`V_(1) mL` of `NaOH` of normality `X` and `V_(2) mL` of `Ba(OH)_(2)` of mormality `Y` are mixed together. The mixture is completely neutralised by `100 mL` of `0.1 N HCl`. If `V_(1)//V_(2)=1/4` and `X/Y=4`, what fraction of the acid is neutralised by `Ba(OH)_(2)`?

Options

A. `0.5`
B. `0.25`
C. `0.33`
D. `0.67`

Explanation

To solve the problem, we need to determine how much of the acid (HCl) is neutralized by Ba(OH)₂ when mixed with NaOH. We will use the information given in the question to find the required fraction.

### Step-by-Step Solution:

1. **Understanding Normality and Milliequivalents**:
- Normality (N) is a measure of concentration equivalent to the number of equivalents of solute per liter of solution.
- For bases like NaOH and Ba(OH)₂, the normality is equal to the molarity multiplied by the number of hydroxide ions (OH⁻) they can provide.
- NaOH provides 1 OH⁻, so its normality is the same as its molarity.
- Ba(OH)₂ provides 2 OH⁻, so its normality is twice its molarity.

2. **Setting Up the Equations**:
- Let’s denote the milliequivalents of NaOH and Ba(OH)₂:
- Milliequivalents of NaOH = \( V_1 \times X \)
- Milliequivalents of Ba(OH)₂ = \( V_2 \times 2Y \)
- The total milliequivalents of HCl neutralized by the mixture is:
- Milliequivalents of HCl = \( 100 \, \text{mL} \times 0.1 \, N = 10 \, \text{meq} \)

3. **Using the Given Ratios**:
- From the question, we have:
- \( \frac{V_1}{V_2} = \frac{1}{4} \) ⇒ \( V_1 = \frac{1}{4} V_2 \)
- \( \frac{X}{Y} = 4 \) ⇒ \( X = 4Y \)

4. **Substituting Values**:
- Substitute \( V_1 \) in terms of \( V_2 \) into the equation for total neutralization:
\[ V_1 \times X + V_2 \times 2Y = 10 \]
- Replacing \( V_1 \) with \( \frac{1}{4} V_2 \) and \( X \) with \( 4Y \):
\[ \left(\frac{1}{4} V_2\right) \times (4Y) + V_2 \times 2Y = 10 \]
- Simplifying:
\[ V_2Y + V_2 \times 2Y = 10 \]
\[ 3V_2Y = 10 \]
\[ V_2Y = \frac{10}{3} \]

5. **Finding the Amount Neutralized by Ba(OH)₂**:
- Milliequivalents of Ba(OH)₂:
\[ \text{Milliequivalents of Ba(OH)₂} = V_2 \times 2Y = 2 \times \frac{10}{3} = \frac{20}{3} \]
- Now, to find the fraction of acid neutralized by Ba(OH)₂:
\[ \text{Fraction} = \frac{\text{Milliequivalents of Ba(OH)₂}}{\text{Total Milliequivalents of HCl}} = \frac{\frac{20}{3}}{10} = \frac{2}{3} \]

### Final Answer:
The fraction of the acid neutralized by Ba(OH)₂ is \( \frac{2}{3} \).