When equal volume of the following solutions are mixed , which of the following gives maximum precipitate ? `(K_(sp) " of " AgCl= 10 ^(-12))`

Options

A. `10^(-4) MAg^+ and 10^(-4) MCl ^(-)`
B. `10^(-3) MAg^+ and 10^(-3) MCl ^(-)`
C. `10^(-5) MAg^+ and 10^(-5) MCl ^(-)`
D. `10^(-6) MAg^+ and 10^(-6) MCl ^(-)`

Answer

To determine which solution gives the maximum precipitate of AgCl when equal volumes of different solutions are mixed, we need to analyze the ionic product of Ag⁺ and Cl⁻ ions in each case. The precipitate will form when the ionic product exceeds the solubility product constant (Ksp) of AgCl, which is given as \( K_{sp} = 10^{-12} \).

### Step-by-Step Solution:

1. **Understand the Concept of Ionic Product**:
The ionic product (IP) is calculated using the formula:
\[ IP = [Ag^+] \times [Cl^-] \]
A precipitate forms when \( IP > K_{sp} \).

2. **Mixing Equal Volumes**:
When equal volumes of two solutions are mixed, the concentration of each ion in the resulting solution is halved. If the initial concentrations of Ag⁺ and Cl⁻ are \( [Ag^+] \) and \( [Cl^-] \), the concentrations after mixing will be:
\[ [Ag^+]_{new} = \frac{[Ag^+]}{2}, \quad [Cl^-]_{new} = \frac{[Cl^-]}{2} \]

3. **Calculate the Ionic Product for Each Case**:
We will calculate the ionic product for each option provided.

- **Option 1**:
- \( [Ag^+] = 10^{-4} \) M, \( [Cl^-] = 10^{-4} \) M
- After mixing:
\[ [Ag^+]_{new} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{M} \]
\[ [Cl^-]_{new} = \frac{10^{-4}}{2} = 5 \times 10^{-5} \, \text{M} \]
\[ IP = (5 \times 10^{-5}) \times (5 \times 10^{-5}) = 2.5 \times 10^{-9} \]

- **Option 2**:
- \( [Ag^+] = 10^{-3} \) M, \( [Cl^-] = 10^{-3} \) M
- After mixing:
\[ [Ag^+]_{new} = \frac{10^{-3}}{2} = 5 \times 10^{-4} \, \text{M} \]
\[ [Cl^-]_{new} = \frac{10^{-3}}{2} = 5 \times 10^{-4} \, \text{M} \]
\[ IP = (5 \times 10^{-4}) \times (5 \times 10^{-4}) = 2.5 \times 10^{-7} \]

- **Option 3**:
- \( [Ag^+] = 10^{-2} \) M, \( [Cl^-] = 10^{-2} \) M
- After mixing:
\[ [Ag^+]_{new} = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{M} \]
\[ [Cl^-]_{new} = \frac{10^{-2}}{2} = 5 \times 10^{-3} \, \text{M} \]
\[ IP = (5 \times 10^{-3}) \times (5 \times 10^{-3}) = 2.5 \times 10^{-5} \]

4. **Compare Ionic Products**:
- Option 1: \( IP = 2.5 \times 10^{-9} \)
- Option 2: \( IP = 2.5 \times 10^{-7} \)
- Option 3: \( IP = 2.5 \times 10^{-5} \)

The highest ionic product is from **Option 2**.

5. **Conclusion**:
Since the ionic product for Option 2 is the highest and exceeds the Ksp of AgCl, it will produce the maximum precipitate.

### Final Answer:
**Option 2** gives the maximum precipitate of AgCl.