When equal volumes of following solution are mixed, precipitation of `AgCl` ?
`(K_(sp)=1.8xx10^(-10))` will occur only with

Options

A. `10^(-4) M (Ag^(+)) and 10^(-4) M (Cl^(-))`
B. `10^(-5) M (Ag^(+)) and 10^(-5) M (Cl^(-))`
C. `10^(-5) M (Ag^(+)) and 10^(-6) M (Cl^(-))`
D. `10^(-10) M (Ag^(+)) and 10^(-10) M (Cl^(-))`

Answer

To determine when precipitation of AgCl will occur upon mixing equal volumes of different solutions, we need to analyze the ionic product (Q) of the solution and compare it with the solubility product constant (Ksp) of AgCl, which is given as \( K_{sp} = 1.8 \times 10^{-10} \).

### Step-by-Step Solution:

1. **Understanding the Condition for Precipitation**:
- Precipitation occurs when the ionic product \( Q \) exceeds the solubility product \( K_{sp} \). Therefore, we need to calculate \( Q \) for each solution and check if \( Q > K_{sp} \).

2. **Formula for Ionic Product**:
- The ionic product \( Q \) for the precipitation of AgCl can be expressed as:
\[ Q = [Ag^+][Cl^-] \]
- When equal volumes of solutions are mixed, the concentrations of ions will be halved.

3. **Calculating Q for Each Option**:
- For each option, we will calculate \( Q \) using the formula:
\[ Q = \frac{m_1 \cdot V_1}{V_1 + V_2} \]
where \( m_1 \) is the initial concentration of the ion and \( V_1 \) and \( V_2 \) are the volumes of the solutions mixed (both are equal).

4. **Option A**:
- Assume the concentration of \( Ag^+ \) is \( 10^{-4} \, M \) and \( Cl^- \) is \( 10^{-4} \, M \):
\[ Q = \left(\frac{10^{-4}}{2}\right) \left(\frac{10^{-4}}{2}\right) = \frac{10^{-4} \times 10^{-4}}{4} = \frac{10^{-8}}{4} = 2.5 \times 10^{-9} \]

5. **Option B**:
- Assume the concentration of \( Ag^+ \) is \( 10^{-10} \, M \) and \( Cl^- \) is \( 10^{-4} \, M \):
\[ Q = \left(\frac{10^{-10}}{2}\right) \left(\frac{10^{-4}}{2}\right) = \frac{10^{-10} \times 10^{-4}}{4} = \frac{10^{-14}}{4} = 2.5 \times 10^{-15} \]

6. **Option C**:
- Assume the concentration of \( Ag^+ \) is \( 10^{-5} \, M \) and \( Cl^- \) is \( 10^{-6} \, M \):
\[ Q = \left(\frac{10^{-5}}{2}\right) \left(\frac{10^{-6}}{2}\right) = \frac{10^{-5} \times 10^{-6}}{4} = \frac{10^{-11}}{4} = 2.5 \times 10^{-12} \]

7. **Option D**:
- Assume the concentration of \( Ag^+ \) is \( 10^{-20} \, M \) and \( Cl^- \) is \( 10^{-4} \, M \):
\[ Q = \left(\frac{10^{-20}}{2}\right) \left(\frac{10^{-4}}{2}\right) = \frac{10^{-20} \times 10^{-4}}{4} = \frac{10^{-24}}{4} = 2.5 \times 10^{-25} \]

8. **Comparing Q with Ksp**:
- Now we compare each calculated \( Q \) with \( K_{sp} = 1.8 \times 10^{-10} \):
- Option A: \( Q = 2.5 \times 10^{-9} > K_{sp} \) (Precipitation occurs)
- Option B: \( Q = 2.5 \times 10^{-15} < K_{sp} \) (No precipitation)
- Option C: \( Q = 2.5 \times 10^{-12} < K_{sp} \) (No precipitation)
- Option D: \( Q = 2.5 \times 10^{-25} < K_{sp} \) (No precipitation)

### Conclusion:
- Precipitation of AgCl will occur only with **Option A**.