`SO_(2)Cl_(2)` (sulphuryl chloride ) reacts with water to given a mixture of `H_(2)SO_(4)` and HCL. What volume of 0.2 M `Ba(OH)_(2)` is needed to completely neutralize 25 mL of 0.2 `MSO_(2)Cl_(2)` solution:

Options

A. 25 mL
B. 50 mL
C. 100 mL
D. 200 mL

Answer

To solve the problem, we need to determine the volume of 0.2 M barium hydroxide (Ba(OH)₂) required to completely neutralize 25 mL of 0.2 M sulphuryl chloride (SO₂Cl₂) solution, which produces a mixture of sulfuric acid (H₂SO₄) and hydrochloric acid (HCl) upon reaction with water.

### Step-by-Step Solution:

1. **Write the reaction of SO₂Cl₂ with water**:
The balanced chemical reaction for the reaction of sulphuryl chloride with water is:
\[ \text{SO}_2\text{Cl}_2 + 2 \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4 + 2 \text{HCl} \]
This shows that 1 mole of SO₂Cl₂ produces 1 mole of H₂SO₄ and 2 moles of HCl.

2. **Calculate the number of moles of SO₂Cl₂**:
Given the molarity (0.2 M) and volume (25 mL) of the SO₂Cl₂ solution, we can calculate the number of moles:
\[ \text{Moles of SO}_2\text{Cl}_2 = \text{Molarity} \times \text{Volume (in L)} \]
Convert 25 mL to liters:
\[ 25 \text{ mL} = 0.025 \text{ L} \]
Now calculate the moles:
\[ \text{Moles of SO}_2\text{Cl}_2 = 0.2 \, \text{mol/L} \times 0.025 \, \text{L} = 0.005 \, \text{mol} = 5 \, \text{mmol} \]

3. **Determine the moles of H₂SO₄ and HCl produced**:
From the balanced equation:
- Moles of H₂SO₄ produced = 5 mmol (1:1 ratio with SO₂Cl₂)
- Moles of HCl produced = 5 mmol × 2 = 10 mmol (2:1 ratio with SO₂Cl₂)

4. **Calculate the total moles of acid to be neutralized**:
The total moles of acid (H₂SO₄ and HCl) that need to be neutralized by Ba(OH)₂:
- Moles of H₂SO₄ = 5 mmol
- Moles of HCl = 10 mmol

The neutralization reactions are:
\[ \text{H}_2\text{SO}_4 + \text{Ba(OH)}_2 \rightarrow \text{BaSO}_4 + 2 \text{H}_2\text{O} \]
\[ \text{HCl} + \text{Ba(OH)}_2 \rightarrow \text{BaCl}_2 + 2 \text{H}_2\text{O} \]

- For H₂SO₄: 1 mole of Ba(OH)₂ neutralizes 1 mole of H₂SO₄.
- For HCl: 1 mole of Ba(OH)₂ neutralizes 2 moles of HCl.

Therefore, the moles of Ba(OH)₂ required:
- Moles of Ba(OH)₂ for H₂SO₄ = 5 mmol
- Moles of Ba(OH)₂ for HCl = 10 mmol / 2 = 5 mmol

Total moles of Ba(OH)₂ required:
\[ \text{Total moles of Ba(OH)}_2 = 5 \, \text{mmol (for H}_2\text{SO}_4) + 5 \, \text{mmol (for HCl)} = 10 \, \text{mmol} \]

5. **Calculate the volume of Ba(OH)₂ solution needed**:
Using the molarity of Ba(OH)₂ (0.2 M):
\[ \text{Volume} = \frac{\text{Number of moles}}{\text{Molarity}} = \frac{10 \, \text{mmol}}{0.2 \, \text{mol/L}} = \frac{10 \times 10^{-3} \, \text{mol}}{0.2 \, \text{mol/L}} = 0.05 \, \text{L} = 50 \, \text{mL} \]

### Final Answer:
The volume of 0.2 M Ba(OH)₂ needed to completely neutralize 25 mL of 0.2 M SO₂Cl₂ solution is **50 mL**.