Which of the following would enable you to remove `SO_(4)^(2-)` ions from a mixture of `C_2O_(4)^(2-)`, and `Cl^(-)` ions?

Options

A. NaOH
B. KOH
C. `Ba(OH)_2`
D. `BaSO_4`

Answer

To solve the question of how to remove \( SO_4^{2-} \) ions from a mixture containing \( C_2O_4^{2-} \) and \( Cl^{-} \) ions, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Ions in the Mixture**:
We have three ions in the mixture:
- \( SO_4^{2-} \) (sulfate ion)
- \( C_2O_4^{2-} \) (oxalate ion)
- \( Cl^{-} \) (chloride ion)

2. **Choose a Suitable Reagent**:
We need a reagent that can selectively precipitate \( SO_4^{2-} \) ions without affecting \( C_2O_4^{2-} \) and \( Cl^{-} \) ions.

3. **Consider Barium Hydroxide**:
Barium hydroxide, \( Ba(OH)_2 \), is a suitable reagent because it reacts with sulfate ions to form barium sulfate, \( BaSO_4 \), which is insoluble in water and precipitates out of the solution.

The reaction can be represented as:
\[ Ba(OH)_2 + SO_4^{2-} \rightarrow BaSO_4 \downarrow + 2OH^{-} \]

4. **Precipitation of Barium Sulfate**:
The formation of barium sulfate precipitate allows for the removal of \( SO_4^{2-} \) ions from the solution. The \( Cl^{-} \) and \( C_2O_4^{2-} \) ions remain in solution because they do not react with barium hydroxide to form insoluble precipitates.

5. **Conclusion**:
Therefore, using barium hydroxide will enable us to effectively remove \( SO_4^{2-} \) ions from the mixture.

### Final Answer:
The reagent that would enable you to remove \( SO_4^{2-} \) ions from a mixture of \( C_2O_4^{2-} \) and \( Cl^{-} \) ions is **Barium Hydroxide (\( Ba(OH)_2 \))**.