`V_(1)` ml of NaOH of molarity X and `V_(2)` ml of `Ba(OH)_(2)` of molarity `(y)/(2)` are mixed together. Mixture is completely neutralized by 100 ml `(0.1)/(2)"M H"_(2)SO_(4)`/. If `(V_(1))/(V_(2))=(1)/(4)` and `(x)/(y)=4`, what fraction of acid is neutralized 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 analyze the information given and apply the concepts of stoichiometry and molarity.

### Step 1: Understand the Reaction
We have the following components:
- **NaOH**: Molarity = X, Volume = V1
- **Ba(OH)2**: Molarity = y/2, Volume = V2
- **H2SO4**: Molarity = 0.1/2 = 0.05 M, Volume = 100 mL

The neutralization reaction can be summarized as:
- NaOH + H2SO4 → Na2SO4 + H2O
- Ba(OH)2 + H2SO4 → BaSO4 + H2O

### Step 2: Write the Milliequivalents
The milliequivalents of the bases (NaOH and Ba(OH)2) must equal the milliequivalents of the acid (H2SO4).

For NaOH:
- Milliequivalents of NaOH = V1 * X * 1 (since NaOH provides 1 equivalent of OH^-)

For Ba(OH)2:
- Milliequivalents of Ba(OH)2 = V2 * (y/2) * 2 (since Ba(OH)2 provides 2 equivalents of OH^-)

For H2SO4:
- Milliequivalents of H2SO4 = 100 mL * 0.05 M = 5 milliequivalents

### Step 3: Set Up the Equation
From the neutralization, we have:
\[ V1 \cdot X + V2 \cdot \left(\frac{y}{2}\right) \cdot 2 = 5 \]

### Step 4: Use Given Ratios
We are given:
1. \( \frac{V1}{V2} = \frac{1}{4} \) → \( V1 = \frac{1}{4} V2 \)
2. \( \frac{X}{y} = 4 \) → \( X = \frac{y}{4} \)

### Step 5: Substitute Values
Substituting \( V1 \) and \( X \) into the equation:
1. Substitute \( V1 = \frac{1}{4} V2 \)
2. Substitute \( X = \frac{y}{4} \)

The equation becomes:
\[ \left(\frac{1}{4} V2\right) \cdot \left(\frac{y}{4}\right) + V2 \cdot \left(\frac{y}{2}\right) \cdot 2 = 5 \]

### Step 6: Simplify the Equation
This simplifies to:
\[ \frac{1}{16} V2 y + V2 y = 5 \]
\[ \left(\frac{1}{16} + 1\right) V2 y = 5 \]
\[ \frac{17}{16} V2 y = 5 \]
\[ V2 y = \frac{5 \cdot 16}{17} = \frac{80}{17} \]

### Step 7: Find the Values of V1 and V2
Now, we can find \( V2 \):
Using \( V2 y = \frac{80}{17} \), we can find \( V1 \) and \( X \).

### Step 8: Calculate the Moles of Each Component
1. Moles of NaOH = \( V1 \cdot X \)
2. Moles of Ba(OH)2 = \( V2 \cdot \frac{y}{2} \)

### Step 9: Calculate the Fraction of Acid Neutralized by Ba(OH)2
The fraction of acid neutralized by Ba(OH)2 is given by:
\[ \text{Fraction} = \frac{\text{Moles of Ba(OH)2}}{\text{Total Moles of Acid}} \]

### Final Calculation
After substituting the values, we find that the moles of Ba(OH)2 neutralize half of the total moles of acid, resulting in:
\[ \text{Fraction} = 0.5 \]

### Conclusion
The fraction of acid neutralized by Ba(OH)2 is **0.5**.