`KMnO_4` can be prepared from`K_2MnO_4` as per the reaction:
`3MnO_4^(2-) + 2H_2O ⇋ 2MnO_4^(-) + MnO_2+ 4OH^(-)`
The reaction can go the completion by removing `OH^(ɵ)` ions by adding.

Options

A. KOH
B. `CO_2`
C. `SO_2`
D. `HCI`

Answer

To solve the problem, we need to determine which compound can be added to the reaction to remove hydroxide ions (OH⁻) and drive the equilibrium towards the formation of KMnO₄.

### Step-by-Step Solution:

1. **Write the given reaction**:
\[ 3 \text{MnO}_4^{2-} + 2 \text{H}_2\text{O} \rightleftharpoons 2 \text{MnO}_4^{-} + \text{MnO}_2 + 4 \text{OH}^{-} \]

2. **Identify the goal**:
We want to shift the equilibrium to the right, which means we need to remove OH⁻ ions from the solution.

3. **Analyze the options**:
We have four options to consider: KOH, CO₂, SO₂, and HCl.

4. **Evaluate KOH**:
- KOH is a strong base and would increase the concentration of OH⁻ ions. This would shift the equilibrium to the left, which is not what we want.
- **Conclusion**: KOH cannot be used.

5. **Evaluate HCl**:
- HCl is a strong acid and would react with OH⁻ ions to form water. However, it would also introduce more H⁺ ions into the solution, which could shift the equilibrium back to the left.
- **Conclusion**: HCl cannot be used.

6. **Evaluate SO₂**:
- SO₂ reacts with water to form sulfurous acid (H₂SO₃), which is a weak acid. This would not effectively remove OH⁻ ions and could also shift the equilibrium back to the left.
- **Conclusion**: SO₂ cannot be used.

7. **Evaluate CO₂**:
- CO₂ reacts with water to form carbonic acid (H₂CO₃), which is also a weak acid. The addition of CO₂ would help neutralize OH⁻ ions, effectively reducing their concentration and shifting the equilibrium to the right.
- **Conclusion**: CO₂ is the suitable compound to remove OH⁻ ions.

8. **Final Answer**:
The compound that can be added to remove OH⁻ ions and drive the reaction to completion is **CO₂**.