Zirconium phosphate, , dissociates into three zirconium cations and four phosphate anions. If the mo — Ionic Equilibrium Chemistry Question

💡 Solution & Explanation

The dissociation is $Zr_3(PO_4)_4 \rightleftharpoons 3Zr^{4+} + 4PO_4^{3-}$. If the solubility is $S$, then $[Zr^{4+}] = 3S$ and $[PO_4^{3-}] = 4S$. The solubility product is $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4 = (3S)^3 (4S)^4 = 27S^3 \times 256S^4 = 6912S^7$. Rearranging for solubility gives $S^7 = K_{sp} / 6912$, which evaluates to $S = (K_{sp} / 6912)^{1/7}$.