CuCl has a Ksp equal to 1.2 x 10-6. A 1.0 L saturated solution of CuCl contains a concentration of Cu+ equal to 1.1 x 10-3 M. If 500 mL of water is evaporated from the solution, what will be the concentration of Cu+ after equilibrium is reestablished? (Hint: Think about Qsp vs Ksp.) 5.5 x 10-4 M 1.1 x 10-3 M 2.2 x 10-3 M 0 M (no ions will be in solution)

Given that Solubility product of CuCl is Ksp =1.2×10-6 Concentration of Cu+ before the evaporation of water is [Cu+] =1.1×10-3 M The concentration of Cl- before the evaporation of water is calculated using the solubility product expression. Ksp =[Cu+] [Cl-] [Cl-] =Ksp [Cu+] [Cl-] =Ksp [Cu+] =1.2×10-61.1×10-3=1.1×10-3 M [Cl-] =1.1×10-3 MAs the water evaporates, the Cu+ and Cl- precipitate in 1:1 ratio to make the solid CuCl(s). When the equilibrium is reestablished after evaporation of water, the product of concentrations of Cu+ and Cl- equals to the solubility product because the following equilibrium is established CuCl(s) ⇔ Cu+(aq) + Cl-(aq) Ksp =[Cu+] [Cl-] ......................(1) In other words, when the equilibrium is reestablished after the evaporation of water, some of the solid CuCl(s) dissolves to maintain the aqueous solution concentration of Cu+ and Cl- such that the product [Cu+] [Cl-] equals to Ksp. Ksp =[Cu+] [Cl-] As per the stoichiometry of equation (1), we have [Cu+] =[Cl-] Ksp =[Cu+] [Cu+] [Cu+] =Ksp [Cu+] =1.2×10-6 =1.1×10-3 M Answer is 1.1×10-3 M