How many millilitres of `0.1 NH_(2)SO_(4)` solution will be required for complete reaction with a solution containing 0.125 g of pure `Na_(2)CO_(3)`?

Options

A. 23.6 mL
B. 25.6 mL
C. 26.3 mL
D. 32.6 mL

Answer

To find out how many milliliters of `0.1 N H₂SO₄` solution will be required for complete reaction with a solution containing `0.125 g` of pure `Na₂CO₃`, we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced reaction between sulfuric acid (H₂SO₄) and sodium carbonate (Na₂CO₃) is:
\[ \text{H}_2\text{SO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{CO}_3 \]

From the balanced equation, we can see that 1 mole of H₂SO₄ reacts with 1 mole of Na₂CO₃.

### Step 2: Calculate the number of moles of Na₂CO₃
To find the number of moles of sodium carbonate, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]

The molar mass of Na₂CO₃ can be calculated as follows:
- Sodium (Na): 23 g/mol × 2 = 46 g/mol
- Carbon (C): 12 g/mol × 1 = 12 g/mol
- Oxygen (O): 16 g/mol × 3 = 48 g/mol

Total molar mass of Na₂CO₃:
\[ 46 + 12 + 48 = 106 \text{ g/mol} \]

Now, calculate the number of moles of Na₂CO₃:
\[ \text{Number of moles of Na}_2\text{CO}_3 = \frac{0.125 \text{ g}}{106 \text{ g/mol}} \]
\[ = 0.001179 \text{ moles} \]

### Step 3: Determine the equivalent factor (n-factor) for Na₂CO₃
Since Na₂CO₃ is a bivalent salt, it can donate 2 moles of Na⁺ ions per mole of Na₂CO₃. Thus, the n-factor for Na₂CO₃ is 2.

### Step 4: Calculate the normality of Na₂CO₃
The equivalent of Na₂CO₃ can be calculated as:
\[ \text{Number of equivalents} = \text{Number of moles} \times \text{n-factor} \]
\[ = 0.001179 \text{ moles} \times 2 = 0.002358 \text{ equivalents} \]

### Step 5: Use the relation N₁V₁ = N₂V₂
Where:
- N₁ = Normality of H₂SO₄ = 0.1 N
- V₁ = Volume of H₂SO₄ (what we want to find)
- N₂ = Normality of Na₂CO₃ = 0.002358 equivalents (calculated)
- V₂ = Volume of Na₂CO₃ (not needed here)

Since we are looking for the volume of H₂SO₄ that will react with the equivalent of Na₂CO₃:
\[ N_1 \times V_1 = N_2 \times \text{Number of equivalents of Na}_2\text{CO}_3 \]
\[ 0.1 \times V_1 = 0.002358 \]

### Step 6: Solve for V₁
\[ V_1 = \frac{0.002358}{0.1} = 0.02358 \text{ L} \]

### Step 7: Convert to milliliters
To convert liters to milliliters:
\[ V_1 = 0.02358 \text{ L} \times 1000 = 23.58 \text{ mL} \]

### Final Answer
The volume of `0.1 N H₂SO₄` required for complete reaction with `0.125 g` of pure `Na₂CO₃` is approximately **23.58 mL**.

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