Question

Zirconium phosphate [$Z{r_3}{(P{O_4})_4}$] dissociates into three zirconium cations of charge (+4) and 4 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) $S = \left\{ {{K_{sp}}/{{(6912)}^{\dfrac{1}{7}}}} \right\}$
(B) $S = {\left\{ {{K_{sp}}/144} \right\}^{\dfrac{1}{7}}}$
(C) $S = {\left\{ {{K_{sp}}/6912} \right\}^{\dfrac{1}{7}}}$
(D) $S = {\left\{ {{K_{sp}}/6912} \right\}^7}$

Answer 1

Hint: First write the dissociation reaction and find out the molar solubilities of zirconium ions and phosphate ions. is a mathematical product of the dissolved ion concentrations that are raised to the power of their stoichiometric coefficients. Now write the solubility product for this reaction using the molar solubilities and solve it further.

Complete step by step solution:
-First let us see what solubility is.
The property of a solute to get dissolved in a solvent to form a solution is known as solubility. The solubility product (${K_{sp}}$) is a type of equilibrium constant for solubility equilibria using molar solubilities of the ions of a solution.
-The dissociation reaction of zirconium phosphate into 3 zirconium ions and 4 phosphate ions can be written as:
                                $Z{r_3}{(P{O_4})_4} \to 3Z{r^{ + 4}} + 4P{O_4}^{ - 3}$
The molar solubility of zirconium phosphate is S and the solubility product is ${K_{sp}}$. After undergoing breakdown the solubility of zirconium ion and phosphate ion becomes 3S and 4S. For zirconium it is 3S because there are 3 ions of zirconium and for phosphate it is 4S because there are 4 ions of phosphate. This can be shown in the following manner also:
                                    $Z{r_3}{(P{O_4})_4} \to 3Z{r^{ + 4}} + 4P{O_4}^{ - 3}$
At t = 0 S 0 0
At t time 0 3S 4S
-Now we will see how to calculate the solubility product of any salt.
${K_{sp}}$ is basically the mathematical product of the dissolved ion concentrations that are raised to the power of their stoichiometric coefficients. For an example let us table any compound like ${M_y}{X_z}$. Its dissociation reaction is:
                                   ${M_y}{X_z}(s) \rightleftarrows y{M^{ + z}}(aq) + z{X^{ - y}}(aq)$
For this reaction the solubility product will be written as:
                                              ${K_{sp}} = {[{M^{ + z}}]^y}{[{X^{ - y}}]^z}$
Where, ${[{M^{ + z}}]^y}$ and ${[{X^{ - y}}]^z}$ are molar solubilities of these ions.
-Hence for zirconium phosphate the solubility product will be:
                                         ${K_{sp}} = {[Z{r^{ + 4}}]^3}{[P{O_4}^{ - 3}]^4}$
                                                = ${[3S]^3}{[4S]^4}$
                                                = $[27{S^3}][256{S^4}]$
                                        ${K_{sp}}$ = $(6912){S^7}$
The above obtained equation can also be written as:
                                        $\dfrac{{{K_{sp}}}}{{6912}} = {S^7}$
                                        S = ${\left( {\dfrac{{{K_{sp}}}}{{6912}}} \right)^{\dfrac{1}{7}}}$
                                        S = ${\left[ {{K_{sp}}{{(6912)}^{ - 1}}} \right]^{\dfrac{1}{7}}}$

So, the correct option is: (C)$S = {\left\{ {{K_{sp}}/6912} \right\}^{\dfrac{1}{7}}}$

Note: Solubility product is an equilibrium constant whose value depends on temperature. As temperature increases, the solubility (S) of the solute in the solvent also increases and thus the solubility product increases.
                                [ Temperature α Solubility α Solubility product ]