1 g of oleum sample is dilute with water. The solution required 54 mL of 0.4 N NaOH for complete neutralization. Find the percentage of free `SO_(3)` in the sample?

Answer

To solve the problem of finding the percentage of free SO₃ in the oleum sample, we can follow these steps:

### Step 1: Understand the Composition of Oleum
Oleum is a solution of sulfur trioxide (SO₃) in sulfuric acid (H₂SO₄). When oleum is diluted with water, it can release SO₃ and H₂SO₄. The reaction with NaOH will neutralize both SO₃ and H₂SO₄.

### Step 2: Write the Neutralization Reaction
The neutralization reactions are:
1. SO₃ + 2 NaOH → Na₂SO₄ + H₂O
2. H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O

### Step 3: Calculate the Milliequivalents of NaOH Used
Given:
- Volume of NaOH solution = 54 mL
- Normality of NaOH = 0.4 N

Milliequivalents of NaOH used can be calculated as:
\[ \text{Milliequivalents of NaOH} = \text{Volume (L)} \times \text{Normality (N)} = \frac{54}{1000} \times 0.4 = 0.0216 \text{ equivalents} \]

### Step 4: Set Up the Equation for Oleum
Let \( x \) be the percentage of SO₃ in the 1 g oleum sample. Therefore, the mass of SO₃ is \( \frac{x}{100} \) g and the mass of H₂SO₄ is \( 1 - \frac{x}{100} \) g.

- Molar mass of SO₃ = 80 g/mol, so equivalent weight = 40 g/equiv (since it reacts with 2 moles of NaOH).
- Molar mass of H₂SO₄ = 98 g/mol, so equivalent weight = 49 g/equiv (since it reacts with 2 moles of NaOH).

Now, we can calculate the milliequivalents contributed by each component:
1. For SO₃:
\[ \text{Milliequivalents of SO₃} = \frac{\frac{x}{100}}{40} \times 1000 = \frac{25x}{100} \text{ meq} \]
2. For H₂SO₄:
\[ \text{Milliequivalents of H₂SO₄} = \frac{1 - \frac{x}{100}}{49} \times 1000 = \frac{1000(1 - \frac{x}{100})}{49} \text{ meq} \]

### Step 5: Set Up the Total Milliequivalents Equation
The total milliequivalents of the oleum must equal the milliequivalents of NaOH used:
\[ \frac{25x}{100} + \frac{1000(1 - \frac{x}{100})}{49} = 21.6 \]

### Step 6: Solve the Equation
Multiply through by 100 to eliminate the fraction:
\[ 25x + \frac{1000(100 - x)}{49} = 2160 \]

Now, multiply everything by 49 to eliminate the denominator:
\[ 1225x + 1000(100 - x) = 105840 \]
\[ 1225x + 100000 - 1000x = 105840 \]
\[ 225x = 5840 \]
\[ x = \frac{5840}{225} \approx 25.93 \text{ (approximately 26)} \]

### Step 7: Calculate the Percentage of SO₃
Thus, the percentage of SO₃ in the oleum sample is approximately 26%.

### Final Answer
The percentage of free SO₃ in the sample is **26%**.

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