Animated Solution for Chemistry - Ionic Equilibrium: To an aqueous solution containing ions such as ${\text{Al}}^{3+}$, ${\text{Zn}}^{2+}$, ${\text{Ca}}^{2+}$, ${\text{Fe}}^{3+}$, ${\text{Ni}}^{2+}$, ${\text{Ba}}^{2+}$ and ${\text{Cu}}^{2+}$ conc. $\text{HCl}$, was added followed by ${\text{H}}_2\text{S}$. The total number of cations precipitated during this reaction is/are

Imagine you are a chemical detective, handed a beaker filled with a mysterious aqueous solution. Inside this solution, a whole party of cations is swimming around: Aluminum (${\text{Al}}^{3+}$), Zinc (${\text{Zn}}^{2+}$), Calcium (${\text{Ca}}^{2+}$), Iron (${\text{Fe}}^{3+}$), Nickel (${\text{Ni}}^{2+}$), Barium (${\text{Ba}}^{2+}$), and Copper (${\text{Cu}}^{2+}$). Your mission is to selectively pull them out of the solution one by one.

To do this, we rely on the principles of Qualitative Salt Analysis, which groups cations based on their precipitation behavior under specific chemical conditions. In this problem, we are adding concentrated Hydrochloric Acid ($\text{HCl}$) followed by bubbling Hydrogen Sulfide (${\text{H}}_2\text{S}$) gas. This specific combination is the classic reagent test for Group II cations.

The Common Ion Effect

Let's look at the chemistry of our reagents. Hydrogen sulfide is a weak diprotic acid. In water, it partially ionizes to give hydrogen ions and sulfide ions:

However, we didn't just add ${\text{H}}_2\text{S}$; we added concentrated $\text{HCl}$ first. Hydrochloric acid is a strong acid that completely dissociates, flooding the solution with ${\text{H}}^{+}$ ions:

According to Le Chatelier's Principle, this massive influx of ${\text{H}}^{+}$ ions pushes the ${\text{H}}_2\text{S}$ equilibrium to the left. This phenomenon is known as the Common Ion Effect. Because the ionization of ${\text{H}}_2\text{S}$ is heavily suppressed, the concentration of sulfide ions ($[{\text{S}}^{2-}]$) in the solution becomes extremely low.

With such a tiny amount of sulfide ions available, only the most desperate cations—those whose sulfides have an incredibly low Solubility Product ($K_{sp}$)—will be able to precipitate.

Let's evaluate our suspects:

1. The Group II Candidate: Copper (${\text{Cu}}^{2+}$) belongs to Group II. Copper(II) sulfide ($\text{CuS}$) has a remarkably low $K_{sp}$. Even the minuscule concentration of ${\text{S}}^{2-}$ provided in this acidic medium is enough to exceed its ionic product, causing it to crash out of the solution as a distinct black precipitate.

2. The Group IV Candidates: Zinc (${\text{Zn}}^{2+}$) and Nickel (${\text{Ni}}^{2+}$) belong to Group IV. While they do form sulfides, their $K_{sp}$ values are much higher than that of $\text{CuS}$. The low sulfide concentration in our acidic beaker is simply not enough to precipitate them. They require a basic medium (like adding ${\text{NH}}_4\text{OH}$) to increase the sulfide concentration sufficiently.

3. The Group III Candidates: Aluminum (${\text{Al}}^{3+}$) and Iron (${\text{Fe}}^{3+}$) belong to Group III. These cations do not form stable sulfides in aqueous solutions; they undergo hydrolysis. In qualitative analysis, they are typically precipitated as hydroxides, but our current medium is highly acidic, so they remain happily dissolved.

4. The Group V Candidates: Calcium (${\text{Ca}}^{2+}$) and Barium (${\text{Ba}}^{2+}$) belong to Group V. Their sulfides are highly soluble in water, meaning they will never precipitate under these conditions regardless of the $\text{pH}$.

Out of the entire mixture of seven cations, the strict conditions created by the common ion effect allow only one to precipitate. Copper (${\text{Cu}}^{2+}$) forms the black $\text{CuS}$ precipitate, while the rest remain in the aqueous phase.

Therefore, the total number of cations precipitated during this reaction is exactly 1.