When 100 " mL of " 0.1 M `Ba(OH)_(2)` is neutralised with a mixture of x " mL of " 0.1 M HCl and y " mL of " 0.2 M `H_(2)SO_(3)` using methyl orange indicator what is value of x and y?

Options

A. 200100
B. 100200
C. 300200
D. 200300

Explanation

To solve the problem, we need to determine the volumes \( x \) and \( y \) of hydrochloric acid (HCl) and sulfurous acid (H₂SO₃) respectively that will neutralize 100 mL of 0.1 M barium hydroxide (Ba(OH)₂).

### Step-by-Step Solution:

1. **Identify the reaction**:
Barium hydroxide (Ba(OH)₂) reacts with hydrochloric acid (HCl) and sulfurous acid (H₂SO₃) to form barium chloride (BaCl₂), water (H₂O), and sulfurous acid (H₂SO₃) will also dissociate to provide additional H⁺ ions.

2. **Calculate the number of equivalents of Ba(OH)₂**:
- Molarity of Ba(OH)₂ = 0.1 M
- Volume of Ba(OH)₂ = 100 mL = 0.1 L
- Basicity of Ba(OH)₂ = 2 (since it can donate 2 OH⁻ ions)
- Number of equivalents (N) = Molarity × Volume × Basicity
\[ N = 0.1 \, \text{mol/L} \times 0.1 \, \text{L} \times 2 = 0.02 \, \text{equivalents} \]

3. **Set up the equation for HCl**:
- Molarity of HCl = 0.1 M
- Basicity of HCl = 1 (since it can donate 1 H⁺ ion)
- Let the volume of HCl be \( x \) mL = \( \frac{x}{1000} \) L
- Number of equivalents from HCl = Molarity × Volume × Basicity
\[ N_{HCl} = 0.1 \, \text{mol/L} \times \frac{x}{1000} \times 1 = 0.0001x \, \text{equivalents} \]

4. **Set up the equation for H₂SO₃**:
- Molarity of H₂SO₃ = 0.2 M
- Basicity of H₂SO₃ = 2 (since it can donate 2 H⁺ ions)
- Let the volume of H₂SO₃ be \( y \) mL = \( \frac{y}{1000} \) L
- Number of equivalents from H₂SO₃ = Molarity × Volume × Basicity
\[ N_{H2SO3} = 0.2 \, \text{mol/L} \times \frac{y}{1000} \times 2 = 0.0004y \, \text{equivalents} \]

5. **Set up the neutralization equation**:
Since the total number of equivalents from HCl and H₂SO₃ must equal the number of equivalents from Ba(OH)₂:
\[ 0.0001x + 0.0004y = 0.02 \]

6. **Solve for \( x \) and \( y \)**:
We can express \( y \) in terms of \( x \) or vice versa. For simplicity, let's express \( y \):
\[ 0.0004y = 0.02 - 0.0001x \]
\[ y = \frac{0.02 - 0.0001x}{0.0004} \]
\[ y = 50 - 0.25x \]

7. **Use the second equation**:
We know that the total volume of the acid mixture can be expressed as:
\[ x + y = 100 \quad \text{(assuming we want a total of 100 mL of acid)} \]
Substitute \( y \) from the previous equation:
\[ x + (50 - 0.25x) = 100 \]
\[ 0.75x = 50 \]
\[ x = \frac{50}{0.75} = 66.67 \, \text{mL} \]

8. **Find \( y \)**:
Substitute \( x \) back into the equation for \( y \):
\[ y = 50 - 0.25(66.67) = 50 - 16.67 = 33.33 \, \text{mL} \]

### Final Values:
- \( x = 66.67 \, \text{mL} \) (volume of HCl)
- \( y = 33.33 \, \text{mL} \) (volume of H₂SO₃)