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 will use the solubility product constant (K_sp) given for AgCl, which is \(1.6 \times 10^{-10}\).

### Step-by-Step Solution:

1. **Write the Dissolution Reaction**:
AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq)

2. **Define the Solubility**:
Let the solubility of AgCl in the solution be \(S\) moles per liter. When AgCl dissolves, it produces \(S\) moles of Ag⁺ and \(S\) moles of Cl⁻.

3. **Account for Cl⁻ from NaCl**:
Since the solution already contains 0.01 M NaCl, which completely dissociates, the concentration of Cl⁻ from NaCl is 0.01 M. Therefore, the total concentration of Cl⁻ in the solution will be:
\[ [Cl⁻] = S + 0.01 \]

4. **Set Up the K_sp Expression**:
The solubility product expression for AgCl is given by:
\[ K_{sp} = [Ag⁺][Cl⁻] \]
Substituting the values we have:
\[ K_{sp} = S \times (S + 0.01) \]
Since \(K_{sp} = 1.6 \times 10^{-10}\), we can write:
\[ 1.6 \times 10^{-10} = S \times (S + 0.01) \]

5. **Assume S is Small**:
Since NaCl contributes a significant amount of Cl⁻ (0.01 M), we can assume that \(S\) is very small compared to 0.01 M. Thus, we can approximate:
\[ S + 0.01 \approx 0.01 \]
Therefore, the equation simplifies to:
\[ 1.6 \times 10^{-10} = S \times 0.01 \]

6. **Solve for S**:
Rearranging the equation gives:
\[ S = \frac{1.6 \times 10^{-10}}{0.01} = 1.6 \times 10^{-8} \]

7. **Conclusion**:
The solubility of AgCl in the solution containing 0.01 M NaCl is:
\[ S = 1.6 \times 10^{-8} \text{ M} \]

### Final Answer:
The solubility of AgCl in the solution containing 0.01 M NaCl is \(1.6 \times 10^{-8} \text{ M}\).

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