Animated Solution for Chemistry - Salt Analysis: The correct match of the group reagents in List-I for precipitating the metal ion given in List-II from solutions, is

The Dance of Group Reagents

Qualitative analysis is like a grand chemical puzzle where we systematically separate and identify different metal cations based on their unique solubility properties. The secret to mastering this lies in understanding the delicate balance of chemical equilibrium, specifically the Common Ion Effect and the role of pH.

Let's break down the logic behind each group reagent and see how they selectively precipitate specific cations.

The Alkaline Sulfides

When we pass ${\text{H}}_2\text{S}$ gas in the presence of ${\text{NH}}_4\text{OH}$, we are creating an alkaline medium. Why is this important? In an alkaline solution, the ${\text{OH}}^{-}$ ions react with the ${\text{H}}^{+}$ ions from ${\text{H}}_2\text{S}$, driving the equilibrium forward and significantly increasing the concentration of sulfide ions ($S^{2-}$).

This high concentration of sulfide ions is exactly what we need to exceed the solubility product ($K_{sp}$) of Group IV cations like ${\text{Mn}}^{2+}$, ${\text{Zn}}^{2+}$, ${\text{Ni}}^{2+}$, and ${\text{Co}}^{2+}$. In our problem, ${\text{Mn}}^{2+}$ reacts to form a beautiful, characteristic pink or buff-colored precipitate of $\text{MnS}$.

The Carbonate Catchers

Moving on to Group V, the reagent is $({\text{NH}}_4{)}_2{\text{CO}}_3$ in the presence of ${\text{NH}}_4\text{OH}$. This combination is designed to precipitate the alkaline earth metals: Barium, Strontium, and Calcium.

The ${\text{NH}}_4\text{OH}$ ensures the solution remains basic, preventing the carbonate ions from reacting with water to form bicarbonate, which would keep the metals in solution. For our specific case, ${\text{Ba}}^{2+}$ reacts with the carbonate ions to form a dense white precipitate of Barium carbonate.

The Common Ion Effect in Action

Group III is where the chemistry gets really elegant. The reagent is ${\text{NH}}_4\text{OH}$ in the presence of ${\text{NH}}_4\text{Cl}$. Here, we want to precipitate cations like ${\text{Fe}}^{3+}$, ${\text{Al}}^{3+}$, and ${\text{Cr}}^{3+}$ as hydroxides, but we don't want to precipitate higher group cations like ${\text{Mg}}^{2+}$.

How do we achieve this? We use the Common Ion Effect. The strong electrolyte ${\text{NH}}_4\text{Cl}$ completely dissociates, flooding the solution with ${\text{NH}}_4^{+}$ ions. These ions suppress the ionization of the weak base ${\text{NH}}_4\text{OH}$, keeping the hydroxide ion (${\text{OH}}^{-}$) concentration very low. This low concentration is just enough to exceed the tiny $K_{sp}$ of Group III hydroxides, but not enough to precipitate ${\text{Mg(OH)}}_2$. Thus, ${\text{Al}}^{3+}$ forms a gelatinous white precipitate of ${\text{Al(OH)}}_3$.

The Acidic Sulfides

Finally, we look at Group II. The reagent is ${\text{H}}_2\text{S}$ gas passed in the presence of dilute $\text{HCl}$. This is the exact opposite of Group IV. The acidic medium provides a high concentration of ${\text{H}}^{+}$ ions, which suppresses the ionization of ${\text{H}}_2\text{S}$ via the common ion effect.

As a result, the sulfide ion concentration is extremely low. Only the sulfides with the absolute lowest solubility products—like those of ${\text{Cu}}^{2+}$, ${\text{Pb}}^{2+}$, and ${\text{Hg}}^{2+}$—can precipitate under these harsh conditions. ${\text{Cu}}^{2+}$ reacts to form a striking black precipitate of Copper sulfide.

By understanding the delicate interplay of pH and equilibrium, qualitative analysis transforms from a memorization exercise into a logical and beautiful chemical narrative.