500 ml of a 0.1 N solution of `AgNO_(3)` added to 500 mlof 0.1 N solution of KCI. The concentration of nitrate ion in the resulting mixture is

Answer

To find the concentration of nitrate ions in the resulting mixture after mixing 500 ml of 0.1 N AgNO₃ with 500 ml of 0.1 N KCl, we can follow these steps:

### Step 1: Calculate the gram equivalents of nitrate ions from AgNO₃
- The normality (N) of the AgNO₃ solution is 0.1 N.
- The volume (V) of the AgNO₃ solution is 500 ml, which is equivalent to 0.5 L.

Using the formula for gram equivalents:
\[ \text{Gram Equivalents} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Gram Equivalents of AgNO₃} = 0.1 \, \text{N} \times 0.5 \, \text{L} = 0.05 \, \text{equivalents} \]

### Step 2: Identify the contribution of nitrate ions
- In AgNO₃, there is one nitrate ion (NO₃⁻) for each formula unit.
- Therefore, the gram equivalents of nitrate ions from AgNO₃ is also 0.05 equivalents.

### Step 3: Calculate the total volume of the mixture
- The total volume after mixing the two solutions is:
\[ \text{Total Volume} = 500 \, \text{ml} + 500 \, \text{ml} = 1000 \, \text{ml} = 1 \, \text{L} \]

### Step 4: Calculate the concentration of nitrate ions in the mixture
- The concentration (C) of nitrate ions can be calculated using the formula:
\[ \text{Concentration (M)} = \frac{\text{Gram Equivalents}}{\text{Total Volume (in L)}} \]
\[ \text{Concentration of NO₃⁻} = \frac{0.05 \, \text{equivalents}}{1 \, \text{L}} = 0.05 \, \text{M} \]

### Final Answer
The concentration of nitrate ions in the resulting mixture is **0.05 M**.

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