100 ml of an acid solution is neutralized by 50 ml of NaOH solution containing 0.2 g NaOH. The concentration of acid solution is

Options

A. 0.1 N
B. 0.05 N and the solution is basic
C. 0.5 N
D. 0.25 N

Explanation

To find the concentration of the acid solution, we can use the formula for neutralization:

\[ N_1 V_1 = N_2 V_2 \]

Where:
- \( N_1 \) = Normality of the acid solution (which we are trying to find)
- \( V_1 \) = Volume of the acid solution (100 ml)
- \( N_2 \) = Normality of the NaOH solution
- \( V_2 \) = Volume of the NaOH solution (50 ml)

### Step 1: Calculate the Normality of NaOH Solution
First, we need to calculate the normality of the NaOH solution. The normality (N) can be calculated using the formula:

\[ N = \frac{\text{mass (g)}}{\text{molar mass (g/mol)} \times \text{volume (L)}} \]

The molar mass of NaOH is approximately 40 g/mol.

Given:
- Mass of NaOH = 0.2 g
- Volume of NaOH solution = 50 ml = 0.050 L

Now, substituting the values into the formula:

\[ N_2 = \frac{0.2 \, \text{g}}{40 \, \text{g/mol} \times 0.050 \, \text{L}} = \frac{0.2}{2} = 0.1 \, \text{N} \]

### Step 2: Substitute Values into the Neutralization Equation
Now we can substitute the values into the neutralization equation:

\[ N_1 \times V_1 = N_2 \times V_2 \]

Substituting the known values:

\[ N_1 \times 100 \, \text{ml} = 0.1 \, \text{N} \times 50 \, \text{ml} \]

### Step 3: Solve for the Normality of the Acid Solution
Now we can solve for \( N_1 \):

\[ N_1 \times 100 = 0.1 \times 50 \]

\[ N_1 \times 100 = 5 \]

Dividing both sides by 100:

\[ N_1 = \frac{5}{100} = 0.05 \, \text{N} \]

### Step 4: Conclusion
The concentration of the acid solution is:

\[ N_1 = 0.05 \, \text{N} \]

### Final Answer
The concentration of the acid solution is **0.05 N**.

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