For AgCl(s) ⇌ Ag+(aq) + Cl-(aq) with Ksp = 1.8×10^-10, if [Ag+] = [Cl-] = 1.0×10^-4 M, is the solution supersaturated, saturated, or unsaturated, and what will happen?

• A. Saturated; no precipitation
• B. Unsaturated; more solid will dissolve
• C. Supersaturated; precipitation occurs until equilibrium
• D. Equilibrium with continuous dissolution and precipitation

Answer: (C)

The main idea is comparing the ionic product to the solubility product to see whether the solution is at, below, or above the equilibrium level.

For AgCl, the solubility product is Ksp = [Ag+][Cl-] = 1.8 × 10^-10. With [Ag+] = [Cl-] = 1.0 × 10^-4 M, the ionic product is [Ag+][Cl-] = (1.0 × 10^-4)(1.0 × 10^-4) = 1.0 × 10^-8. This is larger than Ksp, so the solution is supersaturated with respect to AgCl. As a result, Ag+ and Cl- will combine to form solid AgCl, and precipitation will occur until the concentrations drop to the equilibrium values where [Ag+][Cl-] = Ksp. At equilibrium, with equal concentrations, [Ag+] = [Cl-] = sqrt(Ksp) ≈ sqrt(1.8 × 10^-10) ≈ 1.3 × 10^-5 M. The solid may remain present, maintained by ongoing dissolution and precipitation that keep the system at equilibrium.