Question

A solution contains 0.036 M Ag+ and 0.032 M Pb2+. If you add CI", AgCl and PbCI, will begin to precipitate. What is the concentration of Cl" required, in molarity, when AgCl precipitation begins? concentration of Cl" = What is the concentration of Cl required, in molarity when AgCl precipitation is 99.99% complete? concentration of Cl" = What is the concentration of CI required, in molarity when PbCl, precipitation begins? concentration of Cl" = What is the concentration of Cl" required, in molarity when PbCI, precipitation is 99.99% complete? concentration of Cl" = by precipitation Finally, give the Cl" concentration range in which Agt can be completely separated from Pb Give the lowest Cl" concentration for the separation of Agt from Pb2+. concentration of Cl" = M

Ksp of AgCl = 1.77x10^-10

Ksp of PbCl2 = 1.70x10^-5

thanks!

Answer 1

[Ay+ (04.)] = 0.036M [ P6+2 (09.)] = 0.032M Ksp (Agcl) = 1.77x10-10 Ksp (Agci) = [ Ag+] [CM] [cl - (aq)] = ksp - 1.77816-10 . [Agtea) 0.036 er(oq.)] = 4.92X10-9 M. Concentration of cio - 14.92X10-9 M. # After 99.99% Precipitation Left ammouution Agt of in solution = 0.0368 0.01 = 3.6866-64 Now Ksp (AgCl) = [ Ag+] [CI] [cr ] = Ksp = 1.77X100 = 4.92X/OMF 36x10-6 [c1-(04.)] = 4.92x10-5 M. [Ag+] For PbCla Kep (Pbcle) = [P6+21 [01-12 Tor² = ksp (Pbch) Срь+27

Tr] - 1.70x105-53.125 x 10-5 0.032 [cr] = 5.3125*10-4 M. [cr] = 2.3x10-2M ICT(0%) =2-3X102 M * After 99.99% ammount of Precipitation of PbCl left pbt2 in solution = 0.01% left [pbt2 Coq. ] = 0.032x0.01 = 3.2x10-6 M 100 Kap [ Pbc lg] = [Pb+2] [cr]2 [c1-(09.772 - - Ksp (PP Cly) - 170x10-5 [Pb+2 (89.) I 3.2x10-6 = 17.0 = 5.3125 372 Ler-(09) 7= √5.3125 M = 2.3M. om # Lowest cie concentration for the separation of Agt from Pb+2 = 4.92X10-5M Highest cio concentration for the separation of A from Pf+2 Conch of cia 2.3x10 - 2M