If `Pb[CH_(3)COO]_(2)` and `Na_(2)S` are mixed and dissolved in water and the solution is filtered then the filterate will give test of

Options

A. `Pb^(+2)`
B. `CH_(3)COO^(-)`
C. `S^(-2)`
D. `Na^(+)`

Answer

To solve the question regarding the reaction between lead(II) acetate, \( Pb(CH_3COO)_2 \), and sodium sulfide, \( Na_2S \), we can follow these steps:

### Step-by-Step Solution:

1. **Write the Chemical Equation**:
Start by writing the balanced chemical equation for the reaction between lead(II) acetate and sodium sulfide.
\[ Pb(CH_3COO)_2 (aq) + Na_2S (aq) \rightarrow PbS (s) + 2 CH_3COONa (aq) \]

2. **Identify the Products**:
From the reaction, we can see that lead(II) sulfide, \( PbS \), is formed as a solid precipitate, while sodium acetate, \( CH_3COONa \), remains in the aqueous solution.

3. **Filtration**:
When the mixture is filtered, the solid precipitate \( PbS \) will be removed from the solution. The filtrate will contain the remaining components, which in this case is sodium acetate.

4. **Analyze the Filtrate**:
The filtrate consists of sodium acetate, which dissociates into sodium ions \( Na^+ \) and acetate ions \( CH_3COO^- \) in solution.

5. **Testing the Filtrate**:
To determine what tests can be performed on the filtrate, we can look for the presence of acetate ions \( CH_3COO^- \) and sodium ions \( Na^+ \). The presence of acetate ions can be tested using specific reagents that react with acetate to produce characteristic results.

### Conclusion:
The filtrate will give a test for acetate ions \( CH_3COO^- \) and sodium ions \( Na^+ \).