Question
Zirconium phosphate $$\left[ {Z{r_3}{{\left( {P{O_4}} \right)}_4}} \right]$$ 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.
$$S = \left\{ {\frac{{{K_{sp}}}}{{{{\left( {6912} \right)}^{\frac{1}{7}}}}}} \right\}$$
B.
$$S = {\left\{ {\frac{{{K_{sp}}}}{{144}}} \right\}^{\frac{1}{7}}}$$
C.
$$S = {\left\{ {\frac{{{K_{sp}}}}{{6912}}} \right\}^{\frac{1}{7}}}$$
D.
$$S = {\left\{ {\frac{{{K_{sp}}}}{{6912}}} \right\}^7}$$
Answer :
$$S = {\left\{ {\frac{{{K_{sp}}}}{{6912}}} \right\}^{\frac{1}{7}}}$$
Solution :
$$\eqalign{
& \left[ {Z{r_3}{{\left( {P{O_4}} \right)}_4}} \right] \rightleftharpoons 3\mathop {Zr_{}^{4 + }}\limits_{3S} + 4P\mathop O\limits_{4S} \,_4^{3 - } \cr
& {K_{sp}} = {\left( {3S} \right)^3}{\left( {4S} \right)^4} \cr
& = 27{S^3} \times 256\,{S^4} \cr
& = 6912\,{S^7}. \cr
& \therefore \,\,S = {\left( {\frac{{{K_{sp}}}}{{6912}}} \right)^{\frac{1}{7}}} \cr} $$