Cu is precipitated as CuS while Zn is not precipitated when \[{{H}_{2}}S\] is passes through an acidic solution of \[Cu{{(N{{O}_{3}})}_{2}}\] and \[Zn{{(N{{O}_{3}})}_{2}}\] respectively.
Answer
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Hint: In order to solve this question, you must know the concept of solubility product constant or ${K}_{sp}$. The compound having a greater solubility product constant will dissolve faster. Also, use the concepts of Ionic equilibrium and common ion effect.
Complete step by step answer:
According to the question, when hydrogen sulphide passes through an acidic solution of copper nitrate and zinc nitrate, a precipitate of copper sulphide is formed and there is no precipitate of zinc sulphide.
Hydrogen sulphide in itself is a weak electrolyte. It dissociates as –
\[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}\]
On dissociation of hydrogen sulphide, we get two moles of proton. Therefore, in the presence of an acidic medium, the ionization of hydrogen sulphide decreases due to the common ion effect. This leads to the decrease in the quantity of \[{{S}^{2-}}\] ions.
Also, the solubility product of Copper Sulphate is very low as compared to Zinc Sulphate.
Therefore, a low concentration of sulphide ions can precipitate cupric ions.
Whereas, a high concentration of sulphide ions is required to precipitate zinc ions as Zinc sulphate.
We can represent this as –
\[\begin{align}
& Cu{{(N{{O}_{3}})}_{2}}+Zn{{(N{{O}_{3}})}_{2}}\xrightarrow{{{H}_{2}}S}CuS\downarrow +HN{{O}_{3}}++Zn{{(N{{O}_{3}})}_{2}} \\
& C{{u}^{2+}}+{{S}^{2-}}\to CuS \\
\end{align}\]
Hence, with decreased sulphide ion concentration, only cupric is precipitated as copper sulphate, while zinc sulphate is not.
Additional Information:
The common-ion effect refers to “the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate”.
Note: The solubility product constant, ${K}_{sp}$ , is the equilibrium constant which indicates the solubility of a solid substance dissolving in an aqueous solution. It represents the level at which a solute will dissolve in solution. A compound with higher Ksp dissolves faster.
Complete step by step answer:
According to the question, when hydrogen sulphide passes through an acidic solution of copper nitrate and zinc nitrate, a precipitate of copper sulphide is formed and there is no precipitate of zinc sulphide.
Hydrogen sulphide in itself is a weak electrolyte. It dissociates as –
\[{{H}_{2}}S\rightleftharpoons 2{{H}^{+}}+{{S}^{2-}}\]
On dissociation of hydrogen sulphide, we get two moles of proton. Therefore, in the presence of an acidic medium, the ionization of hydrogen sulphide decreases due to the common ion effect. This leads to the decrease in the quantity of \[{{S}^{2-}}\] ions.
Also, the solubility product of Copper Sulphate is very low as compared to Zinc Sulphate.
Therefore, a low concentration of sulphide ions can precipitate cupric ions.
Whereas, a high concentration of sulphide ions is required to precipitate zinc ions as Zinc sulphate.
We can represent this as –
\[\begin{align}
& Cu{{(N{{O}_{3}})}_{2}}+Zn{{(N{{O}_{3}})}_{2}}\xrightarrow{{{H}_{2}}S}CuS\downarrow +HN{{O}_{3}}++Zn{{(N{{O}_{3}})}_{2}} \\
& C{{u}^{2+}}+{{S}^{2-}}\to CuS \\
\end{align}\]
Hence, with decreased sulphide ion concentration, only cupric is precipitated as copper sulphate, while zinc sulphate is not.
Additional Information:
The common-ion effect refers to “the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate”.
Note: The solubility product constant, ${K}_{sp}$ , is the equilibrium constant which indicates the solubility of a solid substance dissolving in an aqueous solution. It represents the level at which a solute will dissolve in solution. A compound with higher Ksp dissolves faster.
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