100 ml of 0.2N HCl solution is added to 100 ml of 0.2 N `AgNO_(3)` solution. The molarity of nitrate ions in the resulting mixture will be

Options

A. 0.5 M
B. 0.05 M
C. 0.1 M
D. 0.2 M

Explanation

To find the molarity of nitrate ions in the resulting mixture after mixing 100 ml of 0.2 N HCl solution with 100 ml of 0.2 N AgNO₃ solution, we can follow these steps:

### Step 1: Determine the number of moles of HCl and AgNO₃

1. **Calculate moles of HCl:**
- Normality (N) of HCl = 0.2 N
- Volume (V) of HCl = 100 ml = 0.1 L
- Moles of HCl = Normality × Volume = 0.2 N × 0.1 L = 0.02 moles

2. **Calculate moles of AgNO₃:**
- Normality (N) of AgNO₃ = 0.2 N
- Volume (V) of AgNO₃ = 100 ml = 0.1 L
- Moles of AgNO₃ = Normality × Volume = 0.2 N × 0.1 L = 0.02 moles

### Step 2: Identify the reaction between HCl and AgNO₃

The reaction between HCl and AgNO₃ can be represented as:
\[ \text{AgNO}_3 + \text{HCl} \rightarrow \text{AgCl} + \text{HNO}_3 \]

From the reaction, we see that:
- 1 mole of AgNO₃ reacts with 1 mole of HCl to produce 1 mole of AgCl and 1 mole of HNO₃.

### Step 3: Determine the limiting reactant

Since both HCl and AgNO₃ have the same number of moles (0.02 moles), they will completely react with each other:
- Moles of HCl consumed = 0.02 moles
- Moles of AgNO₃ consumed = 0.02 moles

### Step 4: Calculate the moles of nitrate ions produced

The reaction produces nitric acid (HNO₃), which dissociates to give nitrate ions (NO₃⁻):
- Moles of HNO₃ produced = Moles of AgNO₃ reacted = 0.02 moles
- Therefore, moles of nitrate ions (NO₃⁻) = 0.02 moles

### Step 5: Calculate the total volume of the solution

The total volume of the resulting mixture after mixing:
- Volume of HCl = 100 ml
- Volume of AgNO₃ = 100 ml
- Total Volume = 100 ml + 100 ml = 200 ml = 0.2 L

### Step 6: Calculate the molarity of nitrate ions

Molarity (M) is defined as:
\[ \text{Molarity} = \frac{\text{Number of moles}}{\text{Volume in liters}} \]

Using the moles of nitrate ions and the total volume:
\[ \text{Molarity of NO}_3^- = \frac{0.02 \text{ moles}}{0.2 \text{ L}} = 0.1 \text{ M} \]

### Final Answer:
The molarity of nitrate ions in the resulting mixture is **0.1 M**.

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