Question

6. (1 pt) Cd forms a 3:1 complex in three separate steps, with bidentate ligand 1,10-phenanthroline Cphen"). The successive log Kr's are: Log Ki -5.17. Log K:-4.83, Log K 4.25. (a) Write balanced chemical equations showing the three steps in forming the Cd(phen)) complex. the appropriate Kr beside the three equations. Write (b) (1 pt) Calculate Kceveall for the formation of Cd(phen). 6, continued, part (c) (1 pt) Using your balanced overall equation, the Kcovrali and this data, prepare an ICE table and calculate the amount of free [Cd] at equilibrium. (assume plf is such that-100% of phen is available for complexation, ie. a 1.00) Initial conc. of Cd-3.74x 10 Initial conc. of phen 0.00400 M M Free Cd cone at equilibrium

Answer 1

(1) c d t + Phen & [cd (Phon) & log k, = 5.17 =) k, = 1.484105 [cd (Phon) 2] + Phon ke [a (Phom), 24] - log kz = 4.83 7 kg = 6,76x104 [ca (Phenia 2] + then kez (Cd (phen), J. loj ks = 4.25 =) k = 1.78x104 in Overall ovgal Cd2+ + 3 phen - [ Col (Phan) ] Kf=k, x K₂ x kz log ke = loge, + log kt logkez = 14.25 K = 1.788104 41 Ելոուն Մ թր Zadod/610C

K>> 2+ (c) Cd²+ + phen - cd (phen) 0.000374 0.004 0 = 0.004 -0.00377 0.000374 - An = 0.00 30626 sine k is very large, reaction will go to Completion. let remuning Cd2+ = on, In and reaction: Cd (phen) 2+ + Phon 3 Ca (phon 24 0.000374 0.003626 ~ 0 0.003626- 0.000374 0.000374 = 0 = 0.003252 Again, ky is very large, this realhon will also go to completion. 1701

or k >> 27 Nowy 3rd equation Co (phen) 2+ + Phen 0.000374 0.003252 - Ca(phen at 0.003252 - 0000374 = 0.002878 0.000374 =Ddz here also, kz is very large. Now, kz = [cd (phon), 2+] [calphen) 24] [phen =) 1. 78 X107 = 0.000374 Az x 0.002878 - Drz = [ca(Phon) 2+] = 7.3 x 100 m

Use and ₂ amation [co (phen) 2+ ] [co (Phan) 2+] [Phen = 6.76X104 - Diz D12 X 0.002878 x 0.002878 - DY₂ = [ca(Phen) 217 = 3.75X108 m = 7.3 x 156 une 1st equation: K, - red (Phon) 2+] = 12 [2] Phen) Dr, x 0.00287 1.48X105 = 3.758158 an, x 0:002878 on = [G 2+] = 3.75x168 Tennaing 1.48x lor 40.00 2878 - 8.8 x 10 M 7
Final phen concentration will be same in all three equation as these are simultaneous equilibrium, and is equal to 0.002878M .