A solution contains mixture of `H_(2)SO_(4)" and " H_(2)C_(2)O_(4). ` 25 ml of this solution requires `35.5` ml of N/10 for neutralization and `23.45` ml of N/10 `KMnO_(4)` for oxidation, calculate
(i) Normality of `H_(2)C_(2)O_(4) " and " H_(2)SO_(4)`
(ii) Strength of `H_(2)C_(2)O_(4)" and " H_(2)SO_(4)`

Answer

To solve the problem, we need to find the normality and strength of the two acids, \( H_2SO_4 \) and \( H_2C_2O_4 \), in a mixture. We will use the information provided about the neutralization with NaOH and oxidation with KMnO4.

### Step 1: Determine the milli equivalents of NaOH used for neutralization
Given:
- Volume of NaOH = 35.5 mL
- Normality of NaOH = \( \frac{1}{10} \) N

We can calculate the milli equivalents of NaOH using the formula:
\[ \text{milli equivalents} = \text{Normality} \times \text{Volume (in mL)} \]
\[ \text{milli equivalents of NaOH} = \frac{1}{10} \times 35.5 = 3.55 \text{ milli equivalents} \]

### Step 2: Set up the first equation for the acids
Let:
- \( a \) = milli equivalents of \( H_2SO_4 \)
- \( b \) = milli equivalents of \( H_2C_2O_4 \)

From the neutralization reaction, we have:
\[ a + b = 3.55 \quad \text{(Equation 1)} \]

### Step 3: Determine the milli equivalents of KMnO4 used for oxidation
Given:
- Volume of KMnO4 = 23.45 mL
- Normality of KMnO4 = \( \frac{1}{10} \) N

Calculating the milli equivalents of KMnO4:
\[ \text{milli equivalents of KMnO4} = \frac{1}{10} \times 23.45 = 2.345 \text{ milli equivalents} \]

### Step 4: Set up the second equation for the acids
The oxidation of oxalic acid \( H_2C_2O_4 \) can be represented as:
\[ \text{milli equivalents of } H_2C_2O_4 = \text{milli equivalents of KMnO4} \]
Thus, we have:
\[ b = 2.345 \quad \text{(Equation 2)} \]

### Step 5: Solve the equations
Substituting Equation 2 into Equation 1:
\[ a + 2.345 = 3.55 \]
\[ a = 3.55 - 2.345 = 1.205 \]

### Step 6: Calculate the normality of each acid
Normality is calculated using the formula:
\[ \text{Normality} = \frac{\text{milli equivalents}}{\text{Volume (in L)}} \]
Since the total volume of the solution is 25 mL (or 0.025 L):

1. For \( H_2SO_4 \):
\[ \text{Normality of } H_2SO_4 = \frac{1.205}{0.025} = 0.0482 \, \text{N} \]

2. For \( H_2C_2O_4 \):
\[ \text{Normality of } H_2C_2O_4 = \frac{2.345}{0.025} = 0.0938 \, \text{N} \]

### Step 7: Calculate the strength of each acid
The strength can be calculated using the formula:
\[ \text{Strength} = \text{Normality} \times \text{Equivalent Weight} \]

1. For \( H_2SO_4 \):
- Equivalent weight of \( H_2SO_4 = \frac{98}{2} = 49 \, \text{g/equiv} \)
\[ \text{Strength of } H_2SO_4 = 0.0482 \times 49 = 2.362 \, \text{g/L} \]

2. For \( H_2C_2O_4 \):
- Equivalent weight of \( H_2C_2O_4 = \frac{126}{2} = 63 \, \text{g/equiv} \)
\[ \text{Strength of } H_2C_2O_4 = 0.0938 \times 63 = 5.909 \, \text{g/L} \]

### Final Answers:
(i) Normality:
- \( H_2SO_4 \): 0.0482 N
- \( H_2C_2O_4 \): 0.0938 N

(ii) Strength:
- \( H_2SO_4 \): 2.362 g/L
- \( H_2C_2O_4 \): 5.909 g/L