To solve the problem of determining which solution gives the maximum precipitate when mixed, we need to calculate the ionic product (IP) for each solution and compare it with the solubility product (Ksp) of AgCl, which is given as \(10^{-12}\).
### Step-by-Step Solution:
1. **Understanding the Concept of Ionic Product and Ksp**:
- The ionic product (IP) is calculated as the product of the molar concentrations of the ions in solution.
- A precipitate will form when the ionic product exceeds the solubility product (Ksp).
2. **Initial Concentrations**:
- We have four solutions with different concentrations of Ag\(^+\) and Cl\(^-\):
- Solution A: \( [Ag^+] = 10^{-4} \, M \), \( [Cl^-] = 10^{-4} \, M \)
- Solution B: \( [Ag^+] = 10^{-3} \, M \), \( [Cl^-] = 10^{-3} \, M \)
- Solution C: \( [Ag^+] = 10^{-5} \, M \), \( [Cl^-] = 10^{-5} \, M \)
- Solution D: \( [Ag^+] = 10^{-6} \, M \), \( [Cl^-] = 10^{-6} \, M \)
3. **Mixing Equal Volumes**:
- When equal volumes of these solutions are mixed, the concentration of each ion is halved because the total volume doubles.
- Therefore, the new concentrations will be:
- Solution A: \( [Ag^+] = 5 \times 10^{-5} \, M \), \( [Cl^-] = 5 \times 10^{-5} \, M \)
- Solution B: \( [Ag^+] = 5 \times 10^{-4} \, M \), \( [Cl^-] = 5 \times 10^{-4} \, M \)
- Solution C: \( [Ag^+] = 5 \times 10^{-6} \, M \), \( [Cl^-] = 5 \times 10^{-6} \, M \)
- Solution D: \( [Ag^+] = 5 \times 10^{-7} \, M \), \( [Cl^-] = 5 \times 10^{-7} \, M \)
4. **Calculating the Ionic Product (IP)**:
- For each solution, calculate the ionic product:
- **Solution A**:
\[
IP = [Ag^+][Cl^-] = (5 \times 10^{-5})(5 \times 10^{-5}) = 25 \times 10^{-10} = 2.5 \times 10^{-9}
\]
- **Solution B**:
\[
IP = (5 \times 10^{-4})(5 \times 10^{-4}) = 25 \times 10^{-8} = 2.5 \times 10^{-7}
\]
- **Solution C**:
\[
IP = (5 \times 10^{-6})(5 \times 10^{-6}) = 25 \times 10^{-12} = 2.5 \times 10^{-11}
\]
- **Solution D**:
\[
IP = (5 \times 10^{-7})(5 \times 10^{-7}) = 25 \times 10^{-14} = 2.5 \times 10^{-13}
\]
5. **Comparing IP with Ksp**:
- We compare each ionic product with the Ksp of AgCl (\(10^{-12}\)):
- Solution A: \(2.5 \times 10^{-9} > 10^{-12}\) (Precipitate forms)
- Solution B: \(2.5 \times 10^{-7} > 10^{-12}\) (Precipitate forms)
- Solution C: \(2.5 \times 10^{-11} > 10^{-12}\) (Precipitate forms)
- Solution D: \(2.5 \times 10^{-13} < 10^{-12}\) (No precipitate forms)
6. **Conclusion**:
- The maximum ionic product is from Solution B (\(2.5 \times 10^{-7}\)), which is the highest among all calculated values.
- Therefore, the solution that gives the maximum precipitate is **Solution B**.
