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: Understanding the Concept:
NaCl is fully ionised and gives \([\text{Cl}^-] = 0.1\) M. This common ion suppresses the dissolving of AgCl.

Step 2: Key Formula or Approach:
\[ K_{sp} = [\text{Ag}^+][\text{Cl}^-] \]
Chloride from AgCl itself is tiny next to 0.1 M, so we take \([\text{Cl}^-] \approx 0.1\).

Step 3: Detailed Explanation:
\[ S = [\text{Ag}^+] = \frac{1.8\times10^{-10}}{0.1} = 1.8\times10^{-9}\ \text{M} \]
In pure water the solubility would be \(\sqrt{1.8\times10^{-10}} = 1.34\times10^{-5}\) M. The common ion effect reduces it by about four orders of magnitude.
Options (B), (C) and (D) are smaller by a further factor of 10 or more and do not obey the \(K_{sp}\) expression.

Step 4: Final Answer:
\(S = 1.8\times10^{-9}\) M, option (A).

Final Answer:
S is 1.8e-9 M because of the common ion effect. \[ \boxed{\text{(A) }1.8\times10^{-9}\ \text{M}} \]