When 100mL solution of `NaOH and NaCO_(3)` was first titrated with N/10 HCl in presence of HPh, 17.5mL were usedtill end point is obtained. After this end point MeOH was added and 2.5mL of same HCl were required to attain new end point. The amount NaOH in mixture is:

Options

A. 0.06g per 100mL
B. 0.06g per 200mL
C. 0.05 g per 100mL
D. 0.012 g per 200mL

Answer

To solve the problem, we will follow these steps:

### Step 1: Define Variables
Let:
- \( x \) = amount of NaOH in grams
- \( y \) = amount of Na2CO3 in grams

### Step 2: Write the Reactions
1. When NaOH reacts with HCl:
\[ \text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O} \]
2. When Na2CO3 reacts with HCl (half neutralization):
\[ \text{Na}_2\text{CO}_3 + \text{HCl} \rightarrow \text{NaHCO}_3 + \text{NaCl} \]

### Step 3: Calculate Equivalents for the First Titration
From the first titration with phenolphthalein:
- The total volume of HCl used = 17.5 mL = 0.0175 L
- Concentration of HCl = \( \frac{1}{10} \) N = 0.1 N

The equivalents of HCl used:
\[ \text{Equivalents of HCl} = \text{Volume (L)} \times \text{Normality} = 0.0175 \times 0.1 = 0.00175 \text{ equivalents} \]

### Step 4: Set Up the Equation
The total equivalents of NaOH and half the equivalents of Na2CO3 that reacted with HCl:
\[ \frac{x}{40} + \frac{y}{106} \times \frac{1}{2} = 0.00175 \]
Where:
- \( \frac{x}{40} \) is the equivalents of NaOH (molecular weight = 40 g/mol)
- \( \frac{y}{106} \) is the equivalents of Na2CO3 (molecular weight = 106 g/mol)

### Step 5: Simplify the Equation
This can be rewritten as:
\[ \frac{x}{40} + \frac{y}{212} = 0.00175 \quad \text{(Equation 1)} \]

### Step 6: Calculate Equivalents for the Second Titration
After adding methyl orange, 2.5 mL of HCl is used:
- Volume of HCl = 2.5 mL = 0.0025 L

The equivalents of HCl used in the second titration:
\[ \text{Equivalents of HCl} = 0.0025 \times 0.1 = 0.00025 \text{ equivalents} \]

### Step 7: Set Up the Second Equation
The remaining Na2CO3 is fully neutralized:
\[ \frac{y}{106} \times \frac{1}{2} = 0.00025 \]
This can be rewritten as:
\[ \frac{y}{212} = 0.00025 \quad \text{(Equation 2)} \]

### Step 8: Solve for y
From Equation 2:
\[ y = 0.00025 \times 212 = 0.053 \text{ grams} \]

### Step 9: Substitute y into Equation 1
Substituting \( y \) into Equation 1:
\[ \frac{x}{40} + \frac{0.053}{212} = 0.00175 \]
Calculating \( \frac{0.053}{212} \):
\[ \frac{0.053}{212} \approx 0.00025 \]

Thus, we have:
\[ \frac{x}{40} + 0.00025 = 0.00175 \]
\[ \frac{x}{40} = 0.00175 - 0.00025 = 0.0015 \]

### Step 10: Solve for x
\[ x = 0.0015 \times 40 = 0.06 \text{ grams} \]

### Final Answer
The amount of NaOH in the mixture is **0.06 grams**.

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