6.3 g of oxalic acid dihydrate have been dissolved in water to obtain a 250 ml solution. How much volume of 0.1 N NaOH would be required to neutralize 10 mL of this solution ?

Options

A. 40 mL
B. 20 mL
C. 10 mL
D. 4 mL

Explanation

To solve the problem, we will follow these steps:

### Step 1: Calculate the equivalent mass of oxalic acid dihydrate.
- The formula for oxalic acid dihydrate is \( \text{C}_2\text{H}_2\text{O}_4 \cdot 2\text{H}_2\text{O} \).
- The molar mass of oxalic acid dihydrate is calculated as follows:
- Carbon (C): \( 2 \times 12 = 24 \, \text{g} \)
- Hydrogen (H): \( 2 \times 1 + 4 \times 1 + 4 = 6 \, \text{g} \)
- Oxygen (O): \( 4 \times 16 + 2 \times 16 = 64 \, \text{g} \)
- Total = \( 24 + 6 + 64 = 94 \, \text{g} \)

### Step 2: Determine the basicity of oxalic acid.
- Oxalic acid has 2 removable hydrogen ions (H⁺), so its basicity is 2.

### Step 3: Calculate the equivalent mass of oxalic acid.
- The equivalent mass is given by:
\[ \text{Equivalent mass} = \frac{\text{Molar mass}}{\text{Basicity}} = \frac{126 \, \text{g}}{2} = 63 \, \text{g} \]

### Step 4: Calculate the normality of the oxalic acid solution.
- We use the formula:
\[ \text{Normality} (N) = \frac{\text{Given mass}}{\text{Equivalent mass}} \times \frac{1000}{\text{Volume of solution in mL}} \]
- Plugging in the values:
\[ N = \frac{6.3 \, \text{g}}{63 \, \text{g}} \times \frac{1000}{250 \, \text{mL}} = 0.4 \, \text{N} \]

### Step 5: Use the normality to find the volume of NaOH required.
- We apply the formula \( N_1V_1 = N_2V_2 \):
- Where:
- \( N_1 = 0.4 \, \text{N} \) (normality of oxalic acid)
- \( V_1 = 10 \, \text{mL} \) (volume of oxalic acid solution to be neutralized)
- \( N_2 = 0.1 \, \text{N} \) (normality of NaOH)
- \( V_2 = ? \) (volume of NaOH required)

- Rearranging the equation gives:
\[ V_2 = \frac{N_1V_1}{N_2} = \frac{0.4 \times 10}{0.1} = 40 \, \text{mL} \]

### Conclusion:
The volume of 0.1 N NaOH required to neutralize 10 mL of the oxalic acid solution is **40 mL**.

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