Question

Equal volumes of \[0.002{\text{ M}}\] solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate, \[{K_{sp}}{\text{ = 7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{ - 8}}\])

Answer 1

Hint: The concentration of each solution is given which is equally shared. We will find the ionization product of the cupric and iodine ion. If the ionization product comes to be lesser than its solubility product then it is not soluble. If the ionization product comes larger than the solubility product then it will be soluble and form a precipitate.

Complete answer:
We are given with equal volumes of both solution and solubility product of cupric iodate. We will first find the concentration of each ion of cupric iodate. Since the equal volumes of sodium iodate and cupric chlorate are mixed together, the molar concentration of both the solution will be reduced to half. Therefore the concentration will now become \[0.001{\text{ M}}\], which can be shown as
\[NaI{O_3}{\text{ }} \to {\text{ N}}{{\text{a}}^ + }{\text{ + I}}{{\text{O}}_3}^{ - 1}\]
Since the concentration of \[NaI{O_3}\] is \[0.001{\text{ M}}\] then the concentration of \[{\text{N}}{{\text{a}}^ + }\] will be \[0.001{\text{ M}}\]. Similarly for cupric chlorate,
\[{\text{Cu}}{\left( {Cl{O_3}} \right)_2}{\text{ }} \to {\text{ C}}{{\text{u}}^{2 + }}{\text{ + 2Cl}}{{\text{O}}_3}^{ - 1}\]
The concentration of cupric chlorate is \[0.001{\text{ M}}\] then the concentration of \[C{u^{ + 2}}\] will be\[0.001{\text{ M}}\]. We can also say that after mixing both solution,
\[\left[ {I{O_3}^{ - 1}} \right]{\text{ = }}\left[ {NaI{O_3}} \right]{\text{ = 0}}{\text{.001 M}}\]
\[\left[ {C{u^{ + 2}}} \right]{\text{ = }}\left[ {Cu{{\left( {I{O_3}} \right)}_2}} \right]{\text{ = 0}}{\text{.001 M}}\]
We can find the solubility of copper iodate as,
\[Cu{\left( {I{O_3}} \right)_2}{\text{ }} \rightleftarrows {\text{ C}}{{\text{u}}^{ + 2}}{\text{ + 2I}}{{\text{O}}_3}^{ - 1}\]
The ionic product of the reaction can be written as,
Ionic product \[{\text{ = }}\left[ {{\text{C}}{{\text{u}}^{ + 2}}} \right]{\text{ }}{\left[ {{\text{I}}{{\text{O}}_3}^{ - 1}} \right]^2}\]
Since we know the value of the concentration of the ions. On substituting the values we get,
Ionic product \[{\text{ = }}\left[ {0.001} \right]{\text{ }}{\left[ {0.001} \right]^2}\]
Ionic product \[{\text{ = 1}}{{\text{0}}^{ - 9}}\]
The value of \[{K_{sp}}{\text{ = 7}}{\text{.4 }} \times {\text{ 1}}{{\text{0}}^{ - 8}}\] and the value of Ionic product \[{\text{ = 1}}{{\text{0}}^{ - 9}}\]. Since the value of ionic product is less than the value of \[{K_{sp}}\] , therefore no precipitation occurs.

Note:
Since the equal volumes of solution are mixed together, this is why the concentration of ions gets halved. On mixing the two solutions, ions from each solution combine and form a precipitate if the value of their ionic product is greater than their solubility product. We must take care of negative powers while comparing ionic products and solubility products.