In the mixture of `(NaHCO_(3)+Na_(2)CO_(3))` volume of HCl required is x mL with phenolphthalein indicator and then y mL with methyl orange indicator in same titration Hence, volume of HCl for complete reaction of `Na_(2)CO_(3)` is

Options

A. `2x`
B. `y`
C. `(x)/(2)`
D. `(y-x)`

Explanation

To solve the problem step by step, we need to analyze the reactions of sodium bicarbonate (NaHCO₃) and sodium carbonate (Na₂CO₃) with hydrochloric acid (HCl) using two different indicators: phenolphthalein and methyl orange.

### Step 1: Understand the Reactions
1. **NaHCO₃ with HCl**:
- NaHCO₃ reacts with HCl to produce NaCl, CO₂, and H₂O.
- The reaction is:
\[ \text{NaHCO}_3 + \text{HCl} \rightarrow \text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

2. **Na₂CO₃ with HCl**:
- Na₂CO₃ reacts with HCl to produce NaCl, CO₂, and H₂O.
- The reaction is:
\[ \text{Na}_2\text{CO}_3 + 2\text{HCl} \rightarrow 2\text{NaCl} + \text{CO}_2 + \text{H}_2\text{O} \]

### Step 2: Analyze the Titration with Phenolphthalein
- When using phenolphthalein, it indicates the endpoint when Na₂CO₃ has reacted with HCl.
- The volume of HCl required to reach this endpoint is **x mL**.
- At this point, only half of the Na₂CO₃ has reacted because it requires 2 moles of HCl for each mole of Na₂CO₃.

### Step 3: Analyze the Titration with Methyl Orange
- When using methyl orange, both NaHCO₃ and Na₂CO₃ react with HCl.
- The total volume of HCl required is **y mL**.
- Since NaHCO₃ reacts completely with HCl, the volume of HCl used for NaHCO₃ is equal to the volume of HCl used for Na₂CO₃ plus the volume used for NaHCO₃.

### Step 4: Relate the Volumes
- From the first titration (phenolphthalein), we know that:
\[ \text{Volume of HCl for Na}_2\text{CO}_3 = x \text{ mL (half of the required volume)} \]
- From the second titration (methyl orange), we know that:
\[ \text{Volume of HCl for NaHCO}_3 + \text{Volume of HCl for Na}_2\text{CO}_3 = y \text{ mL} \]
- Since NaHCO₃ reacts with the same amount of HCl as Na₂CO₃, we can express this as:
\[ \text{Volume of HCl for NaHCO}_3 = x \text{ mL} \]
- Therefore, the total volume of HCl used in the second titration can be expressed as:
\[ y = x + 2x = 3x \]

### Step 5: Calculate the Volume of HCl for Complete Reaction of Na₂CO₃
- The volume of HCl required for the complete reaction of Na₂CO₃ is:
\[ \text{Volume for complete reaction} = 2x \text{ mL} \]

### Final Answer
The volume of HCl required for the complete reaction of Na₂CO₃ is **2x mL**.

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