Question

The solubility of AgCl in 0.1 M NaCl is S mol/L. If the solubility product of AgCl is \(1.8\times 10^{-10}\), then S is approximately:

• A. \(1.8\times 10^{-9}\) M
• B. \(1.8\times 10^{-10}\) M
• C. \(1.8\times 10^{-11}\) M
• D. \(1.8\times 10^{-12}\) M

Hint:

The common ion Cl- from NaCl is 0.1 M, so S = Ksp / 0.1.

Answer 1

The Correct Option is A

Step 1: Write the ion product:
At saturation $[\text{Ag}^+]\,[\text{Cl}^-] = 1.8\times10^{-10}$.

Step 2: Total chloride:
$[\text{Cl}^-] = 0.1 + S$. Since $S$ is far smaller than 0.1, use $[\text{Cl}^-] = 0.1$.

Step 3: Solve:
$S \times 0.1 = 1.8\times10^{-10}$ gives $S = 1.8\times10^{-9}$ mol/L. That is option (A). Check: $S \ll 0.1$, so the approximation was fine.

Final Answer:
The solubility is 1.8e-9 mol/L. \[ \boxed{\text{(A) }1.8\times10^{-9}\ \text{M}} \]