Question

A mixture containing $C{{u}^{2+}}$ and $N{{i}^{2+}}$ can be separated for identification by:
(A) passing ${{H}_{2}}S$ in acid medium
(B) passing ${{H}_{2}}S$ in alkaline medium
(C) passing ${{H}_{2}}S$ in neutral medium
(D) passing ${{H}_{2}}S$ in dry mixture

Answer 1

Hint: The separation of $C{{u}^{2+}}$ and $N{{i}^{2+}}$ can be separated from a mixture using method of precipitation based on the ${{K}_{sp}}$ value of the given ions. One of the ions among mentioned ions has a low ${{K}_{sp}}$ value which will be responsible for the precipitation of one of the ions in the solution and thus, can be separated easily.

Complete Step by Step Solution:
When a mixture of salts passes with some reagents like hydrogen sulphide or aqueous ammonia in acidic or alkaline medium, then the inorganic metals ions get precipitated. The process is known as precipitation and the reagent used is known as precipitant and the solid formed in the process is known as precipitate.

For the given mixture, when hydrogen sulphide gas is passed in acidic medium i.e., in the presence of hydrochloric acid, the formation of sulphide takes place as a precipitate. As ${{K}_{sp}}$ value of the copper sulphide is less as compared to nickel sulphide which means precipitation of copper sulphide requires low concentration of sulphide which cannot be achieved due to presence of hydrochloric acid. Thus, $N{{i}^{2+}}$ gets precipitated in the form of its sulphide while $C{{u}^{2+}}$ remains soluble in the mixture and hence, can easily be separated.

Therefore, A mixture containing $C{{u}^{2+}}$ and $N{{i}^{2+}}$ can be easily separated for identification by passing ${{H}_{2}}S$ in acid medium. So, the correct answer is option (A).

Note: It is important to note that the ionisation of hydrogen sulphide (${{H}_{2}}S$) is decreased due to presence of dilute hydrochloric acid which is responsible for common ion effect causing reduction in the concentration of sulphide and therefore, letting nickel ion to get precipitated.