If amixture of `Na_(2)CO_(3)` and NaOH in equimolar quantities when reacts with 0.1 M HCl in presence of phenolphthalein indicator consumes 30 ml of the acid. What will be the volume (in mL) of 0.15 M `H_(2)SO_(4)` used in the separate titration of same mixture in presence of methyl orange indicator.

Explanation

To solve the problem step by step, we will analyze the reaction of the mixture of sodium carbonate (Na₂CO₃) and sodium hydroxide (NaOH) with hydrochloric acid (HCl) and sulfuric acid (H₂SO₄) in the presence of different indicators.

### Step 1: Calculate the milli-equivalents of HCl used
The problem states that a mixture of Na₂CO₃ and NaOH in equimolar quantities reacts with 0.1 M HCl and consumes 30 mL of it.

The milli-equivalents of HCl can be calculated using the formula:
\[ \text{milli-equivalents} = \text{Volume (L)} \times \text{Molarity (mol/L)} \times 1000 \]
Substituting the values:
\[ \text{milli-equivalents of HCl} = 30 \, \text{mL} \times 0.1 \, \text{M} = 3 \, \text{milli-equivalents} \]

### Step 2: Determine the contributions of Na₂CO₃ and NaOH
Since Na₂CO₃ and NaOH are in equimolar quantities, let’s denote the amount of each as 'A' moles. The reactions are as follows:
- Na₂CO₃ reacts with HCl:
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \]
This means 1 mole of Na₂CO₃ gives 2 equivalents of H⁺.

- NaOH reacts with HCl:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
This means 1 mole of NaOH gives 1 equivalent of H⁺.

Thus, the total milli-equivalents from the mixture is:
\[ \text{milli-equivalents of Na}_2\text{CO}_3 + \text{milli-equivalents of NaOH} = 2A + A = 3A \]
Setting this equal to the milli-equivalents of HCl:
\[ 3A = 3 \implies A = 1 \, \text{mole} \]

### Step 3: Calculate the milli-equivalents of H₂SO₄
In the presence of methyl orange, we need to calculate the milli-equivalents of H₂SO₄ that will react with the same mixture. The reaction of H₂SO₄ with Na₂CO₃ and NaOH is as follows:
- H₂SO₄ reacts with Na₂CO₃:
\[ \text{Na}_2\text{CO}_3 + \text{H}_2\text{SO}_4 \rightarrow 2\text{Na}_2\text{SO}_4 + \text{H}_2\text{O} + \text{CO}_2 \]
This means 1 mole of Na₂CO₃ gives 1 equivalent of H₂SO₄.

- H₂SO₄ reacts with NaOH:
\[ \text{NaOH} + \text{H}_2\text{SO}_4 \rightarrow \text{Na}_2\text{SO}_4 + \text{H}_2\text{O} \]
This means 1 mole of NaOH gives 1 equivalent of H₂SO₄.

Thus, the total milli-equivalents from the mixture for H₂SO₄ is:
\[ \text{milli-equivalents of Na}_2\text{CO}_3 + \text{milli-equivalents of NaOH} = A + A = 2A \]
Substituting A = 1:
\[ \text{Total milli-equivalents} = 2 \times 1 = 2 \]

### Step 4: Set up the equation for H₂SO₄
Let V be the volume of 0.15 M H₂SO₄ used. The milli-equivalents of H₂SO₄ can be expressed as:
\[ \text{milli-equivalents of H}_2\text{SO}_4 = \text{Volume (L)} \times \text{Molarity (mol/L)} \times 1000 = V \times 0.15 \times 1000 \]
Setting this equal to the milli-equivalents from the mixture:
\[ 0.15 \times V = 2 \]

### Step 5: Solve for V
Now, rearranging the equation gives:
\[ V = \frac{2}{0.15} = \frac{20}{1.5} = 13.33 \, \text{mL} \]

### Step 6: Final Calculation
To find the volume of H₂SO₄ used, we need to consider the total milli-equivalents:
\[ \text{Total milli-equivalents} = 2 \text{ (from Na}_2\text{CO}_3\text{)} + 1 \text{ (from NaOH)} = 3 \]
So, the final volume of 0.15 M H₂SO₄ used is:
\[ V = \frac{3}{0.15} = 20 \, \text{mL} \]

### Conclusion
Thus, the volume of 0.15 M H₂SO₄ used in the titration of the same mixture in the presence of methyl orange is **20 mL**.

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