Three solution of HCl having normalities 12 N, 6N and 2 N are mixed to obtain a solution of 4 N normality. Which among the following volume ratio is correct for the above three components ?

Options

A. `1 : 1: 5`
B. `1 : 2 : 6`
C. `2 : 1 : 9`
D. `1 : 2: 4`

Explanation

To solve the problem of mixing three solutions of HCl with different normalities (12 N, 6 N, and 2 N) to obtain a solution with a normality of 4 N, we will use the principle of conservation of normality.

### Step-by-Step Solution:

1. **Identify the Normalities and Volumes:**
Let:
- \( N_1 = 12 \, \text{N} \) (Volume = \( V_1 \))
- \( N_2 = 6 \, \text{N} \) (Volume = \( V_2 \))
- \( N_3 = 2 \, \text{N} \) (Volume = \( V_3 \))
- Final Normality \( N_f = 4 \, \text{N} \)

2. **Set Up the Equation:**
According to the formula for mixing solutions:
\[ N_1 V_1 + N_2 V_2 + N_3 V_3 = N_f (V_1 + V_2 + V_3) \]

3. **Substitute the Known Values:**
Substitute the normalities into the equation:
\[ 12 V_1 + 6 V_2 + 2 V_3 = 4 (V_1 + V_2 + V_3) \]

4. **Expand the Right Side:**
Expanding the right side gives:
\[ 12 V_1 + 6 V_2 + 2 V_3 = 4 V_1 + 4 V_2 + 4 V_3 \]

5. **Rearrange the Equation:**
Rearranging the equation leads to:
\[ 12 V_1 - 4 V_1 + 6 V_2 - 4 V_2 + 2 V_3 - 4 V_3 = 0 \]
Simplifying this gives:
\[ 8 V_1 + 2 V_2 - 2 V_3 = 0 \]

6. **Simplify Further:**
Dividing the entire equation by 2:
\[ 4 V_1 + V_2 - V_3 = 0 \]
This can be rearranged to:
\[ V_3 = 4 V_1 + V_2 \]

7. **Assume Volume Ratios:**
To find a suitable ratio, we can assume \( V_1 = 1 \) (for simplicity). Then:
\[ V_3 = 4(1) + V_2 \implies V_3 = 4 + V_2 \]

8. **Determine Ratios:**
If we assume \( V_2 = 2 \):
\[ V_3 = 4 + 2 = 6 \]
Thus, the volumes are:
- \( V_1 : V_2 : V_3 = 1 : 2 : 6 \)

9. **Conclusion:**
The correct volume ratio for the three solutions is \( 1 : 2 : 6 \).