`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

Answer

To solve the problem step by step, we will use the information provided in the question and the relationships between the volumes and normalities of the solutions involved.

### Step 1: Write the equation for neutralization
The neutralization reaction can be expressed in terms of milli-equivalents:
\[ \text{milli-equivalents of NaOH} + \text{milli-equivalents of Ba(OH)₂} = \text{milli-equivalents of HCl} \]
This can be mathematically represented as:
\[ X \cdot V_1 + Y \cdot V_2 = 0.1 \cdot 100 \]
Where \(X\) is the normality of NaOH, \(Y\) is the normality of Ba(OH)₂, \(V_1\) is the volume of NaOH, and \(V_2\) is the volume of Ba(OH)₂.

### Step 2: Substitute the known values
From the problem, we know:
- \(V_1/V_2 = 1/4\) implies \(V_1 = \frac{1}{4} V_2\)
- \(X/Y = 4\) implies \(X = 4Y\)

Substituting \(V_1\) and \(X\) in terms of \(V_2\) and \(Y\) into the equation:
\[ (4Y) \cdot \left(\frac{1}{4} V_2\right) + Y \cdot V_2 = 10 \]
This simplifies to:
\[ Y \cdot V_2 + Y \cdot V_2 = 10 \]
\[ 2Y \cdot V_2 = 10 \]

### Step 3: Solve for \(Y \cdot V_2\)
From the equation \(2Y \cdot V_2 = 10\):
\[ Y \cdot V_2 = 5 \]

### Step 4: Find \(X \cdot V_1\)
Now substituting back to find \(X \cdot V_1\):
\[ X \cdot V_1 = 4Y \cdot \left(\frac{1}{4} V_2\right) = Y \cdot V_2 = 5 \]

### Step 5: Calculate the fraction of acid neutralized by Ba(OH)₂
The fraction of acid neutralized by Ba(OH)₂ can be calculated as:
\[ \text{Fraction} = \frac{\text{milli-equivalents of Ba(OH)₂}}{\text{total milli-equivalents of HCl}} = \frac{Y \cdot V_2}{10} \]
Substituting \(Y \cdot V_2 = 5\):
\[ \text{Fraction} = \frac{5}{10} = \frac{1}{2} \]

### Final Answer
Thus, the fraction of the acid that is neutralized by Ba(OH)₂ is:
\[ \frac{1}{2} \text{ or } 0.5 \]