The solubility product of AgCl is `4.0 xx 10^(-10)` at 298 K . The solubility of AgCl in 0.04 M Ca `Cl_(2)` will be

Options

A. `2.0 xx 10^(-5) M`
B. `1.0 xx 10^(-4)M`
C. `5.0 xx 10^(-9) ` M
D. `2. 0 xx 10^(-4) M`

Explanation

To find the solubility of AgCl in a 0.04 M CaCl₂ solution, we will use the concept of the solubility product (Ksp) and the common ion effect.

### Step-by-step Solution:

1. **Write the Dissociation Equation for AgCl:**
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

2. **Define the Solubility Product (Ksp):**
The solubility product constant (Ksp) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given that \( K_{sp} = 4.0 \times 10^{-10} \) at 298 K.

3. **Determine the Concentration of Chloride Ions:**
In a 0.04 M CaCl₂ solution, each formula unit of CaCl₂ produces 2 chloride ions (Cl⁻). Therefore, the concentration of Cl⁻ ions is:
\[ [\text{Cl}^-] = 2 \times 0.04 \, \text{M} = 0.08 \, \text{M} \]

4. **Set Up the Expression for Ksp:**
Let \( s \) be the solubility of AgCl in the presence of the common ion (Cl⁻). The concentration of Ag⁺ ions will be \( s \) and the concentration of Cl⁻ ions will be \( 0.08 \, \text{M} + s \). However, since \( s \) will be very small compared to 0.08 M, we can approximate:
\[ [\text{Cl}^-] \approx 0.08 \, \text{M} \]

5. **Substitute into the Ksp Expression:**
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = s \times 0.08 \]
Therefore, we have:
\[ 4.0 \times 10^{-10} = s \times 0.08 \]

6. **Solve for s:**
\[ s = \frac{4.0 \times 10^{-10}}{0.08} \]
\[ s = 5.0 \times 10^{-9} \, \text{M} \]

Thus, the solubility of AgCl in a 0.04 M CaCl₂ solution is \( 5.0 \times 10^{-9} \, \text{M} \).