Question

The ions which can be precipitated by both $HCl$ and ${{H}_{2}}S$ are:
A.\[\text{P}{{\text{b}}^{2+}}\]
B.\[\text{C}{{\text{u}}^{2+}}\]
C.\[\text{A}{{\text{g}}^{+}}\]
D.\[\text{S}{{\text{n}}^{2+}}\]

Answer 1

Hint: The analysis of cations can be carried out by the addition of certain reagents often called the group reagents. The ion used is a group 14 element.

Complete answer:
The addition of a reagent will precipitate a metal ion only when the solubility product of that salt is low. When lead cations react with hydrochloric acid to form lead chloride, it is precipitated as the white precipitate of lead chloride. Again when hydrogen sulphide gas is passed through it then forms insoluble precipitate of lead disulphide. Hence, it is precipitated by both hydrochloric acid as well as hydrogen sulphide.
Copper ions react to form cupric chloride that is soluble as water and hence not precipitated, but when hydrogen sulphide is passed into it, it forms insoluble cuprous sulphide. Silver form insoluble curdy white precipitate with dilute hydrochloric acid of silver chloride. Lastly stannous ions precipitate out as stannous sulphide on the addition of hydrogen sulphide.
So, among all these metal ions, only lead will be precipitated out by both hydrogen chloride and hydrogen sulphide.

So, the correct answer is option A.

Note: The addition of these reagents, hydrogen chloride and hydrogen sulphide, falls under the qualitative methods of inorganic ions analysis. There are certain rules that need to be obeyed during the qualitative analysis, firstly tap water is not allowed to use as it contains different anions that may disturb the analysis and secondly the pH of the reaction medium is also an important factor.