25mL of a solution containing HCl and `H_2SO_4` required 10 mL of a 1 N NaOH solution for neutralization.20 mL of the same acid mixture on being treated with an excess of `AgNO_3` gives 0.1435 g of AgCl.The normality of the HCl and the normality of the `H_2SO_4` are respectively.

Options

A. 0.40 N and 0.05 N
B. 0.05 N and 0.35 N
C. 0.50 N and 0.25 N
D. 0.40 N and 0.5 N

Explanation

To solve the problem step by step, we will break down the information given and apply the necessary calculations.

### Step 1: Determine the moles of AgCl formed
The mass of AgCl produced is given as 0.1435 g. We need to convert this mass into moles using the molar mass of AgCl.

**Molar mass of AgCl**:
- Ag = 107.87 g/mol
- Cl = 35.45 g/mol
- Molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol

**Calculating moles of AgCl**:
\[ \text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molar mass of AgCl}} = \frac{0.1435 \text{ g}}{143.32 \text{ g/mol}} \approx 0.001 \text{ moles} \]

### Step 2: Relate moles of AgCl to moles of HCl
Since AgCl is formed from HCl, the moles of HCl will be equal to the moles of AgCl because the reaction between HCl and AgNO3 is a 1:1 reaction.

\[ \text{Moles of HCl} = 0.001 \text{ moles} \]

### Step 3: Calculate the normality of HCl
Normality (N) is defined as the number of equivalents per liter of solution. For HCl, which is a monoprotic acid, the number of equivalents is equal to the number of moles.

**Volume of HCl solution used**: 20 mL = 0.020 L

**Calculating normality of HCl**:
\[ N_{HCl} = \frac{\text{moles of HCl}}{\text{volume of solution in L}} = \frac{0.001 \text{ moles}}{0.020 \text{ L}} = 0.05 \text{ N} \]

### Step 4: Calculate equivalents of HCl in 25 mL solution
Now we will calculate the equivalents of HCl in the 25 mL solution.

\[ \text{Equivalents of HCl} = N_{HCl} \times \text{Volume in L} = 0.05 \text{ N} \times 0.025 \text{ L} = 0.00125 \text{ equivalents} = 1.25 \text{ mEq} \]

### Step 5: Determine the equivalents of NaOH used
The problem states that 10 mL of 1 N NaOH is used for neutralization.

\[ \text{Equivalents of NaOH} = N_{NaOH} \times \text{Volume in L} = 1 \text{ N} \times 0.010 \text{ L} = 0.010 \text{ equivalents} = 10 \text{ mEq} \]

### Step 6: Calculate equivalents of H2SO4
The total equivalents of acid (HCl + H2SO4) neutralized by NaOH is equal to the equivalents of NaOH used.

\[ \text{Equivalents of HCl} + \text{Equivalents of H2SO4} = \text{Equivalents of NaOH} \]
\[ 1.25 \text{ mEq} + \text{Equivalents of H2SO4} = 10 \text{ mEq} \]
\[ \text{Equivalents of H2SO4} = 10 - 1.25 = 8.75 \text{ mEq} \]

### Step 7: Calculate the normality of H2SO4
H2SO4 is a diprotic acid, meaning it can donate two protons. Therefore, the number of equivalents is twice the number of moles.

**Volume of H2SO4 solution used**: 25 mL = 0.025 L

**Calculating normality of H2SO4**:
\[ N_{H2SO4} = \frac{\text{Equivalents of H2SO4}}{\text{Volume in L}} = \frac{8.75 \text{ mEq}}{0.025 \text{ L}} = 0.35 \text{ N} \]

### Final Answer
The normality of HCl and H2SO4 are respectively:
- Normality of HCl = 0.05 N
- Normality of H2SO4 = 0.35 N