In equalitative analysis, cations of graph `II` as well as group `IV` both are precipitated in the form of sulphides. Due to low value of `K_(sp)` of group `II` sulphides, group reagent is `H_(2)S` in the presence of dil. `HC1`, and due to high value of `K_(sp)` of group `IV` sulphides, group reagent is `H_(2)S` in the presence of `NH_(4)OH` and `NH_(4)C1`. In a solution containing `0.1M` each of `Sn^(2+), Cd^(2+)`, and `Ni^(2+)` ions, `H_(2)S`gas is passed.
`K_(sp) of SnS = 8 xx 10^(-29), K_(sp) of CdS = 15 10^(-28), K_(sp) of NiS - 3 xx 10^(-21), K_(1) of H_(2)S = 1 xx 10^(-7), K_(2) of H_(2)S = 1 xx 10^(-14)`
If `H_(2)S` is passed into the above mixture in the presence of `HC1`, which ion will be precipitated first?

Options

A. `SnS`
B. `CdS`
C. `NiS`
D. `SnS` and `CdS` (both together)

Explanation

To determine which ion will precipitate first when H₂S gas is passed into a solution containing 0.1 M of Sn²⁺, Cd²⁺, and Ni²⁺ ions in the presence of HCl, we will follow these steps:

### Step-by-Step Solution:

1. **Understand the Precipitation Process**:
- When H₂S is introduced, it will react with the metal cations to form their respective sulfides. The precipitation of these sulfides depends on their solubility product constant (Ksp).

2. **Identify Ksp Values**:
- Ksp for SnS = 8 × 10⁻²⁹
- Ksp for CdS = 1.5 × 10⁻²⁸
- Ksp for NiS = 3 × 10⁻²¹

3. **Calculate the Sulfide Ion Concentration Required for Precipitation**:
- The concentration of sulfide ions ([S²⁻]) required to start precipitation can be calculated using the formula:
\[ [S^{2-}] = \frac{K_{sp}}{[M^{2+}]} \]
- For SnS:
\[ [S^{2-}]_{SnS} = \frac{8 \times 10^{-29}}{0.1} = 8 \times 10^{-28} \, \text{M} \]
- For CdS:
\[ [S^{2-}]_{CdS} = \frac{1.5 \times 10^{-28}}{0.1} = 1.5 \times 10^{-27} \, \text{M} \]
- For NiS:
\[ [S^{2-}]_{NiS} = \frac{3 \times 10^{-21}}{0.1} = 3 \times 10^{-20} \, \text{M} \]

4. **Compare the Required [S²⁻] Concentrations**:
- SnS requires [S²⁻] = 8 × 10⁻²⁸ M
- CdS requires [S²⁻] = 1.5 × 10⁻²⁷ M
- NiS requires [S²⁻] = 3 × 10⁻²⁰ M

5. **Determine the Order of Precipitation**:
- The sulfide that requires the least concentration of [S²⁻] to start precipitating will precipitate first.
- Comparing the values:
- SnS requires the highest concentration of sulfide ions to precipitate.
- CdS requires a moderate concentration.
- NiS requires the lowest concentration of sulfide ions to precipitate.

6. **Conclusion**:
- Since NiS requires the lowest concentration of sulfide ions to precipitate, it will precipitate first when H₂S is passed into the solution.

### Final Answer:
**Ni²⁺ will precipitate first as NiS.**