What is the concentration of nitrate ions if equal volumes of `0.1 M AgNO_(3)` and `0.1 M NaCl` are mixed together?

Options

A. (a)`0.1 N`
B. (b)`0.2 M`
C. (c )`0.05 M`
D. (d)`0.25 M`

Explanation

To find the concentration of nitrate ions (NO₃⁻) after mixing equal volumes of 0.1 M AgNO₃ and 0.1 M NaCl, we can follow these steps:

### Step 1: Determine the initial moles of each solute
- **For AgNO₃**:
- Concentration = 0.1 M
- Volume = 1 L (assuming we take 1 L for simplicity)
- Moles of AgNO₃ = Concentration × Volume = 0.1 mol/L × 1 L = 0.1 mol

- **For NaCl**:
- Concentration = 0.1 M
- Volume = 1 L
- Moles of NaCl = Concentration × Volume = 0.1 mol/L × 1 L = 0.1 mol

### Step 2: Identify the reaction that occurs
When AgNO₃ and NaCl are mixed, they react to form AgCl (a precipitate) and NaNO₃ in solution:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} \downarrow + \text{NaNO}_3 \]

### Step 3: Determine the limiting reagent
In this case, both AgNO₃ and NaCl are present in equal amounts (0.1 mol each), so they will completely react with each other:
- Moles of AgNO₃ used = 0.1 mol
- Moles of NaCl used = 0.1 mol

After the reaction, both will be consumed completely.

### Step 4: Calculate the moles of NaNO₃ produced
- Moles of NaNO₃ produced = Moles of AgNO₃ reacted = 0.1 mol

### Step 5: Calculate the total volume after mixing
- Total volume = Volume of AgNO₃ + Volume of NaCl = 1 L + 1 L = 2 L

### Step 6: Calculate the concentration of nitrate ions (NO₃⁻)
- Moles of NO₃⁻ from NaNO₃ = 0.1 mol (produced from the reaction)
- Concentration of NO₃⁻ = Moles of NO₃⁻ / Total Volume = 0.1 mol / 2 L = 0.05 M

### Final Answer
The concentration of nitrate ions (NO₃⁻) after mixing is **0.05 M**.

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