If 20 mL of 0.4 N NaOH solution completely neutralises 40 mL of a dibasic acid, the molarity of the acid solution is `:`

Options

A. 0.1 M
B. 0.2 M
C. 0.3 M
D. 0.4 M

Explanation

To find the molarity of the dibasic acid solution, we can follow these steps:

### Step 1: Calculate the number of equivalents of NaOH used.
The normality (N) of NaOH is given as 0.4 N, and the volume (V) used is 20 mL.

To find the number of equivalents of NaOH:
\[ \text{Equivalents of NaOH} = N \times V \text{ (in L)} \]
Convert 20 mL to liters:
\[ 20 \text{ mL} = 0.020 \text{ L} \]
Now, substitute the values:
\[ \text{Equivalents of NaOH} = 0.4 \, \text{N} \times 0.020 \, \text{L} = 0.008 \, \text{equivalents} \]

### Step 2: Determine the number of equivalents of the dibasic acid.
Since the acid is dibasic, it can donate 2 protons (H⁺ ions) per molecule. Therefore, the number of equivalents of the dibasic acid is equal to the number of equivalents of NaOH used for neutralization:
\[ \text{Equivalents of acid} = \text{Equivalents of NaOH} = 0.008 \, \text{equivalents} \]

### Step 3: Calculate the molarity of the dibasic acid.
The volume of the dibasic acid solution is given as 40 mL. Convert this to liters:
\[ 40 \text{ mL} = 0.040 \text{ L} \]
Now, we can find the molarity (M) of the dibasic acid using the formula:
\[ \text{Molarity (M)} = \frac{\text{Equivalents of acid}}{\text{Volume of acid in L}} \]
Since the acid is dibasic, we need to divide the equivalents by 2 to find the number of moles:
\[ \text{Moles of acid} = \frac{0.008 \, \text{equivalents}}{2} = 0.004 \, \text{moles} \]
Now, substitute this into the molarity formula:
\[ \text{Molarity (M)} = \frac{0.004 \, \text{moles}}{0.040 \, \text{L}} = 0.1 \, \text{M} \]

### Final Answer:
The molarity of the dibasic acid solution is **0.1 M**.

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