The solubility product of AgCl is `1.8xx10^(-10)`. Precipitation of AgCl will occur only when equal volumes of solutions of :

Options

A. `10^(-8)"M "Ag^(+)and 10^(-8)"M "Cl^(-)` ions
B. `10^(-3)"M "Ag^(+)and 10^(-3)"M "Cl^(-)` ions
C. `10^(-6)"M "Ag^(+)and 10^(-6)"M "Cl^(-)` ions
D. `10^(-10)"M "Ag^(+)and 10^(-10)"M "Cl^(-)` ions

Explanation

To determine when precipitation of AgCl will occur, we need to compare the ionic product (Qsp) with the solubility product (Ksp) of AgCl. The precipitation occurs when Qsp exceeds Ksp.

### Step-by-Step Solution:

1. **Understanding Ksp**:
The solubility product (Ksp) of AgCl is given as \(1.8 \times 10^{-10}\). This value indicates the maximum product of the concentrations of the ions (Ag\(^+\) and Cl\(^-\)) that can exist in solution without precipitating.

2. **Expression for Qsp**:
The ionic product (Qsp) is calculated using the concentrations of the ions in solution:
\[ Qsp = [Ag^+][Cl^-] \]

3. **Analyzing Options**:
We need to evaluate the given options based on their concentrations of Ag\(^+\) and Cl\(^-\) to find when Qsp is greater than Ksp.

- **Option A**:
\[ [Ag^+] = 10^{-8} \, \text{M}, \quad [Cl^-] = 10^{-8} \, \text{M} \]
\[ Qsp = (10^{-8})(10^{-8}) = 10^{-16} \]
Since \(10^{-16} < 1.8 \times 10^{-10}\), no precipitation occurs.

- **Option B**:
\[ [Ag^+] = 10^{-3} \, \text{M}, \quad [Cl^-] = 10^{-3} \, \text{M} \]
\[ Qsp = (10^{-3})(10^{-3}) = 10^{-6} \]
Since \(10^{-6} > 1.8 \times 10^{-10}\), precipitation occurs.

- **Option C**:
\[ [Ag^+] = 10^{-6} \, \text{M}, \quad [Cl^-] = 10^{-6} \, \text{M} \]
\[ Qsp = (10^{-6})(10^{-6}) = 10^{-12} \]
Since \(10^{-12} < 1.8 \times 10^{-10}\), no precipitation occurs.

- **Option D**:
\[ [Ag^+] = 10^{-5} \, \text{M}, \quad [Cl^-] = 10^{-5} \, \text{M} \]
\[ Qsp = (10^{-5})(10^{-5}) = 10^{-10} \]
Since \(10^{-10} = 1.8 \times 10^{-10}\), no precipitation occurs.

4. **Conclusion**:
The only condition where precipitation of AgCl occurs is in **Option B**.