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Question: A solution is 0.1M with respect to \(A{g^ + },C{a^{2 + }},M{g^{2 + }}\) and \(A{l^{3 + }}\) which wi...

A solution is 0.1M with respect to Ag+,Ca2+,Mg2+A{g^ + },C{a^{2 + }},M{g^{2 + }} and Al3+A{l^{3 + }} which will precipitate at lowest concentration of [PO43−]\left[ {P{O_4}^{3 - }} \right] when solution of Na3PO4N{a_3}P{O_4} is added?
A.Ag3PO4(Ksp=1×10−6)A{g_3}P{O_4}(Ksp = 1 \times {10^{ - 6}})
B.Ca3(PO4)2(Ksp=1×10−33)C{a_3}{(P{O_4})_2}(Ksp = 1 \times {10^{ - 33}})
C.Mg3(PO4)2(Ksp=1×10−24)M{g_3}{(P{O_4})_2}(Ksp = 1 \times {10^{ - 24}})
D.AlPO4(Ksp=1×10−20)AlP{O_4}(Ksp = 1 \times {10^{ - 20}})

Explanation

Solution

We will find the concentration of [PO43−]\left[ {P{O_4}^{3 - }} \right] in each case since the solubility product constant, KSP{K_{SP}} is given and we will see in which case the lowest concentration of [PO43−]\left[ {P{O_4}^{3 - }} \right] is needed to precipitate.

Complete step by step answer:
The solubility product constant, KSP{K_{SP}} is the equilibrium constant for a solid substance dissolving in an aqueous solution and to solve the KSP{K_{SP}}, it is necessary to take the molarities or concentrations of the products and multiply them.
In the first equation,
Ag3PO4⇌3Ag++PO43−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
KSP=[Ag]3+[PO4]3−=1×10−6{K_{SP}} = {[Ag]^{3 + }}{[P{O_4}]^{3 - }} = 1 \times {10^{ - 6}}
⇒\Rightarrow [PO4]3−=1×10−6(0.1)3=10−3{[P{O_4}]^{3 - }} = \dfrac{{1 \times 10{}^{ - 6}}}{{{{(0.1)}^3}}} = {10^{ - 3}}
In the second equation,
Ca3(PO4)2⇌3Ca2+C{a_3}{\left( {P{O_4}} \right)_2} \rightleftharpoons 3C{a^{2 + }}

Ksp=[Ca2+]3×[PO43−]2=10−33 ⇒[PO43−]=(10−33(0.1)3)12=10−15  {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}} \\\ Mg3(PO4)2⇌3Mg2++2PO43− ⇒Ksp=[Mg2+]3[2PO43−]2=10−24 ⇒[PO43−]=(10−24(0.1)3)12=10−212  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,

Mg3(PO4)2⇌3Mg2++2PO43− ⇒Ksp=[Mg2+]3[2PO43−]2=10−24 ⇒[PO43−]=(10−24(0.1)3)12=10−212  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,

AlPO4⇌Al3++PO43− ⇒Ksp=[Al3+][PO43−]=10−20 ⇒[PO43−]=10−200.1=10−19  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 PO43−P{O_4}^{3 - } is needed for AlPO4AlP{O_4} to precipitate. So, AlPO4AlP{O_4} will precipitate.

Therefore, the correct answer is option (D).

Note: The reactant is not included in the KSP{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, KSP{K_{SP}} represents the maximum extent that a solid can be dissolved in the solution.