500 mL of `0.1` N solution of `AgNO_(3)` is added to 500 mL of a `0.1` N KCl solution .the concentration of nitrate in the resulting mixture is

Options

A. `0.1` N
B. `0.05` N
C. `0.01` N
D. `0.2 `N

Answer

To solve the problem, we need to determine the concentration of nitrate ions (NO3-) in the resulting mixture after mixing 500 mL of 0.1 N AgNO3 with 500 mL of 0.1 N KCl. Here’s a step-by-step solution:

### Step 1: Calculate the moles of AgNO3
First, we need to calculate the number of moles of AgNO3 in the solution.

- **Normality (N)** of AgNO3 = 0.1 N
- **Volume (V)** of AgNO3 solution = 500 mL = 0.5 L

Using the formula:
\[ \text{Moles of AgNO3} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Moles of AgNO3} = 0.1 \, \text{N} \times 0.5 \, \text{L} = 0.05 \, \text{moles} \]

### Step 2: Calculate the moles of KCl
Next, we calculate the number of moles of KCl in the solution.

- **Normality (N)** of KCl = 0.1 N
- **Volume (V)** of KCl solution = 500 mL = 0.5 L

Using the same formula:
\[ \text{Moles of KCl} = \text{Normality} \times \text{Volume (in L)} \]
\[ \text{Moles of KCl} = 0.1 \, \text{N} \times 0.5 \, \text{L} = 0.05 \, \text{moles} \]

### Step 3: Determine the reaction
The reaction between AgNO3 and KCl can be written as:
\[ \text{AgNO}_3 + \text{KCl} \rightarrow \text{AgCl} + \text{KNO}_3 \]

From the reaction, we can see that 1 mole of AgNO3 reacts with 1 mole of KCl to produce 1 mole of AgCl and 1 mole of KNO3.

### Step 4: Identify the limiting reagent
Since both AgNO3 and KCl have the same number of moles (0.05 moles), they will completely react with each other. Therefore, there will be no limiting reagent, and all of both reactants will be consumed.

### Step 5: Calculate the moles of NO3- produced
From the balanced equation, for every mole of AgNO3 that reacts, one mole of KNO3 is produced. Thus, the moles of NO3- produced will be equal to the moles of AgNO3 used:
\[ \text{Moles of NO3-} = 0.05 \, \text{moles} \]

### Step 6: Calculate the total volume of the solution
The total volume of the resulting mixture is:
\[ \text{Total Volume} = 500 \, \text{mL} + 500 \, \text{mL} = 1000 \, \text{mL} = 1 \, \text{L} \]

### Step 7: Calculate the concentration of NO3-
Finally, we can calculate the concentration of nitrate ions (NO3-) in the resulting mixture:
\[ \text{Concentration of NO3-} = \frac{\text{Moles of NO3-}}{\text{Total Volume (in L)}} \]
\[ \text{Concentration of NO3-} = \frac{0.05 \, \text{moles}}{1 \, \text{L}} = 0.05 \, \text{M} \text{ or } 0.05 \, \text{N} \]

### Final Answer:
The concentration of nitrate in the resulting mixture is **0.05 N**.

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