A solution contains `Na_(2)CO_(3)` and `NaHCO_(3). 10 mL` of the solution required `2.5 mL "of" 0.1M H_(2)SO_(4)` for neutralisation using phenolphthalein as indicator. Methyl orange is then added when a further `2.5 mL "of" 0.2M H_(2)SO_(4)`was required. The amount of `Na_(2)CO_(3)` and `NaHCO_(3)` in `1 "litre"` of the solution is:

Options

A. `5.3 g` and `4.2 g`
B. `3.3 g` and `6.2 g`
C. `4.2 g` and `5.3 g`
D. `6.2 g` and `3.3 g`

Explanation

To solve the problem, we need to determine the amounts of sodium carbonate (Na₂CO₃) and sodium bicarbonate (NaHCO₃) in a 1-liter solution based on the neutralization reactions with sulfuric acid (H₂SO₄) using two different indicators.

### Step 1: Analyze the first neutralization with phenolphthalein
- **Volume of H₂SO₄ used:** 2.5 mL of 0.1 M
- **Moles of H₂SO₄ used:**
\[ \text{Moles of H₂SO₄} = \text{Molarity} \times \text{Volume (L)} = 0.1 \, \text{mol/L} \times 0.0025 \, \text{L} = 0.00025 \, \text{mol} \]

- **N-factor for Na₂CO₃:** 2 (because it can donate 2 moles of Na⁺ ions)
- **Moles of Na₂CO₃ neutralized:**
\[ \text{Moles of Na₂CO₃} = \text{Moles of H₂SO₄} \times \text{N-factor of Na₂CO₃} = 0.00025 \, \text{mol} \times 2 = 0.0005 \, \text{mol} \]

- **Mass of Na₂CO₃ in 10 mL:**
\[ \text{Mass} = \text{Moles} \times \text{Molar mass of Na₂CO₃} = 0.0005 \, \text{mol} \times 106 \, \text{g/mol} = 0.053 \, \text{g} \]

### Step 2: Scale up to 1 liter
- Since 10 mL of solution contains 0.053 g of Na₂CO₃, in 1 liter (1000 mL), it will contain:
\[ \text{Mass in 1 L} = 0.053 \, \text{g} \times 100 = 5.3 \, \text{g} \]

### Step 3: Analyze the second neutralization with methyl orange
- **Volume of H₂SO₄ used:** 2.5 mL of 0.2 M
- **Moles of H₂SO₄ used:**
\[ \text{Moles of H₂SO₄} = 0.2 \, \text{mol/L} \times 0.0025 \, \text{L} = 0.0005 \, \text{mol} \]

- **N-factor for NaHCO₃:** 1 (because it can donate 1 mole of H⁺ ions)
- **Moles of NaHCO₃ neutralized:**
\[ \text{Moles of NaHCO₃} = \text{Moles of H₂SO₄} \times \text{N-factor of NaHCO₃} = 0.0005 \, \text{mol} \times 1 = 0.0005 \, \text{mol} \]

- **Mass of NaHCO₃ in 10 mL:**
\[ \text{Mass} = \text{Moles} \times \text{Molar mass of NaHCO₃} = 0.0005 \, \text{mol} \times 84 \, \text{g/mol} = 0.042 \, \text{g} \]

### Step 4: Scale up to 1 liter
- Since 10 mL of solution contains 0.042 g of NaHCO₃, in 1 liter (1000 mL), it will contain:
\[ \text{Mass in 1 L} = 0.042 \, \text{g} \times 100 = 4.2 \, \text{g} \]

### Final Result
- **Amount of Na₂CO₃ in 1 liter:** 5.3 g
- **Amount of NaHCO₃ in 1 liter:** 4.2 g