25 ml of a solution of `Na_(2)CO_(3)` having a specific gravity of 1.25 required 32.9 ml of a solution of HCl containing 109.5 grams of the acid per litre for complete neutralization. Calculate the volume of 0.84N-`H_(2)SO_(4)` that will be completely neutralized by 125 grams of the `Na_(2)CO_(3)` solution

Options

A. 460 ml
B. 540 ml
C. 480 ml
D. 470 ml

Explanation

To solve the problem step by step, we will follow the sequence of calculations as outlined in the video transcript.

### Step 1: Calculate the mass of Na₂CO₃ in 25 ml of solution
Given:
- Specific gravity of Na₂CO₃ solution = 1.25
- Volume of solution = 25 ml

Using the formula:
\[ \text{Mass} = \text{Specific Gravity} \times \text{Volume} \]
\[ \text{Mass of Na}_2\text{CO}_3 = 1.25 \times 25 \, \text{g} = 31.25 \, \text{g} \]

### Step 2: Calculate the mass of HCl used for neutralization
Given:
- Volume of HCl solution = 32.9 ml
- Concentration of HCl = 109.5 g/L

First, convert the volume of HCl from ml to L:
\[ \text{Volume of HCl in L} = \frac{32.9}{1000} = 0.0329 \, \text{L} \]

Now, calculate the mass of HCl:
\[ \text{Mass of HCl} = \text{Volume in L} \times \text{Concentration} = 0.0329 \times 109.5 \, \text{g} = 3.60 \, \text{g} \]

### Step 3: Calculate the number of moles of HCl
Molar mass of HCl = 36.5 g/mol.
\[ \text{Number of moles of HCl} = \frac{\text{Mass}}{\text{Molar Mass}} = \frac{3.60}{36.5} \approx 0.0987 \, \text{moles} \]

### Step 4: Determine the number of moles of Na₂CO₃
From the neutralization reaction:
\[ \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
This indicates that 1 mole of Na₂CO₃ reacts with 2 moles of HCl.

Thus, the number of moles of Na₂CO₃ can be calculated as:
\[ \text{Number of moles of Na}_2\text{CO}_3 = \frac{0.0987}{2} \approx 0.04935 \, \text{moles} \]

### Step 5: Calculate the number of moles of Na₂CO₃ in 125 g
Using the mass of Na₂CO₃ calculated earlier (31.25 g corresponds to 0.04935 moles):
\[ \text{Moles of Na}_2\text{CO}_3 \text{ in 125 g} = \frac{125 \, \text{g}}{31.25 \, \text{g}} \times 0.04935 \, \text{moles} \approx 0.1974 \, \text{moles} \]

### Step 6: Calculate the equivalent moles of H₂SO₄ required
From the reaction:
\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]
1 mole of Na₂CO₃ reacts with 1 mole of H₂SO₄. Therefore, the number of moles of H₂SO₄ required is:
\[ \text{Moles of H}_2\text{SO}_4 = 0.1974 \, \text{moles} \]

### Step 7: Calculate the equivalent moles of H₂SO₄
Since H₂SO₄ is diprotic, the equivalents of H₂SO₄ are:
\[ \text{Equivalents of H}_2\text{SO}_4 = 0.1974 \times 2 = 0.3948 \, \text{equivalents} \]

### Step 8: Calculate the volume of 0.84 N H₂SO₄ required
Using the formula:
\[ \text{Volume} = \frac{\text{Equivalents}}{\text{Normality}} = \frac{0.3948}{0.84} \approx 0.470 \, \text{L} \]

Convert to ml:
\[ \text{Volume in ml} = 0.470 \times 1000 = 470 \, \text{ml} \]

### Final Answer
The volume of 0.84 N H₂SO₄ that will be completely neutralized by 125 grams of the Na₂CO₃ solution is approximately **470 ml**.