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. (a)23.6 mL
B. (b)25.6 mL
C. (c )26.3 mL
D. (d)32.6 mL

Explanation

To find out how many milliliters of `0.1 N H2SO4` solution are required to completely react with `0.125 g` of pure `Na2CO3`, we can follow these steps:

### Step 1: Write the balanced chemical equation
The reaction between sulfuric acid and sodium carbonate can be represented as:
\[ \text{H}_2\text{SO}_4 + \text{Na}_2\text{CO}_3 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]
This equation shows that 1 mole of sulfuric acid reacts with 1 mole of sodium carbonate.

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

The molar mass of `Na2CO3` is 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

Adding these together:
\[ \text{Molar mass of Na2CO3} = 46 + 12 + 48 = 106 \text{ g/mol} \]

Now, substituting the values:
\[ \text{Number of moles of Na2CO3} = \frac{0.125 \text{ g}}{106 \text{ g/mol}} \approx 0.00118 \text{ moles} \]

### Step 3: Determine the equivalent of Na2CO3
Since sodium carbonate is a bivalent salt, it has an n-factor of 2. Thus, the number of equivalents of `Na2CO3` is:
\[ \text{Equivalents of Na2CO3} = \text{Number of moles} \times \text{n-factor} = 0.00118 \text{ moles} \times 2 \approx 0.00236 \text{ equivalents} \]

### Step 4: Relate equivalents of H2SO4 to Na2CO3
From the balanced equation, we know that 1 equivalent of `H2SO4` reacts with 1 equivalent of `Na2CO3`. Therefore, the equivalents of `H2SO4` required will also be `0.00236`.

### Step 5: Calculate the volume of H2SO4 solution needed
Using the formula for normality:
\[ \text{Normality (N)} = \frac{\text{Equivalents}}{\text{Volume (L)}} \]
Rearranging gives us:
\[ \text{Volume (L)} = \frac{\text{Equivalents}}{\text{Normality}} \]

Substituting the values:
\[ \text{Volume (L)} = \frac{0.00236 \text{ equivalents}}{0.1 \text{ N}} = 0.0236 \text{ L} \]

### Step 6: Convert volume to milliliters
To convert liters to milliliters:
\[ \text{Volume (mL)} = 0.0236 \text{ L} \times 1000 = 23.6 \text{ mL} \]

### Final Answer
The volume of `0.1 N H2SO4` solution required is **23.6 mL**.

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