The solubility of AgCl in 0.2 M NaCl is `[K_(sp) AgCl =1.8 xx 10^(-10)]`

Options

A. `1.8 xx 10^(-11)M`
B. `9.0 xx 10^(-10)M`
C. `6.5 xx 10^(12)M`
D. `5.6 xx 10^(-11)M`

Explanation

To find the solubility of AgCl in a 0.2 M NaCl solution, we can follow these steps:

### Step 1: Understand the Dissociation of AgCl
AgCl dissociates in water according to the following equation:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Set Up the Expression for Ksp
The solubility product constant (Ksp) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Given that \( K_{sp} = 1.8 \times 10^{-10} \).

### Step 3: Define the Solubility in the Presence of NaCl
Let the solubility of AgCl in the 0.2 M NaCl solution be \( S \). In this case, the concentration of Cl⁻ ions from NaCl is already 0.2 M. Therefore, the total concentration of Cl⁻ ions in the solution will be:
\[ [\text{Cl}^-] = 0.2 + S \]
However, since \( S \) will be very small compared to 0.2 M, we can approximate:
\[ [\text{Cl}^-] \approx 0.2 \, \text{M} \]

### Step 4: Substitute into the Ksp Expression
Now, substituting the concentrations into the Ksp expression:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \times 0.2 \]
Thus,
\[ 1.8 \times 10^{-10} = S \times 0.2 \]

### Step 5: Solve for S
To find the solubility \( S \):
\[ S = \frac{1.8 \times 10^{-10}}{0.2} \]
\[ S = 9.0 \times 10^{-10} \, \text{M} \]

### Conclusion
The solubility of AgCl in 0.2 M NaCl is:
\[ S = 9.0 \times 10^{-10} \, \text{M} \]