Equal volumes of 0.1M AgNO 3 ​ and 0.2M NaCl are mixed. The concentration of NO 3 − ​ ions in the mixture will be:

Options

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

Answer

To find the concentration of \( \text{NO}_3^- \) ions in the mixture after mixing equal volumes of 0.1 M \( \text{AgNO}_3 \) and 0.2 M \( \text{NaCl} \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction between silver nitrate and sodium chloride is:
\[ \text{AgNO}_3 + \text{NaCl} \rightarrow \text{AgCl} + \text{NaNO}_3 \]

2. **Define the Volumes**:
Let the volume of each solution (both \( \text{AgNO}_3 \) and \( \text{NaCl} \)) be \( V \).

3. **Calculate Moles of Reactants**:
- Moles of \( \text{AgNO}_3 \):
\[ \text{Moles of } \text{AgNO}_3 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{M} \times V = 0.1V \]
- Moles of \( \text{NaCl} \):
\[ \text{Moles of } \text{NaCl} = \text{Molarity} \times \text{Volume} = 0.2 \, \text{M} \times V = 0.2V \]

4. **Determine the Limiting Reagent**:
The stoichiometry of the reaction shows that 1 mole of \( \text{AgNO}_3 \) reacts with 1 mole of \( \text{NaCl} \). Since we have:
- \( 0.1V \) moles of \( \text{AgNO}_3 \)
- \( 0.2V \) moles of \( \text{NaCl} \)

\( \text{AgNO}_3 \) is the limiting reagent because it is present in a smaller amount.

5. **Calculate Moles After Reaction**:
- Moles of \( \text{AgNO}_3 \) consumed: \( 0.1V \) (all of it)
- Moles of \( \text{NaCl} \) consumed: \( 0.1V \) (since it reacts in a 1:1 ratio)
- Moles of \( \text{NaCl} \) remaining:
\[ 0.2V - 0.1V = 0.1V \]
- Moles of \( \text{NaNO}_3 \) produced: \( 0.1V \)
- Moles of \( \text{AgCl} \) produced: \( 0.1V \)

6. **Total Volume of the Mixture**:
The total volume after mixing both solutions:
\[ \text{Total Volume} = V + V = 2V \]

7. **Calculate Concentration of \( \text{NO}_3^- \) Ions**:
The moles of \( \text{NO}_3^- \) ions come from \( \text{NaNO}_3 \):
\[ \text{Moles of } \text{NO}_3^- = 0.1V \]
The concentration of \( \text{NO}_3^- \) ions is given by:
\[ \text{Concentration} = \frac{\text{Moles}}{\text{Total Volume}} = \frac{0.1V}{2V} = 0.05 \, \text{M} \]

### Final Answer:
The concentration of \( \text{NO}_3^- \) ions in the mixture is **0.05 M**.

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