`Cu^(2+)` and `Ag^(+)` are both present in the same solution.
To precipitate one of the ions and leaves the other in solution, add

Options

A. `H_(2)S(aq)`
B. `HCl(aq)`
C. `HNO_(3)(aq)`
D. `NH_(4)NO_(3)(aq)`

Explanation

To solve the problem of selectively precipitating either \( \text{Cu}^{2+} \) or \( \text{Ag}^{+} \) from a solution containing both ions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ions**: We have two ions in the solution: \( \text{Cu}^{2+} \) (Copper ion) and \( \text{Ag}^{+} \) (Silver ion).

2. **Choose a Reagent**: To selectively precipitate one of the ions, we need to choose a suitable reagent. In this case, we can use hydrochloric acid (HCl).

3. **Add HCl to the Solution**: When HCl is added to the solution, it reacts with \( \text{Ag}^{+} \) ions to form silver chloride (AgCl), which is an insoluble precipitate:
\[ \text{Ag}^{+} + \text{Cl}^{-} \rightarrow \text{AgCl} \, (s) \]
This means that \( \text{Ag}^{+} \) will precipitate out of the solution.

4. **Leave \( \text{Cu}^{2+} \) in Solution**: Since \( \text{Cu}^{2+} \) does not precipitate with HCl, it remains in the solution.

5. **Further Precipitation of \( \text{Cu}^{2+} \)**: If we want to precipitate \( \text{Cu}^{2+} \) after removing \( \text{Ag}^{+} \), we can pass hydrogen sulfide (H2S) gas through the solution. This will lead to the formation of copper sulfide (CuS), which is also an insoluble precipitate:
\[ \text{Cu}^{2+} + \text{H}_2\text{S} \rightarrow \text{CuS} \, (s) \]

### Conclusion:
To precipitate \( \text{Ag}^{+} \) first and leave \( \text{Cu}^{2+} \) in solution, you should add **HCl** to the solution. Afterward, if you wish to precipitate \( \text{Cu}^{2+} \), you can pass H2S gas.

### Final Answer:
Add **HCl** to the solution.