91Ó°ÊÓ

\(CaC{O_3}(s) \to C{a^{2 + }}(aq) + CO_3^{2 - }(aq)\)

First, we calculate the value of solubility product constant:

\({K_{sp}} = \left( {C{a^{2 + }}} \right)\left( {CO_3^{2 - }} \right) = 4.8 \times 1{0^{ - 9}}\)

Then we calculate the ion product:

\(Q = \left( {C{a^{2 + }}} \right)\left( {CO_3^{2 - }} \right) = 0.003 \times 0.003 = 9 \times 1{0^{ - 6}}\)

and compare it against the value of\({K_{sp}}\):

\(4.8 \times 1{0^{ - 9}} < 9 \times 1{0^{ - 6}};{K_{sp}} < Q\)

The ion product value is higher than the value of solubility product constant, therefore\(CaC{O_3}\)does precipitate

\(Co{(OH)_2}(s) \to C{o^{2 + }}(aq) + 2O{H^ - }(aq)\)

First, we calculate the value of solubility product constant:

\({K_{sp}} = \left( {C{o^{2 + }}} \right){\left( {2 \times O{H^ - }} \right)^2} = 2 \times 1{0^{ - 16}}\)

Then we calculate the ion product:

\(\begin{array}{l}Q = \left( {C{o^{2 + }}} \right){(2 \times OH)^2}\\Q = 0.01 \times {\left( {2 \times 1{0^{ - 7}}} \right)^2}\\Q\; = 4 \times 1{0^{ - 16}}\end{array}\)

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

\(2 \times 1{0^{ - 16}} > 1 \times 1{0^{ - 16}};{K_{sp}} > Q\)

The ion product value is lower than the value of solubility product constant, therefore\(Co{(OH)_2}\)does not precipitate.

\(CaHP{O_4}(s) \to C{a^{2 + }}(aq) + HPO_4^{2 - }(aq)\)

First, we calculate the value of solubility product constant:

\({K_{sp}} = \left( {C{a^{2 + }}} \right)\left( {HPO_4^{2 - }} \right) = 5 \times 1{0^{ - 6}}\)

Then we calculate the ion product:

\(Q = \left( {C{a^{2 + }}} \right)\left( {HPO_4^{2 - }} \right) = 0.01 \times 2 \times 1{0^{ - 6}} = 2 \times 1{0^{ - 8}}\)

and compare it against the value of\({K_{sp}}\):

\(5 \times 1{0^{ - 6}} > 2 \times 1{0^{ - 8}};{K_{sp}} > Q\)

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

\(P{b_3}{\left( {P{O_4}} \right)_2}(s) \to 3\;P{b^{2 + }}(aq) + 2PO_4^{3 - }(aq)\)

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

\({K_{sp}} = {\left( {P{b^{2 + }}} \right)^3}{\left( {PO_4^{3 - }} \right)^2} = 3 \times 1{0^{ - 44}}\)

Then we calculate the ion product:

\(\begin{array}{l}Q = {\left( {3 \times P{b^{2 + }}} \right)^3}{\left( {2 \times PO_4^{3 - }} \right)^2}\\Q\; = {(3 \times 0.01)^3} \times {\left( {2 \times 1{0^{ - 13}}} \right)^2}\\Q\; = \;(27 \times 0.03)\; \times \;\left( {4 \times 1{0^{ - 26}}} \right)\\Q\; = \;3.24\; \times {10^{ - 24}}\end{array}\)

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

\(3 \times 1{0^{ - 44}} < 1 \times 1{0^{ - 32}};{K_{sp}} < Q\)

The ion product value is higher than the value of solubility product constant, therefore \(P{b_3}{\left( {P{O_4}} \right)_2}\) does precipitate.