The molar solubility (s) of zirconium phospha | vedclass

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(A) The dissociation of zirconium phosphate is represented as: $Zr_3(PO_4)_4(s) \rightleftharpoons 3Zr^{4+}(aq) + 4PO_4^{3-}(aq)$Let the molar solubil

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$(\frac{K_{sp}}{6912})^{\frac{1}{7}}$

Vedclass · Answer

(A) The dissociation of zirconium phosphate is represented as: $Zr_3(PO_4)_4(s) ightleftharpoons 3Zr^{4+}(aq) + 4PO_4^{3-}(aq)$Let the molar solubility be $s$. Then,the concentration of $Zr^{4+}$ is $3s$ and the concentration of $PO_4^{3-}$ 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) 	imes (256s^4) = 6912s^7$Solving for $s$: $s^7 = \frac{K_{sp}}{6912}$Therefore,$s = (\frac{K_{sp}}{6912})^{\frac{1}{7}}$

The molar solubility $(s)$ of zirconium phosphate with molecular formula $(Zr^{4+})_3(PO_4^{3-})_4$ is given by the relation:

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