Zirconium phosphate $\left[\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}ight)_{4}ight]$ dissociates into three…

Zirconium phosphate $\left[\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}ight)_{4}ight]$ dissociates into three zirconium cations of charge $+4$ and four phosphate anions of charge $-3$. If molar solubility of zirconium phosphate is denoted by S and its solubility product by $\mathrm{K}_{\mathrm{sp}}$ then which of the following relationship between $\mathrm{S}$ and $\mathrm{K}_{\mathrm{sp}}$ is correct?
  1. $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} /(6912)^{1 / 7}ight\}$
  2. $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} / 144ight\}^{1 / 7}$
  3. $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} / 6912ight\}^{1 / 7}$
  4. $\mathrm{S}=\left\{\mathrm{K}_{\mathrm{sp}} / 6912ight\}^{7}$

Solution

$\begin{aligned}\left[\mathrm{Zr}_{3}\left(\mathrm{PO}_{4}ight)_{4}ight] ightleftharpoons 3 \mathrm{Zr}^{4+}+4 \mathrm{PO}_{4}^{3-} \\ K_{s p} &=(3 S)^{3}(4 S)^{4} \\ &=27 S^{3} \times 256 S^{4} \\ &=6912 S^{7} \\ \therefore \quad S=&\left(\frac{\mathrm{K}_{\mathrm{sp}}}{6912}ight)^{1 / 7} \end{aligned}$ *

Asked in: JEE-TOPICTESTS-CHEMISTRY

Practice more EQUILIBRIUM questions on Aicharya