Question

The expression of solubility product of mercurous iodide is:-
A.${\left[ {{\text{2H}}{{\text{g}}^{\text{ + }}}} \right]^{\text{2}}} \times {\text{2}}{\left[ {{{\text{I}}^{\text{ - }}}} \right]^{\text{2}}}$
B.\[{\left[ {{\text{2H}}{{\text{g}}^{{\text{ + + }}}}} \right]^{\text{2}}} \times {\text{ }}{\left[ {{\text{2}}{{\text{I}}^{\text{ - }}}} \right]^{\text{2}}}\]
C.\[\left[ {{\text{H}}{{\text{g}}_{\text{2}}}^{{\text{2 + }}}} \right]{\text{ }} \times {\text{ }}{\left[ {{{\text{I}}^{\text{ - }}}} \right]^{\text{2}}}\]
D.\[{\left[ {{\text{H}}{{\text{g}}^{{\text{2 + }}}}} \right]^{\text{2}}} \times {\left[
{{{\text{I}}^{\text{ - }}}} \right]^{\text{2}}}\]

Answer 1

Hint:The solubility product of a substance is the equilibrium constant for the dissolution of a moderately insoluble substance into a solution. Here, the concentration of the undissociated substance is considered to be much higher that the concentration of the ions present in the solution.

Complete step by step answer:
Mercurous Iodide dissolves very less in water due to the strong bond between the mercurous ions and the iodide ions and hence we get the solubility product of it instead of the equilibrium constant. The molecular formula of mercurous iodide is: ${\text{H}}{{\text{g}}_{\text{2}}}{{\text{I}}_{\text{2}}}$ and it slightly dissociates in water in the following way:
\[{\text{H}}{{\text{g}}_{\text{2}}}{{\text{I}}_{\text{2}}} \rightleftharpoons {\left[ {{\text{H}}{{\text{g}}_{\text{2}}}} \right]^{{\text{2 + }}}}{\text{ + 2}}{\left[ {\text{I}} \right]^{\text{ - }}}\]
Therefore the equilibrium product for the above can be expressed as,
 K =\[\dfrac{{\left[ {{\text{H}}{{\text{g}}_{\text{2}}}^{{\text{2 + }}}} \right] \times {{\left[ {{{\text{I}}^{\text{ - }}}} \right]}^{\text{2}}}}}{{\left[ {{\text{H}}{{\text{g}}_{\text{2}}}{{\text{I}}_{\text{2}}}} \right]}}\]
Considering the volume of ${\text{H}}{{\text{g}}_{\text{2}}}{{\text{I}}_{\text{2}}}$ to be relatively unchanged we get, the solubility product of mercurous iodide, \[{{\text{K}}_{{\text{sp}}}}{\text{ = }}\left[ {{\text{H}}{{\text{g}}_{\text{2}}}^{{\text{2 + }}}} \right] \times {\left[ {{{\text{I}}^{\text{ - }}}} \right]^{\text{2}}}\]

So, the correct answer is option C.

Note:
Mercury has two oxidation states, $\left( { + 1} \right)$ and $\left( { + 2} \right)$, the first oxidation state is called mercurous while the second is called mercuric.
The reason for the low dissociation of mercurous iodide in water is the covalent nature of the bond between the mercurous ions and the iodide ions. It has a linear structure with the iodide groups at the two sides while the mercurous ion in the centre.
Mercurous Iodide was used as a drug in the 19th century and was used to cure a number of diseases including syphilis, kidney disease, and acne.