Zirconium phosphate \(\mathrm{[Zr_3(PO_4)_4]}\) 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 \(\mathrm{K_{sp}}\) then which of the following relationship between S and \(\mathrm{K_{sp}}\) is correct ?
Show Hint
- \(S=\{K_{sp}/(6912)^{1/7}\}\)
- \(S=\{K_{sp}/144\}^{1/7}\)
- \(S=(K_{sp}/6912)^{1/7}\)
- \(S=\{K_{sp}/6912\}^{7}\)
The Correct Option is C
Solution and Explanation
If molar solubility is S, \([\mathrm{Zr^{4+}}]=3S\) and \([\mathrm{PO_4^{3-}}]=4S\).
\(K_{sp}=(3S)^3(4S)^4=27S^3\times 256S^4=6912\,S^7\).
So \(S^7=K_{sp}/6912\), i.e. \(S=(K_{sp}/6912)^{1/7}\).
Answer: (3) \(S=(K_{sp}/6912)^{1/7}\)
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