Hydrolic acid solution A and B have concentration of 0.5 N and 0.1N respectively. The volume of solutions A and B required to make 2 litres of 0.2 N hydrochloric are

Options

A. `0.5 l "of" A + 1.5 l "of" B`
B. `1.5 l "of" A + 0.5 l "of" B`
C. `1.0 l "of" A + 1.0 l "of" B`
D. `0.75 l "of" A + 1.25 l "of" B`

Answer

To solve the problem of determining the volumes of hydrochloric acid solutions A and B required to create 2 liters of a 0.2 N hydrochloric acid solution, we can follow these steps:

### Step 1: Define the known values
- Normality of solution A (N1) = 0.5 N
- Normality of solution B (N2) = 0.1 N
- Final normality (Nf) = 0.2 N
- Final volume (Vf) = 2 liters

### Step 2: Set up the equations
We can use the formula for mixing solutions, which states that the total normality times the total volume of the final solution equals the sum of the normalities times their respective volumes:

\[ N_1 \times V_1 + N_2 \times V_2 = N_f \times V_f \]

Where:
- \( V_1 \) = volume of solution A
- \( V_2 \) = volume of solution B

### Step 3: Express \( V_2 \) in terms of \( V_1 \)
Since the total volume of the final solution is 2 liters, we can express \( V_2 \) as:

\[ V_2 = V_f - V_1 = 2 - V_1 \]

### Step 4: Substitute \( V_2 \) into the equation
Now we can substitute \( V_2 \) into the mixing equation:

\[ 0.5 \times V_1 + 0.1 \times (2 - V_1) = 0.2 \times 2 \]

### Step 5: Simplify the equation
Now, simplify the equation:

\[ 0.5V_1 + 0.2 - 0.1V_1 = 0.4 \]

Combine like terms:

\[ (0.5 - 0.1)V_1 + 0.2 = 0.4 \]

This simplifies to:

\[ 0.4V_1 + 0.2 = 0.4 \]

### Step 6: Solve for \( V_1 \)
Now, isolate \( V_1 \):

\[ 0.4V_1 = 0.4 - 0.2 \]

\[ 0.4V_1 = 0.2 \]

\[ V_1 = \frac{0.2}{0.4} = 0.5 \text{ liters} \]

### Step 7: Calculate \( V_2 \)
Now, substitute \( V_1 \) back to find \( V_2 \):

\[ V_2 = 2 - V_1 = 2 - 0.5 = 1.5 \text{ liters} \]

### Conclusion
The volumes required are:
- Volume of solution A (V1) = 0.5 liters
- Volume of solution B (V2) = 1.5 liters

Thus, the correct answer is **0.5 liters of A and 1.5 liters of B**.