
\[{H_2}S\;\] gas when passed through a solution of cations containing\[\;HCl\] precipitates the cations of second group of qualitative analysis but not those belonging to the fourth group it is because
(A) Presence of \[\;HCl\] decreases the sulphide ion concentration
(B) Solubility product of group $II$sulphates is more than that of group $IV$ sulphides
(C) Presence of \[\;HCl\] increases the sulphide ion concentration
(D) Sulphides of group $IV$cations are unstable in \[\;HCl\]
Answer
162.6k+ views
Hint: Second group of cations are precipitated on passing \[{H_2}S\;\] gas in an acidic medium say\[\;HCl\]. While the fourth group cations are precipitated in presence of basic medium. These groups are classified on the basis of selective precipitation of sparingly soluble salts.
Complete Step by Step Solution:
Hydrogen sulphide is a weak acid which dissociates as the following reaction
${H_2}S\underset {} \leftrightarrows {H^ + } + {S^{2 - }}$
As it is a weak acid so the reaction is in both directions.
Dissociation of \[\;HCl\]can be considered as follow:
$HCl\xrightarrow{{}}{H^ + } + C{l^ - }$
\[\;HCl\]is a strong acid hence is largely ionized to give ${H^ + }$ions. Thus, concentration of hydronium ions is increased and consequently the concentration of sulphide ions produced by ionisation of \[{H_2}S\;\]is sufficiently decreased due to common ion effects. As a result of which the sulphide ion concentration is sufficient enough to exceed the solubility product of sulphides of group $2$ cations. Since Ksp of sulphides of group $3$ and $4$cations is very high so these cations are not precipitated.
Thus, it is a common ion effect that suppresses the dissociation of\[{H_2}S\;\]to achieve precipitation conditions. This is due to the presence of $HCl$ that decreases the sulphide ion concentration.
Hence, the correct answer is A.
Note: Any other acid cannot be used in place of \[\;HCl\]. Weak acids like acetic acid are themselves partially dissociating that cannot suppress the dissociation of \[{H_2}S\;\]. Sulphuric acid will lead to precipitation of sulphates of group $5$. Nitric acid is a powerful oxidising agent that will oxidise \[{H_2}S\;\]to sulphur. Even concentrated hydrochloric acid cannot be used because it will suppress dissociation of \[{H_2}S\;\]to a large extent.
Complete Step by Step Solution:
Hydrogen sulphide is a weak acid which dissociates as the following reaction
${H_2}S\underset {} \leftrightarrows {H^ + } + {S^{2 - }}$
As it is a weak acid so the reaction is in both directions.
Dissociation of \[\;HCl\]can be considered as follow:
$HCl\xrightarrow{{}}{H^ + } + C{l^ - }$
\[\;HCl\]is a strong acid hence is largely ionized to give ${H^ + }$ions. Thus, concentration of hydronium ions is increased and consequently the concentration of sulphide ions produced by ionisation of \[{H_2}S\;\]is sufficiently decreased due to common ion effects. As a result of which the sulphide ion concentration is sufficient enough to exceed the solubility product of sulphides of group $2$ cations. Since Ksp of sulphides of group $3$ and $4$cations is very high so these cations are not precipitated.
Thus, it is a common ion effect that suppresses the dissociation of\[{H_2}S\;\]to achieve precipitation conditions. This is due to the presence of $HCl$ that decreases the sulphide ion concentration.
Hence, the correct answer is A.
Note: Any other acid cannot be used in place of \[\;HCl\]. Weak acids like acetic acid are themselves partially dissociating that cannot suppress the dissociation of \[{H_2}S\;\]. Sulphuric acid will lead to precipitation of sulphates of group $5$. Nitric acid is a powerful oxidising agent that will oxidise \[{H_2}S\;\]to sulphur. Even concentrated hydrochloric acid cannot be used because it will suppress dissociation of \[{H_2}S\;\]to a large extent.
Recently Updated Pages
Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Which of the following is most stable A Sn2+ B Ge2+ class 11 chemistry JEE_Main

The enolic form of acetone contains a 10sigma bonds class 11 chemistry JEE_Main

The specific heat of metal is 067 Jg Its equivalent class 11 chemistry JEE_Main

The increasing order of a specific charge to mass ratio class 11 chemistry JEE_Main

Which one of the following is used for making shoe class 11 chemistry JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Types of Solutions

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Chemistry In Hindi Chapter 1 Some Basic Concepts of Chemistry

NCERT Solutions for Class 11 Chemistry Chapter 7 Redox Reaction

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Verb Forms Guide: V1, V2, V3, V4, V5 Explained
