The solubility of AgCl in 0.2 M NaCl solution (`K_(sp)` for `AgCl = 1.20 xx 10^(-10)`) is

Options

A. 0.2 M
B. `1.2 xx 10^(-10) M`
C. `0.2 xx 10^(-10) M`
D. `6.0 xx 10^(-10) M`

Explanation

To find the solubility of AgCl in a 0.2 M NaCl solution, we can use the concept of the solubility product constant (Ksp) and the common ion effect. Here are the steps to solve the problem:

### Step 1: Write the dissociation equation for AgCl
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

### Step 2: Write the expression for Ksp
The solubility product constant (Ksp) for AgCl can be expressed as:
\[ K_{sp} = [Ag^+][Cl^-] \]

### Step 3: Define the solubility in the presence of NaCl
Let the solubility of AgCl in the 0.2 M NaCl solution be 's' mol/L. In this case, since NaCl completely dissociates, the concentration of Cl⁻ ions from NaCl is already 0.2 M. Therefore, the total concentration of Cl⁻ ions in the solution will be:
\[ [Cl^-] = 0.2 + s \approx 0.2 \]
(s is negligible compared to 0.2 M)

### Step 4: Substitute into the Ksp expression
Now, we can substitute the concentrations into the Ksp expression:
\[ K_{sp} = [Ag^+][Cl^-] = s \cdot 0.2 \]

### Step 5: Substitute the value of Ksp
Given that \( K_{sp} = 1.20 \times 10^{-10} \), we can write:
\[ 1.20 \times 10^{-10} = s \cdot 0.2 \]

### Step 6: Solve for 's'
Now, we can solve for 's':
\[ s = \frac{1.20 \times 10^{-10}}{0.2} \]
\[ s = 6.0 \times 10^{-10} \]

### Conclusion
The solubility of AgCl in 0.2 M NaCl solution is \( 6.0 \times 10^{-10} \) M.

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