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6-2.Equilibrium-II (Ionic Equilibrium)
hard
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
$S = {\{ {K_{sp}}/6912\} ^7}$
(JEE MAIN-2014)
Solution
$[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}$
$\therefore \,S = {\left( {\frac{{{K_{sp}}}}{{6912}}} \right)^{1/2}}$
Standard 11
Chemistry
