_{4}^{3-}ight)_{4}$ is given by relation\

Professor PC Thomas Classes · Answer

(d) :\ $\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}ight)_{4} ightarrow \underset{3 s}{3 \mathrm{Zr}^{4+}}+\underset{4 s}{4 \mathrm{PO}_{4}^{3-}}$\ $K_{s p}=(3 s)^{3} 	imes(4 s)^{4}=6912 s^{7}$\ $	herefore s=\left(\frac{K_{s p}}{6912}ight)^{\frac{1}{7}}$\