Question

In qualitative analysis, in order to detect second group basic radical, ${{\text{H}}_2}{\text{S}}$ gas is passed in the presence of dilute ${\text{HCl}}$ to:
A.increase the dissociation of ${{\text{H}}_2}{\text{S}}$
B.decrease the dissociation of salt solution
C.decrease the dissociation of ${{\text{H}}_2}{\text{S}}$
D.increase the dissociation of salt solution

Answer 1

The presence of the diluted ${\text{HCl}}$ is intended to weaken the shell of the cation. ${{\text{H}}_2}{\text{S}}$ is a weak acid, donating both one and two hydrogen ions in neutralization reactions, forming ${\text{H}}{{\text{S}}^{\text{ - }}}{\text{ and }}{{\text{S}}^{{\text{ - 2}}}}$ ions.

Complete step by step answer:

${\text{HCl}}$ decreases the ionization of ${{\text{H}}_2}{\text{S}}$ due to common ion effect so that the concentration of sulfide ion in the solution is very low,
Group I: ${\text{A}}{{\text{g}}^{\text{ + }}}{\text{, P}}{{\text{b}}^{{\text{2 + }}}}{\text{, H}}{{\text{g}}^{{\text{2 + }}}}$ cations produce insoluble chlorides so they form precipitates with dilute ${\text{HCl}}$, while all other cations remain in the solution.
Group II: ${\text{C}}{{\text{u}}^{{\text{2 + }}}}{\text{, B}}{{\text{i}}^{{\text{3 + }}}}{\text{, C}}{{\text{d}}^{{\text{2 + }}}}{\text{, H}}{{\text{g}}^{{\text{2 + }}}}{\text{, A}}{{\text{s}}^{{\text{3 + }}}}{\text{, S}}{{\text{b}}^{{\text{3 + }}}}{\text{, S}}{{\text{n}}^{{\text{4 + }}}}$ cations produce very insoluble sulfides which also get precipitated by low amounts, this can be achieved by adding an acidic solution of ${{\text{H}}_2}{\text{S}}$.
Group III: ${\text{A}}{{\text{l}}^{{\text{3 + }}}}{\text{, C}}{{\text{r}}^{{\text{3 + }}}}{\text{, F}}{{\text{e}}^{{\text{3 + }}}}{\text{, Z}}{{\text{n}}^{{\text{2 + }}}}{\text{, N}}{{\text{i}}^{{\text{2 + }}}}{\text{, C}}{{\text{o}}^{{\text{2 + }}}}{\text{, M}}{{\text{n}}^{{\text{2 + }}}}$ cations produce slightly soluble sulfides such that it gets precipitated by relatively high amounts of sulfide ion, this can be achieved by adding a basic solution of ${{\text{H}}_2}{\text{S}}$.
Group IV: ${\text{M}}{{\text{g}}^{{\text{2 + }}}}{\text{, C}}{{\text{a}}^{{\text{2 + }}}}{\text{, S}}{{\text{r}}^{{\text{2 + }}}}{\text{, B}}{{\text{a}}^{{\text{2 + }}}}$ cations, as well as all of the above groups, produce insoluble carbonates which get precipitated by the addition of carbonate.
Group V: ${\text{N}}{{\text{a}}^{\text{ + }}}{\text{, }}{{\text{K}}^{\text{ + }}}{\text{, N}}{{\text{H}}^{{\text{4 + }}}}$ cations do not precipitate with any of the mentioned reagents.
Hence the correct option is C.

Additional information- When ${{\text{H}}_2}{\text{S}}$ gas is passed through ${\text{HCl}}$ containing aqueous solution of ${\text{CuC}}{{\text{l}}_{{\text{2}}}}{\text{,HgC}}{{\text{l}}_{{\text{2}}}}{\text{,BiC}}{{\text{l}}_{{\text{2}}}}{\text{, and CoC}}{{\text{l}}_{\text{2}}}$, it does not precipitate out ${\text{CoS}}$ as it belongs to IV group radical.
Group 2 cations produce insoluble sulfides which can only react with cations or elements of group 2 and not that of group 3, group 4 or group 5. Precipitates of ${\text{HgS, PbS, B}}{{\text{i}}_2}{{\text{S}}_3}{\text{, CuS, CdS, A}}{{\text{s}}_2}{{\text{S}}_3}{\text{, S}}{{\text{b}}_2}{{\text{S}}_3}$ and ${\text{SnS}}$ are formed
Example- ${\text{CuC}}{{\text{l}}_2} + {{\text{H}}_2}{\text{S}} \to {\text{CuS}} \downarrow {\text{blackppt}}$
Once we have separated a mixture into major groups a variety of reagents can be added to distinguish subgroups or individual ions. These reagents are chosen because they react with specific ions which do not disturb general properties (reactions can involve precipitation, acid-base neutralization, oxidation-reduction, or complex ion formation).

Note:
For a successful qualitative analysis with these five groups, the order of reagent addition is very important and must be followed in the order (i.e. first group, second group, and so on). For example, addition of carbonate first will precipitate all group elements but the group V ions or addition of basic ${{\text{H}}_{\text{2}}}{\text{S}}$ first would precipitate all group II and group III ions.