A solution contains 0.002 M Pb^2+ and 0.002 M Ag^+. When NaCl(s) is added to raise Cl− to 0.01 M, which substance precipitates first or at all?

• A. Neither PbCl2 nor AgCl will precipitate.
• B. Both PbCl2 and AgCl precipitate.
• C. Only PbCl2 precipitates.
• D. Only AgCl precipitates.

Answer: (D)

The key idea is solubility product and ion-product thresholds. For a salt to precipitate, the product of the ion concentrations in solution must reach or exceed its Ksp.

AgCl has a very small Ksp (about 1.8 × 10^-10). With 0.002 M Ag+, the minimum chloride concentration needed to start precipitation is [Cl-] = Ksp/[Ag+] ≈ 1.8×10^-10 / 0.002 ≈ 9×10^-8 M. That threshold is extremely low, so even a tiny amount of Cl- will cause AgCl to begin precipitating. At 0.01 M Cl-, the ion product is [Ag+][Cl-] = 0.002 × 0.01 = 2×10^-5, which is far above Ksp, confirming that AgCl will precipitate.

PbCl2 has a much larger Ksp (about 1.7 × 10^-5). To reach saturation with Pb2+ at 0.002 M, you’d need [Cl-] such that [Pb2+][Cl-]^2 = Ksp, giving [Cl-] ≈ sqrt(Ksp/[Pb2+]) ≈ sqrt(1.7×10^-5 / 0.002) ≈ 0.092 M. At 0.01 M Cl-, the ion product is [Pb2+][Cl-]^2 = 0.002 × (0.01)^2 = 2×10^-7, well below Ksp, so PbCl2 will not precipitate at this chloride level.

Thus, as chloride is added to 0.01 M, AgCl precipitates first (and PbCl2 remains dissolved at this level).