If 25mL of a `H_(2)SO_(4)` solution reacts completely with `1.06g` of pure `Na_(2)CO_(3)` , what is the normality of this acid solution :

Options

A. 1 N
B. 0.5 N
C. 1.8 N
D. 0.8 N

Answer

To find the normality of the `H₂SO₄` solution that reacts with `Na₂CO₃`, we can follow these steps:

### Step 1: Identify the reaction
The reaction between sulfuric acid (`H₂SO₄`) and sodium carbonate (`Na₂CO₃`) can be represented as:
\[ H_2SO_4 + Na_2CO_3 \rightarrow Na_2SO_4 + H_2O + CO_2 \]

### Step 2: Calculate the number of milli equivalents of `Na₂CO₃`
To find the number of milli equivalents of `Na₂CO₃`, we use the formula:
\[ \text{Number of milli equivalents} = \frac{\text{Weight (g)}}{\text{Equivalent weight (g/equiv)}} \times 1000 \]

#### Step 2.1: Calculate the equivalent weight of `Na₂CO₃`
The equivalent weight of a compound can be calculated using the formula:
\[ \text{Equivalent weight} = \frac{\text{Molar mass}}{n} \]
Where `n` is the number of acidic protons (or basic hydroxides) that can react.

- The molar mass of `Na₂CO₃` is calculated as follows:
- Na: 23 g/mol × 2 = 46 g/mol
- C: 12 g/mol = 12 g/mol
- O: 16 g/mol × 3 = 48 g/mol
- Total = 46 + 12 + 48 = 106 g/mol

- For `Na₂CO₃`, `n` (the number of replaceable H⁺ ions) is 2 since it can neutralize 2 moles of acid.

Thus, the equivalent weight of `Na₂CO₃` is:
\[ \text{Equivalent weight} = \frac{106 \text{ g/mol}}{2} = 53 \text{ g/equiv} \]

#### Step 2.2: Calculate the number of milli equivalents of `Na₂CO₃`
Now substituting the values into the formula:
\[ \text{Number of milli equivalents of } Na_2CO_3 = \frac{1.06 \text{ g}}{53 \text{ g/equiv}} \times 1000 \]

Calculating this gives:
\[ \text{Number of milli equivalents of } Na_2CO_3 = \frac{1.06}{53} \times 1000 \approx 20 \text{ milli equivalents} \]

### Step 3: Set up the equation for normality
According to the principle of neutralization:
\[ \text{Number of milli equivalents of acid} = \text{Number of milli equivalents of base} \]

Let `N` be the normality of the `H₂SO₄` solution. The number of milli equivalents of `H₂SO₄` can be calculated as:
\[ \text{Number of milli equivalents of } H_2SO_4 = N \times \text{Volume (mL)} \]
\[ = N \times 25 \]

### Step 4: Equate the milli equivalents
Setting the milli equivalents of acid equal to those of the base:
\[ N \times 25 = 20 \]

### Step 5: Solve for normality (N)
Now solve for `N`:
\[ N = \frac{20}{25} = 0.8 \]

### Conclusion
The normality of the `H₂SO₄` solution is **0.8 N**.

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