Question

A saturated solution of \[{H_{\mathbf{2}}}S\] in \[0.1M\] \[HCl\] at \[25^\circ C\] contains \[{S^{{\mathbf{2}} - }}\] ion concentration of \[{10^{ - {\mathbf{23}}}}mol{L^{ - {\mathbf{1}}}}\] . The solubility product of some sulphides are \[CuS = {10^{ - 44}}\], \[FeS = {10^{ - 14}}\], \[MnS = {10^{ - 15}}\], \[CdS = {10^{ - 25}}\]. If \[0.01M\]solution of these salts in \[1M\]\[HCl\] are saturated with \[{H_{\mathbf{2}}}S\], which of these will be precipitated?
(A) All
(B) All except \[MnS\]
(C) All except \[MnS\] and \[FeS\]
(D) Only \[CuS\]

Answer 1

Hint: We need to know the concept of solubility of salts and solubility product in an equilibrium. Considering the equilibrium between a sparingly soluble ionic salt and its saturated aqueous solution, solubility product can be explained. Let us consider a salt such as Barium Sulphate (BaSO4) in a saturated aqueous solution. The equilibrium between the undissolved salt and its dissolved ions in an aqueous saturated solution is represented by the equation:
$BaS{O_4}\overset {{\text{saturated solution}}} \leftrightarrows Ba_{\left( {aq} \right)}^{2 + } + SO_{\left( {aq} \right)}^{2 - }$

Complete answer:
In the given question, \[{S^{{\mathbf{2}} - }}\] ion concentration, \[\left[ {{S^{{\mathbf{2}} - }}} \right] = {10^{ - 23}}mol{L^{ - 1}}\]
Since \[0.01M\]solution of \[CuS\],\[FeS\],\[MnS\] and \[CdS\]are added in \[1M\]\[HCl\]saturated with \[{H_{\mathbf{2}}}S\],Therefore the cation in this reaction can be considered as\[{M^{2 + }}\],where \[{M^{2 + }}\]+ is\[C{u^{2 + }}\],\[F{e^{2 + }}\] ,\[M{n^{2 + }}\] and \[C{d^{2 + }}\] the concentration of \[{M^{2 + }}\],\[\left[ {{\text{ }}{M^{2 + }}} \right]\] is \[0.01M\]that is \[{10^{ - 2}}M\].
Therefore, Solubility product \[{K_{sp}} = \left[ {{S^{ - 2}}} \right] \times \left[ {{M^{ + 2}}} \right]\]
\[ = {10^{ - {\mathbf{23}}}} \times {10^{ - 2}}\]
\[ = {10^{ - 25}}\]
Since the solubility product of undissolved \[CuS = {10^{ - 44}}\] which is lesser than the solubility product when dissolved which is \[{10^{ - 25}}\], it is likely to have lesser solubility and hence precipitate out.
Since the solubility product of undissolved \[FeS = {10^{ - 14}}\], which is more than the solubility product when dissolved which is \[{10^{ - 25}}\], it is likely to have more solubility and hence will not precipitate out.
Since the solubility product of undissolved \[MnS = {10^{ - 15}}\], which is more than the solubility product when dissolved which is \[{10^{ - 25}}\],it is likely to have more solubility and hence will not precipitate out.
Since the solubility product of undissolved \[CdS\] \[ = \]\[{10^{ - {\mathbf{25}}}}\], which is equal to the solubility product when dissolved which is \[{10^{ - 25}}\],it is likely to have equal solubility and hence will not precipitate out.
Therefore, the correct option is option (D).

Additional note:
By the definition of equilibrium constant which states that at a given temperature, the product of the concentrations of the reaction products raised to the respective stoichiometric coefficients in the balanced chemical equation divided by the product of the concentrations of the reactants raised to the respective stoichiometric coefficients has a constant value known as the Equilibrium Constant.
 Thus, for the above reaction the equilibrium constant is given by the equation:
\[K{\text{ }} = {\text{ }}\dfrac{{\left[ {B{a^{2 + }}} \right]\left[ {S{O_4}^{2 - }} \right]}}{{\left[ {BaS{O_4}} \right]}}\]
For a pure solid substance or a pure salt which is completely soluble, the equilibrium constant is known Solubility Product Constant or simply Solubility Product, denoted as $K_{sp}$ and is given by the equation:
\[{K_{sp}} = \left[ {B{a^{2 + }}} \right]\left[ {S{O_4}^{2 - }} \right]\]
Greater is the solubility product, greater will be the solubility of the substance in the saturated solution.

Note:
It must be noted that Solubility product is a constant value hence it has no units. To predict whether the salt dissolved in a saturated solution will precipitate or not, the basic rule to follow is that the solubility product of the undissolved salt must be greater or equal to the solubility product when dissolved in the saturated solution. If it is lesser, it is likely to precipitate out.