Consider a solution containing 0.181 m lead (ii) ions and 0.174 m mercury(ii) ions. calculate the maximum concentration of sulfide ions that can be in solution without precipitating any lead ions. ksp: lead sulfide is 3.4 × 10−28; mercury(ii) sulfide is 4.0 × 10−53.

Consider the reduction reactions and their equilibrium constants. cu+(aq)+e−↽−−⇀cu(s)pb2+(aq)+2e−↽−−⇀pb(s)fe3+(aq)+3e−↽−−⇀fe(=6.2×108=4.0×10−5=9.3×10−3 cu + ( aq ) + e − ↽ − − ⇀ cu ( s ) k =6.2× 10 8 pb 2 + ( aq ) +2 e − ↽ − − ⇀ pb ( s ) k =4.0× 10 − 5 fe 3 + ( aq ) +3 e − ↽ − − ⇀ fe ( s ) k =9.3× 10 − 3 arrange these ions from strongest to weakest oxidizing agent.