A solution containing `Na_(2)CO_(3)` and `NaOH` requires `300 ml` of `0.1 N HCl` using phenolphthalein as an indicator. Methyl orange is then added to the above titrated solution when a further `25 ml` of `0.2 N HCl` is required. The amount of `NaOH` present in solution is `(NaOH=40, Na_(2)CO_(3)=106)`

Options

A. (a)`0.6 g`
B. (b)`1.0 g`
C. (c )`1.5 g`
D. (d)`2.0 g`

Explanation

To solve the problem step by step, we will analyze the titration process involving sodium carbonate (Na₂CO₃) and sodium hydroxide (NaOH) with hydrochloric acid (HCl).

### Step 1: Understanding the Reaction
When Na₂CO₃ reacts with HCl, it first forms sodium bicarbonate (NaHCO₃) and then can react further to form NaCl and water. The reaction can be represented as:
1. \( \text{Na}_2\text{CO}_3 + 2 \text{HCl} \rightarrow 2 \text{NaCl} + \text{H}_2\text{O} + \text{CO}_2 \)
2. \( \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \)

### Step 2: Calculate the Milliequivalents of HCl Used
The total volume of HCl used in the first titration is 300 mL of 0.1 N HCl. The milliequivalents of HCl can be calculated as follows:
\[ \text{Milliequivalents of HCl} = \text{Volume (mL)} \times \text{Normality (N)} = 300 \, \text{mL} \times 0.1 \, \text{N} = 30 \, \text{meq} \]

### Step 3: Calculate the Additional HCl Required
After the first titration with phenolphthalein, when methyl orange is added, an additional 25 mL of 0.2 N HCl is required. The milliequivalents for this additional HCl are:
\[ \text{Milliequivalents of additional HCl} = 25 \, \text{mL} \times 0.2 \, \text{N} = 5 \, \text{meq} \]

### Step 4: Total Milliequivalents of HCl
The total milliequivalents of HCl used for complete neutralization is:
\[ \text{Total meq of HCl} = 30 \, \text{meq} + 5 \, \text{meq} = 35 \, \text{meq} \]

### Step 5: Determine the Milliequivalents of Na₂CO₃ and NaOH
The first 30 meq of HCl neutralizes both Na₂CO₃ and NaOH. Since Na₂CO₃ requires 2 equivalents of HCl to be fully neutralized, we can let \( x \) be the milliequivalents of Na₂CO₃ and \( y \) be the milliequivalents of NaOH:
\[ x + y = 30 \quad \text{(1)} \]
Since Na₂CO₃ requires 2 equivalents of HCl:
\[ 2x + y = 35 \quad \text{(2)} \]

### Step 6: Solve the Equations
From equation (1):
\[ y = 30 - x \]
Substituting \( y \) in equation (2):
\[ 2x + (30 - x) = 35 \]
\[ 2x - x + 30 = 35 \]
\[ x + 30 = 35 \]
\[ x = 5 \, \text{meq (of Na₂CO₃)} \]
Substituting \( x \) back into equation (1):
\[ 5 + y = 30 \Rightarrow y = 25 \, \text{meq (of NaOH)} \]

### Step 7: Calculate the Amount of NaOH
Since the normality of NaOH is 1 (N factor = 1), the number of moles of NaOH can be calculated as:
\[ \text{Moles of NaOH} = \frac{25 \, \text{meq}}{1000} = 0.025 \, \text{moles} \]
The mass of NaOH can be calculated using its molar mass (40 g/mol):
\[ \text{Mass of NaOH} = 0.025 \, \text{moles} \times 40 \, \text{g/mol} = 1 \, \text{gram} \]

### Final Answer
The amount of NaOH present in the solution is **1 gram**.