A solution of `0.01M Fe^(2+)` in a saturated `H_(2)S` solution and (i) `0.2M of H^(o+) (ii) 0.001M of H^(o+).(K_(1) xx K_(2) of H_(2)S = 10^(-21), K_(sp) FeS = 3.7 xx 10^(-19))`. Which of the following statements is//are correct

Options

A. `FeS` will precipitate in solution (i).
B. `FeS` will not precipitate in solution (i).
C. `FeS` will precipitate in solution (ii).
D. `FeS` will precipitate in solution (ii).

Explanation

To solve the problem, we need to analyze the precipitation of iron(II) sulfide (FeS) in two different solutions containing varying concentrations of hydrogen ions (H⁺) in a saturated H₂S solution. We will use the given dissociation constants and solubility product constant (K_sp) to determine whether FeS will precipitate in each case.

### Step-by-Step Solution:

1. **Understanding the Dissociation of H₂S:**
H₂S dissociates in water according to the following reactions:
- \( H_2S \rightleftharpoons H^+ + HS^- \) (K₁)
- \( HS^- \rightleftharpoons H^+ + S^{2-} \) (K₂)

The overall dissociation can be represented as:
\[ H_2S \rightleftharpoons 2H^+ + S^{2-} \]
The product of the dissociation constants is given as:
\[ K_{1} \times K_{2} = 10^{-21} \]

2. **Analyzing the First Solution (0.2 M H⁺):**
- The concentration of H⁺ ions is 0.2 M.
- Using the dissociation constants, we can find the concentration of sulfide ions (S²⁻):
\[ [S^{2-}] = \frac{K_{1} \times K_{2}}{[H^+]^2} = \frac{10^{-21}}{(0.2)^2} = \frac{10^{-21}}{0.04} = 2.5 \times 10^{-20} \, \text{M} \]

3. **Calculating the Ionic Product (IP) for FeS:**
- The ionic product (IP) for FeS is given by:
\[ IP = [Fe^{2+}][S^{2-}] = (0.01)(2.5 \times 10^{-20}) = 2.5 \times 10^{-22} \]
- Comparing this with the K_sp of FeS (3.7 × 10⁻¹⁹):
\[ IP < K_{sp} \quad \Rightarrow \quad \text{FeS will not precipitate.} \]

4. **Analyzing the Second Solution (0.001 M H⁺):**
- The concentration of H⁺ ions is 0.001 M.
- Again, we calculate the concentration of sulfide ions (S²⁻):
\[ [S^{2-}] = \frac{K_{1} \times K_{2}}{[H^+]^2} = \frac{10^{-21}}{(0.001)^2} = \frac{10^{-21}}{10^{-6}} = 1 \times 10^{-15} \, \text{M} \]

5. **Calculating the Ionic Product (IP) for FeS in the Second Solution:**
- The ionic product (IP) for FeS is:
\[ IP = [Fe^{2+}][S^{2-}] = (0.01)(1 \times 10^{-15}) = 1 \times 10^{-17} \]
- Comparing this with the K_sp of FeS:
\[ IP > K_{sp} \quad \Rightarrow \quad \text{FeS will precipitate.} \]

### Conclusion:
- In the first solution (0.2 M H⁺), FeS does not precipitate.
- In the second solution (0.001 M H⁺), FeS does precipitate.

### Correct Statements:
- FeS will not precipitate in the first solution (True).
- FeS will precipitate in the second solution (True).