Question

A solution is 0.1M with respect to $A{g^ + },C{a^{2 + }},M{g^{2 + }}$ and $A{l^{3 + }}$ which will precipitate at lowest concentration of \[\left[ {P{O_4}^{3 - }} \right]\] when solution of $N{a_3}P{O_4}$ is added?
A.\[A{g_3}P{O_4}(Ksp = 1 \times {10^{ - 6}})\]
B.\[C{a_3}{(P{O_4})_2}(Ksp = 1 \times {10^{ - 33}})\]
C.\[M{g_3}{(P{O_4})_2}(Ksp = 1 \times {10^{ - 24}})\]
D.\[AlP{O_4}(Ksp = 1 \times {10^{ - 20}})\]

Answer 1

Hint: We will find the concentration of \[\left[ {P{O_4}^{3 - }} \right]\] in each case since the solubility product constant, \[{K_{SP}}\] is given and we will see in which case the lowest concentration of \[\left[ {P{O_4}^{3 - }} \right]\] is needed to precipitate.

Complete step by step answer:
The solubility product constant, \[{K_{SP}}\] is the equilibrium constant for a solid substance dissolving in an aqueous solution and to solve the \[{K_{SP}}\], it is necessary to take the molarities or concentrations of the products and multiply them.
In the first equation,
$A{g_3}P{O_4} \rightleftharpoons 3A{g^ + } + P{O_4}^{3 - }$
To solve the KSP, it is necessary to take the molarities or concentrations of the products and multiply them
\[{K_{SP}} = {[Ag]^{3 + }}{[P{O_4}]^{3 - }} = 1 \times {10^{ - 6}}\]
$\Rightarrow$ ${[P{O_4}]^{3 - }} = \dfrac{{1 \times 10{}^{ - 6}}}{{{{(0.1)}^3}}} = {10^{ - 3}}$
In the second equation,
$C{a_3}{\left( {P{O_4}} \right)_2} \rightleftharpoons 3C{a^{2 + }}$
\[
  {K_{sp}} = {[C{a^{2 + }}]^3} \times {[P{O_4}^{3 - }]^2} = {10^{ - 33}} \\
   \Rightarrow [P{O_4}^{3 - }] = {\left( {\dfrac{{{{10}^{ - 33}}}}{{{{(0.1)}^3}}}} \right)^{\dfrac{1}{2}}} = {10^{ - 15}} \\
 \]

\[
  M{g_3}{\left( {P{O_4}} \right)_2} \rightleftharpoons 3M{g^{2 + }} + 2P{O_4}^{3 - } \\
   \Rightarrow {K_{sp}} = {[M{g^{2 + }}]^3}{\left[ {2P{O_4}^{3 - }} \right]^2} = {10^{ - 24}} \\
   \Rightarrow \left[ {P{O_4}^{3 - }} \right] = {\left( {\dfrac{{{{10}^{ - 24}}}}{{{{(0.1)}^3}}}} \right)^{\dfrac{1}{2}}} = {10^{ - \dfrac{{21}}{2}}} \\
 \]
In the third equation,

$
  M{g_3}{\left( {P{O_4}} \right)_2} \rightleftharpoons 3M{g^{2 + }} + 2P{O_4}^{3 - } \\
   \Rightarrow {K_{sp}} = {\left[ {M{g^{2 + }}} \right]^3}{\left[ {2P{O_4}^{3 - }} \right]^2} = {10^{ - 24}} \\
   \Rightarrow \left[ {P{O_4}^{3 - }} \right] = {\left( {\dfrac{{{{10}^{ - 24}}}}{{{{(0.1)}^3}}}} \right)^{\dfrac{1}{2}}} = {10^{\dfrac{{ - 21}}{2}}} \\
 $

In the fourth equation,

$
  AlP{O_4} \rightleftharpoons A{l^{3 + }} + P{O_4}^{3 - } \\
   \Rightarrow {K_{sp}} = [A{l^{3 + }}][P{O_4}^{3 - }] = {10^{ - 20}} \\
   \Rightarrow [P{O_4}^{3 - }] = \dfrac{{{{10}^{ - 20}}}}{{0.1}} = {10^{ - 19}} \\
 $

Since, the lowest concentration of \[P{O_4}^{3 - }\] is needed for \[AlP{O_4}\] to precipitate. So, \[AlP{O_4}\] will precipitate.

Therefore, the correct answer is option (D).

Note: The reactant is not included in the \[{K_{SP}}\] equation but only products are multiplied. Solids are not included when we calculate the equilibrium constant expressions, because their concentrations do not change the expression. Hence, \[{K_{SP}}\] represents the maximum extent that a solid can be dissolved in the solution.