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Question: Zirconium phosphate \([Z{r_3}{(P{O_4})_4}]\) dissociates into three zirconium cations of charge +4 a...

Zirconium phosphate [Zr3(PO4)4][Z{r_3}{(P{O_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 Ksp{K_{sp}} then which of the following relationship between S and Ksp{K_{sp}} is correct?
A. S=Ksp/(6912)1/7S = \\{ {K_{sp}}/{(6912)^{1/7}}\\}
B. S=Ksp/1441/7S = {\\{ {K_{sp}}/144\\} ^{1/7}}
C. S=Ksp/69121/7S = {\\{ {K_{sp}}/6912\\} ^{1/7}}
D. S=Ksp/69127S = {\\{ {K_{sp}}/6912\\} ^7}

Explanation

Solution

The solubility of a compound in the solvent is defined as the total amount of solute which can form a saturated solution on dissolving at equilibrium. The solubility product constant is defined as an equilibrium constant which shows the equilibrium condition between the solute and its constituents ions formed after dissociation.

Complete step by step answer:
When zirconium phosphate is dissolved in the solvent, it dissociates into its constituent ions.
The dissociation reaction for zirconium phosphate is shown below.
[Zr3(PO4)4]→3Zr+4+4(PO4)−3[Z{r_3}{(P{O_4})_4}] \to 3Z{r^{ + 4}} + 4{(P{O_4})^{ - 3}}
In this reaction, one mole of zirconium phosphate dissociates into 3 zirconium ion Zr+4Z{r^{ + 4}} and 4 phosphate ion PO4−3PO_4^{ - 3}.
Given, the molar solubility of zirconium phosphate is denoted by S and its solubility product by Ksp{K_{sp}}.
Then, [Zr3(PO4)4][Z{r_3}{(P{O_4})_4}] is denoted as S, Zr+4Z{r^{ + 4}} is denoted as 3S and PO4−3PO_4^{ - 3} is denoted as 4S.
The solubility product constant (Ksp{K_{sp}}) tells that when the compound is in equilibrium with the saturated solution the product of the concentration of both the ions formed after the dissociation of the compound is equal to the solubility product constant.

The Ksp{K_{sp}} is given as
Ksp=(SZr+4)3×(SPO4−3)4{K_{sp}} = {(SZ{r^{ + 4}})^3} \times {(SPO_4^{ - 3})^4}
⇒Ksp=(3s)3×(4s)4\Rightarrow {K_{sp}} = {(3s)^3} \times {(4s)^4}
⇒Ksp=27S3×256S4\Rightarrow {K_{sp}} = 27{S^3} \times 256{S^4}
⇒Ksp=6912S7\Rightarrow {K_{sp}} = 6912{S^7}
⇒S7=Ksp6912\Rightarrow {S^7} = \dfrac{{{K_{sp}}}}{{6912}}
⇒S=(Ksp6912)17\Rightarrow S = {\left( {\dfrac{{{K_{sp}}}}{{6912}}} \right)^{\dfrac{1}{7}}}
Thus, the relation between solubility S and solubility product constant Ksp{K_{sp}} is given as S=(Ksp6912)17S = {\left( {\dfrac{{{K_{sp}}}}{{6912}}} \right)^{\dfrac{1}{7}}}.

Therefore, the correct option is C .

Note: The value of the solubility product constant is dependent on the temperature. As the solubility of solute in the solvent increases, the temperature also increases with the increase in the value of solubility product constant.