If 20ml of 0.4 N NaOH solution completely neutralise 40ml of a dibasic acid, the molarity of the acid solution is:

Explanation

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

### Step 1: Understand the given information
We have:
- Volume of NaOH solution (V1) = 20 mL
- Normality of NaOH solution (N1) = 0.4 N
- Volume of dibasic acid solution (V2) = 40 mL
- The acid is dibasic, meaning it can donate 2 protons (H⁺ ions).

### Step 2: Calculate the number of equivalents of NaOH used
Using the formula for equivalents:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in L)} \]
Convert the volume from mL to L:
\[ V1 = 20 \text{ mL} = 0.020 \text{ L} \]
Now, calculate the equivalents of NaOH:
\[ \text{Equivalents of NaOH} = 0.4 \, \text{N} \times 0.020 \, \text{L} = 0.008 \, \text{equivalents} \]

### Step 3: Relate the equivalents of NaOH to the dibasic acid
Since the dibasic acid is neutralized by the NaOH, the equivalents of the acid will also be equal to the equivalents of NaOH:
\[ \text{Equivalents of acid} = 0.008 \, \text{equivalents} \]

### Step 4: Calculate the normality of the dibasic acid
The normality (N2) of the dibasic acid can be calculated using the formula:
\[ \text{Normality} = \frac{\text{Number of equivalents}}{\text{Volume (in L)}} \]
Convert the volume of the acid from mL to L:
\[ V2 = 40 \text{ mL} = 0.040 \text{ L} \]
Now, calculate the normality of the acid:
\[ N2 = \frac{0.008 \, \text{equivalents}}{0.040 \, \text{L}} = 0.2 \, \text{N} \]

### Step 5: Calculate the molarity of the dibasic acid
Since the acid is dibasic, its n-factor is 2. The molarity (M2) can be calculated using the relationship:
\[ \text{Molarity} = \frac{\text{Normality}}{\text{n-factor}} \]
Substituting the values:
\[ M2 = \frac{0.2 \, \text{N}}{2} = 0.1 \, \text{M} \]

### Conclusion
The molarity of the dibasic acid solution is:
\[ \text{Molarity} = 0.1 \, \text{M} \]