The only cations present in a slightly acidic solution are `Fe^(3+),Zn^(2+)` and `Cu^(2+)` . The regent that when added in excess to this solution would identify and separate `Fe^(3+)` in one step is

Options

A. 2 M HCl
B. 6 M `NH_(3)`
C. 6 M NaOH
D. `H_(2)S` gas

Answer

To solve the problem of identifying and separating `Fe^(3+)` from a slightly acidic solution containing `Fe^(3+)`, `Zn^(2+)`, and `Cu^(2+)`, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Cations Present**: The cations in the solution are `Fe^(3+)`, `Zn^(2+)`, and `Cu^(2+)`. We need to focus on separating `Fe^(3+)` from these cations.

2. **Consider the Reagents**: We have four options to consider as reagents:
- 2M HCl
- 6M NH3
- 6M NaOH
- H2S gas

3. **Analyze the Effect of Each Reagent**:
- **2M HCl**: Adding hydrochloric acid would not precipitate `Fe^(3+)` and would keep all cations in solution.
- **6M NH3**: Ammonia can react with `Fe^(3+)` to form a precipitate of `Fe(OH)3`, while `Zn^(2+)` and `Cu^(2+)` form soluble complexes with ammonia.
- **6M NaOH**: Sodium hydroxide would precipitate `Fe(OH)3`, but it may also precipitate `Zn(OH)2` and `Cu(OH)2`, making it less selective for `Fe^(3+)`.
- **H2S gas**: Hydrogen sulfide would precipitate `FeS`, but it would also precipitate `CuS`, making it non-selective.

4. **Choose the Best Reagent**: The reagent that selectively identifies and separates `Fe^(3+)` in one step is **6M NH3**. When added in excess, it forms a brown precipitate of `Fe(OH)3`, while `Zn^(2+)` and `Cu^(2+)` remain in solution as soluble complexes.

5. **Write the Reaction**: The reaction can be represented as:
\[ Fe^{3+} + 6NH_3 + 3H_2O \rightarrow Fe(OH)_3 (s) + [Zn(NH_3)_4]^{2+} + [Cu(NH_3)_4]^{2+} \]
Here, `Fe(OH)3` precipitates out, while `Zn` and `Cu` remain in solution.

### Conclusion:
The reagent that, when added in excess to the solution, would identify and separate `Fe^(3+)` is **6M NH3**.

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