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 break it down into steps:

### Step 1: Understand the reaction
The reaction between sodium hydroxide (NaOH) and hydrochloric acid (HCl) is a neutralization reaction. Sodium carbonate (Na2CO3) can also react with HCl, but it will first react with NaOH before the HCl is added.

### Step 2: Calculate moles of HCl used in the first titration
We know that 17.5 mL of N/10 HCl was used in the first titration.

1. Convert the volume of HCl used to liters:
\[ 17.5 \, \text{mL} = 0.0175 \, \text{L} \]

2. Calculate the number of moles of HCl:
\[ \text{Molarity} = \frac{\text{moles}}{\text{volume in L}} \]
Since the concentration is N/10 (which is 0.1 N for HCl):
\[ \text{Moles of HCl} = 0.1 \, \text{N} \times 0.0175 \, \text{L} = 0.00175 \, \text{moles} \]

### Step 3: Determine the amount of NaOH and Na2CO3 neutralized
In the first titration, the total moles of NaOH and Na2CO3 will react with the HCl. The reaction can be summarized as:
- NaOH + HCl → NaCl + H2O
- Na2CO3 + 2HCl → 2NaCl + H2O + CO2

Let \( x \) be the moles of NaOH and \( y \) be the moles of Na2CO3 in the solution. The equations based on the reactions are:
1. For NaOH: \( x \) moles of NaOH react with \( x \) moles of HCl.
2. For Na2CO3: \( y \) moles of Na2CO3 react with \( 2y \) moles of HCl.

Thus, the total moles of HCl used in the first titration can be expressed as:
\[ x + 2y = 0.00175 \quad \text{(1)} \]

### Step 4: Calculate moles of HCl used in the second titration
After the first titration, 2.5 mL of the same N/10 HCl is used to reach a new endpoint after adding MeOH.

1. Convert the volume of HCl used in the second titration to liters:
\[ 2.5 \, \text{mL} = 0.0025 \, \text{L} \]

2. Calculate the number of moles of HCl used:
\[ \text{Moles of HCl} = 0.1 \, \text{N} \times 0.0025 \, \text{L} = 0.00025 \, \text{moles} \]

### Step 5: Determine the amount of Na2CO3 neutralized in the second titration
The second titration involves the reaction of HCl with Na2CO3, which means that the moles of HCl used in the second titration will correspond to the moles of Na2CO3:
\[ 2y = 0.00025 \quad \text{(2)} \]

### Step 6: Solve the equations
From equation (2):
\[ y = \frac{0.00025}{2} = 0.000125 \, \text{moles} \]

Substituting \( y \) back into equation (1):
\[ x + 2(0.000125) = 0.00175 \]
\[ x + 0.00025 = 0.00175 \]
\[ x = 0.00175 - 0.00025 = 0.0015 \, \text{moles} \]

### Step 7: Calculate the mass of NaOH
To find the mass of NaOH, we use its molar mass (approximately 40 g/mol):
\[ \text{Mass of NaOH} = \text{moles} \times \text{molar mass} = 0.0015 \, \text{moles} \times 40 \, \text{g/mol} = 0.06 \, \text{g} \]

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

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