Equal volumes of 0.1 M `AgNO_(3)` and 0.2 M NaCl solutions are mixed. The concentration of nitrate ions in the resultant mixture will be

Options

A. 0.1 M
B. 0.2 M
C. 0.05 M
D. 0.15 M

Explanation

To find the concentration of nitrate ions in the resultant mixture after mixing equal volumes of 0.1 M AgNO₃ and 0.2 M NaCl solutions, we can follow these steps:

### Step 1: Write the balanced chemical equation
When AgNO₃ reacts with NaCl, it forms AgCl and NaNO₃:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \]

### Step 2: Determine the number of moles of each reactant
Assume we mix equal volumes \( V \) liters of both solutions.

- Moles of AgNO₃:
\[ \text{Moles of AgNO}_3 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times V \, \text{L} = 0.1V \, \text{mol} \]

- Moles of NaCl:
\[ \text{Moles of NaCl} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{mol/L} \times V \, \text{L} = 0.2V \, \text{mol} \]

### Step 3: Identify the limiting reagent
From the balanced equation, we see that AgNO₃ and NaCl react in a 1:1 ratio. Since we have 0.1V moles of AgNO₃ and 0.2V moles of NaCl, AgNO₃ is the limiting reagent because it will be completely consumed first.

### Step 4: Calculate the remaining moles after the reaction
- Moles of AgNO₃ consumed: 0.1V (all of it)
- Moles of NaCl consumed: 0.1V (since it reacts in a 1:1 ratio with AgNO₃)

Remaining moles of NaCl:
\[ \text{Remaining moles of NaCl} = 0.2V - 0.1V = 0.1V \, \text{mol} \]

### Step 5: Calculate the moles of NaNO₃ produced
Since 0.1V moles of AgNO₃ reacted, 0.1V moles of NaNO₃ will be produced.

### Step 6: Determine the total volume of the solution
The total volume after mixing the two solutions is:
\[ \text{Total volume} = V + V = 2V \, \text{liters} \]

### Step 7: Calculate the concentration of nitrate ions
The concentration of nitrate ions (from NaNO₃) in the resultant mixture is given by:
\[ \text{Concentration of NO}_3^- = \frac{\text{Moles of NaNO}_3}{\text{Total Volume}} = \frac{0.1V}{2V} = \frac{0.1}{2} = 0.05 \, \text{M} \]

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

---