Zirconium phosphate [Zr_3(PO_4)_4] dissociate | vedclass

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(C) The dissociation of zirconium phosphate is given by: $Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq)$ If $S$ is the molar s

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

Vedclass · Answer

(C) The dissociation of zirconium phosphate is given by: $Zr_3(PO_4)_4 (s) \rightleftharpoons 3 Zr^{4+} (aq) + 4 PO_4^{3-} (aq)$ If $S$ is the molar solubility,then $[Zr^{4+}] = 3S$ and $[PO_4^{3-}] = 4S$. The solubility product expression is: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$ $K_{sp} = (3S)^3 (4S)^4$ $K_{sp} = (27 S^3) (256 S^4)$ $K_{sp} = 6912 S^7$ Therefore,$S^7 = K_{sp} / 6912$ $S = (K_{sp} / 6912)^{1/7}$

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