Question

A 1.00 L solution contains 5.00x106-5 M Cu(NO3)2 and 1.00x10^-3 M ethylenediamine. What is the concentration of Cu2+ (aq) in the solution

Answer 1

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

                                           = 5*10^-5 *1 = 0.00005 moles

no of moles of en                 = molarity * volume in L

                                          = 1*10^-3 *1   = 0.001 moles

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

I              0.00005       0.001                        0

C            -0.00005       -2*0.00005                0.00005

E           0                     0.0009                     0.00005

            x

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

   1*10^20   = 0.00005/x*(0.0009)^2

x                = 0.00005/10^20*(0.0009)^2    = 6.17*10^-19 M

[Cu^2+]   = x = 6.17*10^-19 M >>>>answer