A solution is 0.15 M in both Pb²⁺ and Ag⁺. If Cl^- is added to this solution, what is [Ag⁺] when PbCl₂ begins to precipitate?

The precipitation of a compound may occur when its concentration exceeds its solubility. This can be due to temperature changes, solvent evaporation, or by mixing solvents. Precipitation occurs more rapidly from a stronglysupersaturated solution.

PbCl₂(s)⇌Pb²⁺(aq) + 2Cl^- (aq)

AgCl (s)⇌Ag⁺ (aq) + Cl^- (aq)

\(PbC{l_2}:{K_{sp}} = 1.6 \times 1{0^{ - 5}}\)

\({\rm{AgCl}}:{{\rm{K}}_{np}} = 1.6 \cdot {10^{ - 10}}\)

\(\left( {{\rm{A}}{{\rm{g}}^ + }} \right) = 0.15{\rm{M\;and\;}}\left( {{\rm{P}}{{\rm{b}}^{2 + }}} \right) = 0.15{\rm{M}}\)

\begin{aligned}{{K_{sp}}=\left[{{\rm{P}}{{\rm{b}}^{2+}}}\right]{{\left[{{\rm{C}}{{\rm{l}}^-}}\right]}^2}}\\{1.6\cdot{{10}^{-5}}=0.15\cdot{{\left[{{\rm{C}}{{\rm{l}}^-}}\right]}^2}}\\{\left[{{\rm{C}}{{\rm{l}}^-}}\right]=\sqrt{\frac{{1.6\cdot{{10}^{-5}}}}{{0.15}}}}\\{\left[{{\rm{C}}{{\rm{l}}^-}}\right]=1.03\cdot{{10}^{-2}}}\end{aligned}

\begin{aligned}{{{\rm{K}}_{sp}}=\left[{{\rm{A}}{{\rm{g}}^+}}\right]\left[{{\rm{C}}{{\rm{l}}^-}}\right]}\\{1.6\cdot{{10}^{-10}}=\left[{{\rm{A}}{{\rm{g}}^+}}\right]\cdot1.03\cdot{{10}^{-2}}}\\{\left[{{\rm{A}}{{\rm{g}}^+}}\right]=\frac{{1.6\cdot{{10}^{-10}}}}{{1.03\cdot{{10}^{-2}}}}}\\{\left[{{\rm{A}}{{\rm{g}}^+}}\right]=1.55\cdot {{10}^{-8}}}\end{aligned}

When PbCl₂starts to precipitate, the concentration of Ag⁺ is 1.55.10^-8.