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)`
At what value of `pH, NiS` will start to precipitate?

Options

A. `12.76`
B. `7`
C. `1.24`
D. 4`

Explanation

To determine the pH at which NiS will start to precipitate in a solution containing 0.1 M each of Sn²⁺, Cd²⁺, and Ni²⁺ ions when H₂S gas is passed, we need to follow these steps:

### Step 1: Understand the Precipitation Conditions
NiS will precipitate when the concentration of sulfide ions (S²⁻) exceeds the solubility product (Ksp) of NiS. The Ksp of NiS is given as 3 × 10⁻²¹.

### Step 2: Write the Expression for Ksp
The Ksp expression for NiS is:
\[ K_{sp} = [Ni^{2+}][S^{2-}] \]
Given that the concentration of Ni²⁺ is 0.1 M, we can rewrite the Ksp expression:
\[ 3 \times 10^{-21} = (0.1)[S^{2-}] \]

### Step 3: Solve for [S²⁻]
Rearranging the equation to find the concentration of sulfide ions:
\[ [S^{2-}] = \frac{3 \times 10^{-21}}{0.1} = 3 \times 10^{-20} \, \text{M} \]

### Step 4: Relate [S²⁻] to pH
The sulfide ions (S²⁻) are produced from the dissociation of H₂S:
1. \( H_2S \rightleftharpoons H^+ + HS^- \) (with K₁ = 1 × 10⁻⁷)
2. \( HS^- \rightleftharpoons H^+ + S^{2-} \) (with K₂ = 1 × 10⁻¹⁴)

The overall reaction can be combined to find the relationship between [H⁺] and [S²⁻]:
\[ K_1 \cdot K_2 = [H^+]^2 [S^{2-}] \]

Substituting the values of K₁ and K₂:
\[ (1 \times 10^{-7})(1 \times 10^{-14}) = [H^+]^2 [S^{2-}] \]
\[ 1 \times 10^{-21} = [H^+]^2 (3 \times 10^{-20}) \]

### Step 5: Solve for [H⁺]
Rearranging to find [H⁺]:
\[ [H^+]^2 = \frac{1 \times 10^{-21}}{3 \times 10^{-20}} \]
\[ [H^+]^2 = \frac{1}{3} \times 10^{-1} \]
\[ [H^+] = \sqrt{\frac{1}{3} \times 10^{-1}} \]
\[ [H^+] \approx 5.77 \times 10^{-11} \, \text{M} \]

### Step 6: Calculate pH
Now, we can find the pH:
\[ pH = -\log[H^+] \]
\[ pH = -\log(5.77 \times 10^{-11}) \]
\[ pH \approx 10.24 \]

### Conclusion
The value of pH at which NiS will start to precipitate is approximately **10.24**.