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 will analyze the two solutions separately, considering the concentrations of \( \text{H}^+ \) ions and the implications for the solubility of \( \text{FeS} \).

### Step 1: Analyze the first solution (0.2 M \( \text{H}^+ \))

1. **Given Data**:
- \( [\text{Fe}^{2+}] = 0.01 \, \text{M} \)
- \( [\text{H}^+] = 0.2 \, \text{M} \)
- \( K_1 \times K_2 = 10^{-21} \) (for \( \text{H}_2\text{S} \))
- \( K_{sp}(\text{FeS}) = 3.7 \times 10^{-19} \)

2. **Calculate \( [\text{S}^{2-}] \)**:
Using the equilibrium expression for \( \text{H}_2\text{S} \):
\[ K_1 \times K_2 = \frac{[\text{H}^+]^2 [\text{S}^{2-}]}{[\text{H}_2\text{S}]} \]
Rearranging gives:
\[ [\text{S}^{2-}] = \frac{K_1 \times K_2 \times [\text{H}_2\text{S}]}{[\text{H}^+]^2} \]
Substituting the values:
\[ [\text{S}^{2-}] = \frac{10^{-21} \times 0.1}{(0.2)^2} = \frac{10^{-21} \times 0.1}{0.04} = 2.5 \times 10^{-21} \, \text{M} \]

3. **Calculate \( Q_{sp} \)**:
\[ Q_{sp} = [\text{Fe}^{2+}][\text{S}^{2-}] = (0.01)(2.5 \times 10^{-21}) = 2.5 \times 10^{-23} \]

4. **Compare \( Q_{sp} \) with \( K_{sp} \)**:
Since \( Q_{sp} (2.5 \times 10^{-23}) < K_{sp} (3.7 \times 10^{-19}) \), no precipitation occurs.

### Step 2: Analyze the second solution (0.001 M \( \text{H}^+ \))

1. **Given Data**:
- \( [\text{H}^+] = 0.001 \, \text{M} \)

2. **Calculate \( [\text{S}^{2-}] \)**:
Using the same equilibrium expression:
\[ [\text{S}^{2-}] = \frac{K_1 \times K_2 \times [\text{H}_2\text{S}]}{[\text{H}^+]^2} \]
Substituting the values:
\[ [\text{S}^{2-}] = \frac{10^{-21} \times 0.1}{(0.001)^2} = \frac{10^{-21} \times 0.1}{10^{-6}} = 10^{-16} \, \text{M} \]

3. **Calculate \( Q_{sp} \)**:
\[ Q_{sp} = [\text{Fe}^{2+}][\text{S}^{2-}] = (0.01)(10^{-16}) = 10^{-18} \]

4. **Compare \( Q_{sp} \) with \( K_{sp} \)**:
Since \( Q_{sp} (10^{-18}) < K_{sp} (3.7 \times 10^{-19}) \), precipitation occurs.

### Conclusion:
- In the first solution, \( \text{FeS} \) does not precipitate.
- In the second solution, \( \text{FeS} \) precipitates.

### Correct Statements:
- Statement B: \( \text{FeS} \) precipitate will not be formed in solution 1 (Correct).
- Statement C: \( \text{FeS} \) precipitate will be formed in solution 2 (Correct).