Two acids A and B are titrated separately each time with 25 ml of 1N `Na_2 CO_3` solution and require 10 ml and 40 ml respectively for complete neutralization. What volume of acids. A and B will we need to mix to produce a 1 litre 1 N solution?

Options

A. A = 100 ml , B = 200 ml
B. A = 300 ml , B = 200 ml
C. A = 200 ml , B = 800 ml
D. A = 400 ml , B = 400 ml

Answer

To solve the problem step by step, we will first determine the normalities of acids A and B based on the titration data provided. Then, we will use these normalities to find the required volumes of acids A and B to create a 1-liter 1N solution.

### Step 1: Calculate the Normality of Acid A
Given:
- Volume of Na2CO3 solution = 25 ml = 0.025 L
- Normality of Na2CO3 (N1) = 1N
- Volume of Acid A used (V2) = 10 ml = 0.010 L

Using the formula for normality:
\[ N_1 \times V_1 = N_2 \times V_2 \]
Where:
- \(N_2\) is the normality of Acid A
- \(V_1\) is the volume of Na2CO3

Substituting the values:
\[ 1 \times 0.025 = N_2 \times 0.010 \]
\[ N_2 = \frac{1 \times 0.025}{0.010} = 2.5N \]

### Step 2: Calculate the Normality of Acid B
Given:
- Volume of Acid B used (V3) = 40 ml = 0.040 L

Using the same formula:
\[ N_1 \times V_1 = N_3 \times V_3 \]
Where:
- \(N_3\) is the normality of Acid B

Substituting the values:
\[ 1 \times 0.025 = N_3 \times 0.040 \]
\[ N_3 = \frac{1 \times 0.025}{0.040} = 0.625N \]

### Step 3: Set Up the Equation for Mixing Acids A and B
Let:
- \(V_A\) = volume of Acid A to be mixed
- \(V_B\) = volume of Acid B to be mixed

We need to create a 1-liter (1000 ml) solution with a normality of 1N. Therefore, we can set up the following equations:

1. **Total Volume Equation**:
\[ V_A + V_B = 1000 \quad \text{(1)} \]

2. **Normality Equation**:
The total equivalents from both acids must equal the equivalents in 1 liter of 1N solution:
\[ N_A \times V_A + N_B \times V_B = 1 \times 1 \quad \text{(2)} \]
Substituting the normalities:
\[ 2.5V_A + 0.625V_B = 1 \quad \text{(3)} \]

### Step 4: Solve the Equations
From equation (1):
\[ V_B = 1000 - V_A \]

Substituting \(V_B\) in equation (3):
\[ 2.5V_A + 0.625(1000 - V_A) = 1000 \]
Expanding this:
\[ 2.5V_A + 625 - 0.625V_A = 1000 \]
Combining like terms:
\[ (2.5 - 0.625)V_A = 1000 - 625 \]
\[ 1.875V_A = 375 \]
\[ V_A = \frac{375}{1.875} = 200 \text{ ml} \]

Now, substituting \(V_A\) back into equation (1) to find \(V_B\):
\[ V_B = 1000 - 200 = 800 \text{ ml} \]

### Final Answer
The volumes of acids A and B needed to produce a 1-liter 1N solution are:
- Volume of Acid A (\(V_A\)) = 200 ml
- Volume of Acid B (\(V_B\)) = 800 ml