In the study of titration of NaOH and `Na_(2)CO_(3)`. NaOH and `NaHCO_(3)`, `Na_(2)CO_(3)` and `NaHCO_(3)`, phenophthalein and methyl orange are used as indicators.
(a). When phenolphthalein is used as an indicator for the above mixture:
(i). It indicates complete neutralisation of NaOH or KOH
(ii). It indicates half neutralisation of `Na_(2)CO_(3)` because `NaHCO_(3)` is formed at the end point.
(b). When methyl orange is used as an indicator for the above mixture
(i). It indicates complete neutralisation of `NaOH` or KOH
(ii). It indicates half neutralisation of `Na_(2)CO_(3)` because `NaCl` is formed at the end point.
Q. 1 L solution of `Na_(2)CO_(3)` and `NaOH` was made in `H_(2)O`. 100 " mL of " this solution required 20 " mL of " 0.4 M HCl in the presence of phenolphthalein however, another 100 mL sample of the same solution required 25 " mL of " the same acid in the presence of methyl orange as indicator. What is the molar ratio of `Na_(2)CO_(3)` and `NaOH` in the original mixture.

Options

A. `3:2`
B. `3:1`
C. `1:3`
D. `1:1`

Explanation

To solve the problem, we need to determine the molar ratio of Na₂CO₃ to NaOH in the original mixture based on the titration results with two different indicators: phenolphthalein and methyl orange.

### Step-by-Step Solution:

1. **Understand the Reaction and Indicators**:
- Na₂CO₃ reacts with HCl to form NaHCO₃ and NaCl.
- NaOH reacts with HCl to form NaCl and water.
- Phenolphthalein indicates the endpoint when NaOH is completely neutralized, and Na₂CO₃ is half neutralized (forming NaHCO₃).
- Methyl orange indicates the endpoint when NaOH and Na₂CO₃ are completely neutralized.

2. **Calculate the Milliequivalents of HCl Used**:
- For phenolphthalein:
- Volume of HCl = 20 mL
- Molarity of HCl = 0.4 M
- Milliequivalents of HCl = Volume (mL) × Molarity (mol/L) × 1 (n-factor for HCl)
- Milliequivalents of HCl = 20 mL × 0.4 mol/L × 1 = 8 milliequivalents.

- For methyl orange:
- Volume of HCl = 25 mL
- Molarity of HCl = 0.4 M
- Milliequivalents of HCl = 25 mL × 0.4 mol/L × 1 = 10 milliequivalents.

3. **Set Up the Equations**:
- Let A be the equivalents of Na₂CO₃ and B be the equivalents of NaOH.
- From the titration with phenolphthalein:
\[ \frac{A}{2} + B = 8 \quad \text{(Equation 1)} \]
- From the titration with methyl orange:
\[ A + B = 10 \quad \text{(Equation 2)} \]

4. **Solve the Equations**:
- From Equation 2, we can express B in terms of A:
\[ B = 10 - A \]
- Substitute B into Equation 1:
\[ \frac{A}{2} + (10 - A) = 8 \]
\[ \frac{A}{2} - A + 10 = 8 \]
\[ -\frac{A}{2} + 10 = 8 \]
\[ -\frac{A}{2} = -2 \]
\[ A = 4 \]
- Substitute A back to find B:
\[ B = 10 - 4 = 6 \]

5. **Determine the Molar Ratio**:
- The molar ratio of Na₂CO₃ to NaOH is:
\[ \text{Ratio} = \frac{A}{B} = \frac{4}{6} = \frac{2}{3} \]
- Therefore, the molar ratio of Na₂CO₃ to NaOH in the original mixture is **2:3**.

### Final Answer:
The molar ratio of Na₂CO₃ to NaOH in the original mixture is **2:3**.