Question

At 298 K, the conductivity of a saturated solution of AgCl in water is \( 2.6 \times 10^{-5} \, \text{S cm}^{-1} \). Its solubility product at 298 K is

• A. \( 2.0 \times 10^{-5} \, \text{M}^2 \)
• B. \( 4.0 \times 10^{-10} \, \text{M}^2 \)
• C. \( 4.0 \times 10^{-8} \, \text{M}^2 \)
• D. \( 2.0 \times 10^{-6} \, \text{M}^2 \)

Hint:

The solubility product can be calculated using the conductivity and solubility of the salt in solution.

Answer 1

The Correct Option is C

Step 1: Conductivity and Solubility Product.
The conductivity \( \kappa \) of a saturated solution of AgCl can be related to its solubility product \( K_{sp} \). The solubility product \( K_{sp} \) is given by: \[ K_{sp} = s^2 \] where \( s \) is the solubility of AgCl. Using the conductivity, we can calculate \( s \) and thus \( K_{sp} \).
Step 2: Conclusion.
The correct answer is (C), \( 4.0 \times 10^{-8} \, \text{M}^2 \).