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)`
Which of the following sulphides is more soluble in pure water?

Options

A. `CdS`
B. `NiS`
C. `SnS`
D. Equal solubility for all

Explanation

To determine which of the sulfides (SnS, CdS, or NiS) is more soluble in pure water, we will analyze the solubility product constant (Ksp) values provided for each sulfide.

### Step-by-Step Solution:

1. **Understanding Ksp and Solubility**:
The solubility product constant (Ksp) is a measure of the solubility of a compound in water. The higher the Ksp value, the more soluble the compound is. The relationship between solubility (S) and Ksp for a salt that dissociates into ions can be expressed as:
\[ K_{sp} = [M^{n+}][S^{m-}]^m \]
For sulfides like SnS, CdS, and NiS, they dissociate as follows:
\[ \text{M}_2\text{S} \rightleftharpoons \text{M}^{2+} + \text{S}^{2-} \]
If we let S be the solubility of the sulfide, then:
\[ K_{sp} = S^2 \]

2. **Calculating Solubility from Ksp**:
We can express solubility in terms of Ksp:
\[ S = \sqrt{K_{sp}} \]
Now, we will calculate the solubility for each sulfide using their Ksp values.

3. **Given Ksp Values**:
- Ksp of SnS = \(8 \times 10^{-29}\)
- Ksp of CdS = \(15 \times 10^{-28}\)
- Ksp of NiS = \(3 \times 10^{-21}\)

4. **Calculating Solubility**:
- For SnS:
\[ S_{SnS} = \sqrt{8 \times 10^{-29}} \approx 2.83 \times 10^{-15} \]
- For CdS:
\[ S_{CdS} = \sqrt{15 \times 10^{-28}} \approx 3.87 \times 10^{-14} \]
- For NiS:
\[ S_{NiS} = \sqrt{3 \times 10^{-21}} \approx 5.48 \times 10^{-11} \]

5. **Comparing Solubility Values**:
Now, we compare the calculated solubility values:
- Solubility of SnS: \(2.83 \times 10^{-15}\)
- Solubility of CdS: \(3.87 \times 10^{-14}\)
- Solubility of NiS: \(5.48 \times 10^{-11}\)

From these calculations, we can see that:
\[ S_{NiS} > S_{CdS} > S_{SnS} \]

6. **Conclusion**:
The sulfide that is more soluble in pure water is **NiS**.

### Final Answer:
**NiS is the most soluble sulfide in pure water.**