15 ml of 0.2 N alkali is required to complete neutralization of 30 ml acid solution. Concentration of the acid solution is

Options

A. 0.1 N
B. 0.3 N
C. 0.15 N
D. 0.4 N

Explanation

To find the concentration of the acid solution, we can use the concept of normality and the neutralization reaction between the acid and the alkali. Here’s the step-by-step solution:

### Step 1: Identify the given values
- Volume of alkali (V1) = 15 ml
- Normality of alkali (N1) = 0.2 N
- Volume of acid (V2) = 30 ml
- Normality of acid (N2) = ?

### Step 2: Use the neutralization formula
The relationship between the normalities and volumes of the acid and alkali can be expressed by the formula:
\[ N_1 \times V_1 = N_2 \times V_2 \]

### Step 3: Substitute the known values into the formula
Substituting the known values into the formula gives:
\[ 0.2 \, N \times 15 \, ml = N_2 \times 30 \, ml \]

### Step 4: Solve for N2
Rearranging the equation to solve for N2:
\[ N_2 = \frac{0.2 \, N \times 15 \, ml}{30 \, ml} \]

### Step 5: Calculate N2
Calculating the right side:
\[ N_2 = \frac{0.2 \times 15}{30} \]
\[ N_2 = \frac{3}{30} \]
\[ N_2 = 0.1 \, N \]

### Conclusion
The concentration of the acid solution is:
\[ N_2 = 0.1 \, N \]

### Final Answer
The concentration of the acid solution is **0.1 N**.

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