Question: 30. Which of the following compounds precipitates from a solution that has the concentrations indicated? (See Appendix J for \({K_{sp}}\) values.)

(a) \(KCl{O_4}:\left( {{K^ + }} \right) = 0.01{M^ - }\left( {ClO_4^ - } \right) = 0.01M\)

(b) \({K_2}PtC{l_6}:\left( {{K^ + }} \right) = 0.01M,\left( {PtC{l_6}^{2 - }} \right) = 0.01M\) \(\)

(c) \(Pb{I_2}:\left( {P{b^{2 + }}} \right) = 0.003M,\left( {{I^ - }} \right) = 1.3 \times 1{0^{ - 3}}M\)

(d) \(A{g_2}\;S:\left( {A{g^ + }} \right) = 1 \times 1{0^{ - 10}}M,\left( {{S^{2 - }}} \right) = 1 \times 1{0^{ - 13}}M\)

\(KCl{O_4}(s) \to {K^ + }(aq) + ClO_4^ - (aq)\)

First, we calculate the value of the solubility product constant:

\(\begin{array}{l}{K_{sp}} = \left( {{K^ + }} \right)\left( {ClO_4^ - } \right)\\{K_{sp}} = 1.05 \times 1{0^{ - 2}}\end{array}\)

Then we calculate the ion product:

\(\begin{array}{l}Q = \left( {{K^ + }} \right)\left( {ClO_4^ - } \right)\\Q = 0.01 \times 0.01\\Q = 1 \times 1{0^{ - 4}}\end{array}\)

and compare it against the value of\({{\rm{K}}_{{\rm{sp}}}}\):

\(1.05 \times 1{0^{ - 2}} > 1 \times 1{0^{ - 4}};{K_{sp}} > Q\)

The ion product value is lower than the value of the solubility product constant, therefore\(KCl{O_4}\)does not precipitate

\({K_2}PtC{l_6}({\rm{s}}) \to 2{K^ + }({\rm{aq}}) + PtCl_6^{2 - }({\rm{aq}})\)

First, we calculate the value of solubility product constant:

\(\begin{array}{l}{K_{sp}}\; = \;\left( {2{K^ + }} \right)\;{\left( {PtC{l_6}} \right)^{2 - }}\\{K_{sp\;}}\; = \;7.48 \times \;{10^{ - 6}}\end{array}\)

Then we calculate the ion product:

\(\begin{array}{l}Q = {(2 \times {K^ + }]^2}\;\;\;\left( {PtC{l_6}^{2 - }} \right)\;\\Q\; = \;{0.01^2}\;\; \times \;0.01\\Q\; = \;1\; \times \;{10^{ - 6}}\end{array}\)

and compare it against the value of\({{\rm{K}}_{{\rm{sp}}}}\):

\(7.48 \times 1{0^{ - 6}} > 1 \times 1{0^{ - 6}};{K_{sp}} > Q\)

The ion product value is lower than the value of solubility product constant, therefore\({K_2}PtC{l_6}\)does not precipitate.

\(Pb{I_2}(s) \to P{b^{2 + }}(aq) + 2{I^ - }(aq)\)

First, we calculate the value of solubility product constant:

\(\begin{array}{l}{K_{sp}} = \left( {P{b^{2 + }}} \right){\left( {2 \times {I^ - }} \right)^2}\\{K_{sp}} = 1.4 \times 1{0^{ - 8}}\end{array}\)

Then we calculate the ion product:

\(\begin{array}{l}Q = \left( {P{b^{2 + }}} \right){\left( {2 \times {I^ - }} \right)^2}\\Q = \left( {0.003} \right) \times {\left( {2 \times 21.3 \times 1{0^{ - 3}}} \right)^2}\\Q = 5.07 \times 1{0^{ - 9}}\end{array}\)

and compare it against the value of\({{\rm{K}}_{{\rm{sp}}}}\) :

\(1.4 \times 1{0^{ - 8}} > 5.07 \times 1{0^{ - 9}};{K_{sp}} > Q\)

The ion product value is higher than the value of solubility product constant, therefore\(Pb{I_2}\)does not precipitate.

\(A{g_2}S(s) \to 2A{g^ + }(aq) + {S^{2 - }}(aq)\)

First, we calculate the value of solubility product constant:

\(\begin{array}{l}{K_{sp}} = {\left( {2 \times A{g^ + }} \right)^2}\left( {{S^{2 - }}} \right)\\{K_{sp}} = 1.6 \times 1{0^{ - 49}}\end{array}\)

Then we calculate the ion product:

\(\begin{array}{l}Q = {\left( {2 \times A{g^ + }} \right)^2}\left( {{S^{2 - }}} \right)\\Q = {\left( {2 \times 1{0^{ - 10}}} \right)^2} \times 1 \times 1{0^{ - 13}}\\Q = 1 \times 1{0^{ - 33}}\end{array}\)

and compare it against the value of\({{\rm{K}}_{{\rm{sp}}}}\):

\(1.6 \times 1{0^{ - 49}} < 1 \times 1{0^{ - 33}};{K_{{\rm{sp}}}} < Q\)

The ion product value is higher than the value of solubility product constant, therefore it does precipitate.