Question

When $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas is passed in a metal sulphate solution in presence of$\text{ N}{{\text{H}}_{\text{4}}}\text{OH }$ , a white precipitate is produced. The metal is/are not identified as:
The question has multiple correct options
A) $\text{ Zn }$
B) $\text{ Fe }$
C) $\text{ Pb }$
D) $\text{ Hg }$

Answer 1

Hint: The metals ion in the solution reacts with the hydrogen sulphide gas and precipitated as the metal sulphide. The reaction is as shown below,
$\text{ }{{\text{M}}^{\text{2+}}}\text{ + }{{\text{H}}_{\text{2}}}\text{S }\rightleftharpoons \text{ MS (s) + 2}{{\text{H}}^{\text{+}}}\text{ }$
Where the metal sulphide is a precipitate. The group II and III cation measures were identified by passing the hydrogen sulphide gas through the metal ion solution. The colour of precipitate depends on the metal ion solution.

Complete Solution :
Now, let us talk about the metals given in the options step by step.
- The first we have $\text{ Zn }$ , so when zinc will react in the form of zinc sulphate solution with the $\text{ }{{\text{H}}_{2}}\text{S }$ gas, and it will lead to the formation of zinc sulphide, and the by-product sulphuric acid. The reaction between the zinc sulphate and the hydrogen sulphide gas is given as follows,
$\text{ }\begin{matrix}
   \text{ZnS}{{\text{O}}_{\text{4}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{S} & \rightleftharpoons & \text{ZnS} & \text{+} & {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \\
   {} & {} & {} & {} & (\text{white ppt)} & {} & {} \\
\end{matrix}$
As we know that zinc sulphide precipitates are white in colour.

The second we have $\text{ Fe }$ , thus, when ferrous sulphate will react with $\text{ }{{\text{H}}_{\text{2}}}\text{S }$ gas, then it will also form $\text{ FeS }$ i.e. iron sulphide. The reaction between the ferrous sulphate and the hydrogen sulphide is as shown below,
$\text{ }\begin{matrix}
   \text{FeS}{{\text{O}}_{\text{4}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{S} & \rightleftharpoons & \text{FeS} & \text{+} & {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \\
   {} & {} & {} & {} & (\text{Black ppt)} & {} & {} \\
\end{matrix}$

 The $\text{ FeS }$ precipitates are black in colour.
Talking about the third i.e. Pb, so when lead sulphate reacts with $\text{ }{{\text{H}}_{2}}\text{S }$ gas, it also forms $\text{ PbS }$ (lead sulphide); that is black in colour. The reaction is as shown below,
$\text{ }\begin{matrix}
   \text{PbS}{{\text{O}}_{\text{4}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{S} & \rightleftharpoons & \text{PbS} & \text{+} & {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \\
   {} & {} & {} & {} & (\text{Black ppt)} & {} & {} \\
\end{matrix}$

- Now, the last we have Hg, it will also form$\text{ HgS }$ , i.e. mercuric sulphide. The mercury sulphite reacts in presence of ammonium hydroxide, the reaction is as shown below,
$\text{ }\begin{matrix}
   \text{HgS}{{\text{O}}_{\text{4}}} & \text{+} & {{\text{H}}_{\text{2}}}\text{S} & \rightleftharpoons & \text{HgS} & \text{+} & {{\text{H}}_{\text{2}}}\text{S}{{\text{O}}_{\text{4}}} \\
   {} & {} & {} & {} & (\text{Red ppt)} & {} & {} \\
\end{matrix}$
$\text{ HgS }$ Is also red in colour.
So, the iron, lead, and mercury are not identified as the white precipitate.
So, the correct answer is “Option B, C and D”.

Note: There are very little sulphide ions $\text{ }{{\text{S}}^{\text{2}-}}\text{ }$ in the solution. The extent of the formation of a product (metal sulphide) depends on the medium. The acidic medium prevents the precipitation of the metal. Here, the metal sulphide has low solubility precipitate but sulphides that have high solubility do not. Thus, we use the basic medium such as $\text{ N}{{\text{H}}_{\text{4}}}\text{OH }$ to precipitate the metal.