Animated Solution for Chemistry - Salt Analysis: The pair(s) of ions where BOTH the ions are precipitated upon passing ${\text{H}}_2\text{S}$ gas in presence of dilute $\text{HCl}$, is(are)

Imagine you are a chemical detective, standing in a laboratory with a test tube containing a mystery mixture of metal ions. Your goal? To selectively separate and identify them. This is the beautiful art of qualitative salt analysis.

In this problem, our primary weapon is hydrogen sulfide gas (${\text{H}}_2\text{S}$) passed through a solution acidified with dilute hydrochloric acid ($\text{HCl}$). This specific combination is not random; it is the legendary group reagent for Group II cations. But why do we need the acid?

Controlling the Reagent

When we bubble ${\text{H}}_2\text{S}$ into water, it acts as a weak diprotic acid, dissociating slightly to give hydrogen ions (${\text{H}}^{+}$) and sulfide ions (${\text{S}}^{2-}$).

However, by adding dilute $\text{HCl}$, we flood the solution with a massive amount of ${\text{H}}^{+}$ ions. According to Le Chatelier's Principle, this common ion effect violently pushes the ${\text{H}}_2\text{S}$ equilibrium backward, drastically suppressing its ionization. The result? The concentration of sulfide ions ($[{\text{S}}^{2-}]$) plummets to an extremely low value.

For any metal sulfide to precipitate, its ionic product must exceed its solubility product ($K_{sp}$). Because our available $[{\text{S}}^{2-}]$ is so tiny, only those metal sulfides with an exceptionally low $K_{sp}$ will precipitate. These elite, highly insoluble sulfides belong exclusively to the Group II cations (like ${\text{Cu}}^{2+}$, ${\text{Pb}}^{2+}$, ${\text{Hg}}^{2+}$, and ${\text{Bi}}^{3+}$).

Option by Option

Let's interrogate our options one by one to see who survives this acidic test.

Option A: We have Barium (${\text{Ba}}^{2+}$) and Zinc (${\text{Zn}}^{2+}$). Barium is a Group V cation, and its sulfide is completely soluble in water. Zinc is a Group IV cation. While it does form a precipitate ($\text{ZnS}$), its $K_{sp}$ is relatively high. It demands a much higher sulfide concentration to precipitate, which can only be achieved in a basic medium. Thus, neither will precipitate here.

Option B: Here we have Bismuth (${\text{Bi}}^{3+}$) and Iron (${\text{Fe}}^{3+}$). Bismuth is a proud member of Group II and will readily form a dark brown/black precipitate of ${\text{Bi}}_2{\text{S}}_3$. But watch out for the trap! Iron (${\text{Fe}}^{3+}$) is a Group III cation and a strong oxidizing agent. Instead of precipitating, it reacts with ${\text{H}}_2\text{S}$ in a redox battle. The ${\text{H}}_2\text{S}$ reduces ${\text{Fe}}^{3+}$ to ${\text{Fe}}^{2+}$, while being oxidized to elemental sulfur (a white/yellow turbidity). No iron sulfide precipitate is formed.

Option C: Copper (${\text{Cu}}^{2+}$) and Lead (${\text{Pb}}^{2+}$). Both of these are classic Group II cations. They both possess incredibly low solubility products. Copper forms a black precipitate of $\text{CuS}$, and Lead forms a black precipitate of $\text{PbS}$. Both ions successfully precipitate!

Option D: Mercury (${\text{Hg}}^{2+}$) and Bismuth (${\text{Bi}}^{3+}$). Mercury is another heavy hitter from Group II, forming a black precipitate of $\text{HgS}$. As we established earlier, Bismuth also precipitates. Therefore, this pair is also a winner.

By mastering the interplay of the common ion effect and solubility products, we can confidently conclude that the pairs in Option C and Option D will both fully precipitate under these acidic conditions.

This problem is a brilliant reminder that chemistry isn't just about memorizing groups; it's about understanding the delicate balance of equilibria and redox reactions that govern the physical world!