Question

A $0.3M$ $HCl$ solution contains the following ions $H{g^{2 + }},C{d^{2 + }}, S{r^{2 + }}, F{e^{2 + }}, C{u^{2 + }}.$ The addition of ${H_2}S$ to above solution will precipitate:
(A) $Cd,Cu \ and \ Hg$
(B) $Cd,Fe \ and \ Sr$
(C) $Hg,Cu \ and \ Fe$
(D) $Cu,Sr \ and \ Fe$

Answer 1

Hint: This question is a part of solubility equilibrium so before starting the question we discuss about what is solubility equilibrium, the maximum amount of solid that can be dissolved in the aqueous solution which is called by the name of solubility equilibrium so let’s start the solution before further delays.

Complete Step by Step Solution:
As in the question it is said that $HCl$ solution has many types of ions through which we have to react with them with ${H_2}S$ And what precipitate will be formed by doing that all reactions.

When ions are reacted with ${H_2}S$ in presence of $HCl$ solution we get that,
As by taking all the reactions with ${H_2}S$ We get that the following ions as get precipitated
In result, we get that when ${H_2}S$ is added with them $Cd,Cu \ and \ Hg$ got as been precipitated,
As a result the answer of the solution is $Cd,Cu \ and \ Hg$ .
Hence, the correct option is (A).

Note: When we have to do multiple reactions in a single question beyond this if we know the nature of many elements without doing the reaction we can easily fix the MCQs with the nature and capability of that element and the separation of molecules and elements from which the question of JEE and NEET can be done faster in less time and it is very useful in many competition exams as done in various forms and very important to do higher work study in less time.