Zirconium phosphate [Zr_3 (PO_ 4)_4] dissociates into three zirco… - Vidyadip

MCQ

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
$S = {\{ {K_{sp}}/144\} ^{1/7}}$
✓
$S = {\{ {K_{sp}}/6912\} ^{1/7}}$
D
$S = {\{ {K_{sp}}/6912\} ^7}$

✓

Answer

Correct option: C.

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

c
$[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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