The solubility product of AgCl at 298 K is — Electrochemistry Chemistry Question

💡 Solution & Explanation

The dissolution reaction AgCl(s) <=> Ag^+(aq) + Cl^–(aq) is the reverse of the formation reaction. Thus, its standard free energy change is ΔG° = +57 kJ/mol = +57,000 J/mol. Using ΔG° = -2.303 RT log(Ksp), we get 57,000 = -2.303 × 8.314 × 298 × log(Ksp) = -5705.8 × log(Ksp). Therefore, log(Ksp) = 57,000 / -5705.8 ≈ -10. This gives Ksp = 10^-10. Therefore, correct answer is B.