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 bys and the solubility product by $K_{sp}$ then which of the following relationship betweens and $K_{sp}$ is correct?

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$s = K_{sp} / (6912)^{1 / 7}$

$s = {(K_{sp} / 6912)}^{7}$

$s = {(K_{sp} / 144)}^{1 / 7}$

$s = {[K_{sp} / 6912]}^{1 / 7}$

answer is C.

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Detailed Solution

$\begin{aligned} {Zr}_{3} {({PO}_{4})}_{4} ⇌ 3 {Zr}^{4 +} + 4 {PO}_{4}^{3 -} \\ (3 s) (4 s) \\ K_{sp} = {[{Zr}^{4 +}]}^{3} {[{PO}_{4}^{3 -}]}^{4} = (3 s)^{3} (4 s)^{4} = 6912 s^{7} \\ Hence, s = {(K_{sp} / 6912)}^{1 / 7} \end{aligned}$

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