Solubility of AgCl in the solution containing 0.01 M NaCl is `[given, K_sp (AgCL)= 1.6 x 10^-10]`

Options

A. `8 x 10^-9`
B. `1.6 x 10^-9`
C. `2 x 10^-8`
D. `1.6 x 10^-8`

Answer

To find the solubility of AgCl in a solution containing 0.01 M NaCl, we can follow these steps:

### Step 1: Write the dissociation equation for AgCl
AgCl is a sparingly soluble salt that dissociates in water as follows:
\[ \text{AgCl (s)} \rightleftharpoons \text{Ag}^+ (aq) + \text{Cl}^- (aq) \]

### Step 2: Define the solubility (S)
Let the solubility of AgCl in the solution be \( S \) mol/L. This means that the concentration of Ag\(^+\) ions will be \( S \) and the concentration of Cl\(^-\) ions contributed by AgCl will also be \( S \).

### Step 3: Consider the contribution of Cl\(^-\) from NaCl
Since we have a 0.01 M NaCl solution, NaCl will dissociate completely into Na\(^+\) and Cl\(^-\):
\[ \text{NaCl (s)} \rightleftharpoons \text{Na}^+ (aq) + \text{Cl}^- (aq) \]
Thus, the concentration of Cl\(^-\) from NaCl is 0.01 M.

### Step 4: Write the expression for Ksp
The solubility product constant \( K_{sp} \) for AgCl is given by:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Substituting the concentrations, we have:
\[ K_{sp} = S \cdot (0.01 + S) \]
However, since \( S \) is expected to be very small compared to 0.01 M, we can approximate:
\[ K_{sp} \approx S \cdot 0.01 \]

### Step 5: Substitute the given Ksp value
We know that \( K_{sp} = 1.6 \times 10^{-10} \). Therefore, we can write:
\[ 1.6 \times 10^{-10} = S \cdot 0.01 \]

### Step 6: Solve for S
Rearranging the equation to solve for \( S \):
\[ S = \frac{1.6 \times 10^{-10}}{0.01} \]
\[ S = 1.6 \times 10^{-8} \]

### Conclusion
The solubility of AgCl in a 0.01 M NaCl solution is \( 1.6 \times 10^{-8} \) M.