Question 17.SE.17: A solution contains 1.0 × 10^-2 M Ag^+(aq) and 2.0 × 10^-2 M...
A solution contains 1.0 × 10^{-2} M Ag^+(aq) and 2.0 × 10^{-2} M Pb^{2+}(aq).When Cl^-(aq) is added, both AgCl (K_{sp} = 1.8 × 10^{-10}) and PbCl_2 (K_{sp} = 1.7 × 10^{-5}) can precipitate. What concentration of Cl^-(aq) is necessary to begin the precipitation of each salt? Which salt precipitates first?
Analyze We are asked to determine the concentration of Cl^-(aq) necessary to begin the precipitation from a solution containing Ag^+(aq) and Pb^{2 +}(aq) ions, and to predict which metal chloride will begin to precipitate first.
Plan We are given K_{sp} values for the two precipitates. Using these and the metal ion concentrations, we can calculate what Cl^-(aq) concentration is necessary to precipitate each salt. The salt requiring the lower Cl^-(aq) ion concentration precipitates first.
Solve For AgCl we have K_{s p}=\left[ Ag ^{+}\right]\left[ Cl ^{-}\right]=1.8 \times 10^{-10}.
Because \left[ Ag ^{+}\right]=1.0 \times 10^{-2}, the greatest concentration of Cl^-(aq) that can be present without causing precipitation of AgCl can be calculated from the K_{sp} expression:
\begin{aligned}K_{s p}&=\left(1.0 \times 10^{-2}\right)\left[ Cl ^{-}\right]=1.8 \times 10^{-10}\\{\left[ Cl ^{-}\right]}&=\frac{1.8 \times 10^{-10}}{1.0 \times 10^{-2}}=1.8 \times 10^{-8}M\end{aligned}
Any Cl^-(aq) in excess of this very small concentration will cause AgCl to precipitate from solution. Proceeding similarly for PbCl_2, we have
\begin{gathered}K_{s p}=\left[ Pb ^{2+}\right]\left[ Cl ^{-}\right]^2=1.7 \times 10^{-5}\\\left(2.0 \times 10^{-2}\right)\left[ Cl ^{-}\right]^2=1.7 \times 10^{-5}\\{\left[ Cl ^{-}\right]^2=\frac{1.7 \times 10^{-5}}{2.0 \times 10^{-2}}=8.5 \times 10^{-4}}\\{\left[ Cl ^{-}\right]=\sqrt{8.5 \times 10^{-4}}=2.9 \times 10^{-2}\,M}\end{gathered}
Thus, a concentration of Cl^-(aq) in excess of 2.9 × 10^{-2} M causes PbCl_2 to precipitate.
Comparing the Cl^-(aq) concentration required to precipitate each salt, we see that as Cl^-(aq) is added, AgCl precipitates first because it requires a much smaller concentration of Cl^-.
Thus, Ag^+(aq) can be separated from Pb^{2+}(aq) by slowly adding Cl^-(aq) so that the chloride ion concentration remains between 1.8 \times 10^{-8}\,M and 2.9 \times 10^{-2}\,M
Comment Precipitation of AgCl will keep the Cl^-(aq) concentration low until the number of moles of Cl^-(aq) added exceeds the number of moles of Ag^+(aq) in the solution. Once past this point, [ Cl^-] rises sharply and PbCl_{2} will soon begin to precipitate.