Question

A 1.00 L solution contains 2.00×10-4 M Cu(NO3)2 and 2.500×10-3 M ethylenediamine (en). The Kf for Cu(en)22+ is 1.00×1020. What is the concentration of Cu2+(aq ) in the solution?

Answer 1

no of moles of Cu(NO3)2   = molarity * volume in L

                                          = 2*10^-4 *1   = 2*10^-4 moles

no of moles of en               = molarity * volume in L

                                          = 2.5*10^-3 *1

                                           = 2.5*10^-3 moles

The Kf for Cu(en)22+ is 1.00×10^20 . Kf value is very large

at equilibriumno of moles of Cu^2+ [Cu^2+]   = x

                         [en]      = 2.5*10^-3 -2*10^-4     = 0.0023 moles

                                                            [ Cu(en)2]^2+   = 2*10^-4 moles

Cu^2+ (aq) + 2en ---------------->[ Cu(en)2]^2+ (aq)

                       Kf   = [ Cu(en)2]^2+/[Cu^2+][en]^2

                          1*10^20 = 2*10^-4/x*0.0023

                            x              = 2*10^4/(10^20*0.0023)

                                            = 8.7*10^-14M

                           [Cu^2+]   = x   = 8.7*10^-14mole

                                                     = 8.7*10^-14/1 = 8.7*10^-14 M