(C) The dissociation reaction is: $[Zr_3(PO_4)_4] \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$ If $S$ is the molar solubility,the concentration of $Zr^{4

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(C) The dissociation reaction is: $[Zr_3(PO_4)_4] \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$ If $S$ is the molar solubility,the concentration of $Zr^{4+}$ is $3S$ and $PO_4^{3-}$ is $4S$. The solubility product expression is: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$ Substituting the values: $K_{sp} = (3S)^3 (4S)^4$ $K_{sp} = (27S^3) \times (256S^4)$ $K_{sp} = 6912S^7$ Therefore,$S = (K_{sp} / 6912)^{1/7}$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Difficult

Zirconium phosphate $[Zr_3(PO_4)_4]$ 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 $K_{sp}$,then which of the following relationships between $S$ and $K_{sp}$ is correct?

JEE MAIN 2014 All JEE MAIN

A
$S = \{K_{sp} / (6912)^{1/7}\}$
B
$S = \{K_{sp} / 144\}^{1/7}$
C
$S = \{K_{sp} / 6912\}^{1/7}$
D
$S = \{K_{sp} / 6912\}^{7}$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

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MediumJEE MAIN 2025

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