JEE Mains · Chemistry · STD 11 - 6.2. Equilibrium - II (icon Equilibrium)
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 by \(S\) and its solubility product by \(K_{sp}\) then which of the following relationship between \(S\) and \(K_{sp}\) is correct?
- A \(S = \{ {K_{sp}}/{\left( {6912} \right)^{1/7}}\}\)
- B \(S = {\{ {K_{sp}}/144\} ^{1/7}}\)
- C \(S = {\{ {K_{sp}}/6912\} ^{1/7}}\)
- D \(S = {\{ {K_{sp}}/6912\} ^7}\)
Answer & Solution
Correct Answer
(C) \(S = {\{ {K_{sp}}/6912\} ^{1/7}}\)
Step-by-step Solution
Detailed explanation
\([Z{r_3}{(P{O_4})_4}] \rightleftharpoons \mathop {3Z{r^{4 + }}}\limits_{3S} + \mathop {4P{O_4}^{3 - }}\limits_{4S} \) \({K_{sp}} = {(3S)^3}{(4S)^4}\) \( = 27{S^3} \times 256{S^4}\) \( = 6912{S^7}\) \(\therefore \,S = {\left( {\frac{{{K_{sp}}}}{{6912}}} \right)^{1/2}}\)
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