Question

Calculate pCd2+ at each of the given points in the titration of 55.00 mL of 0.0080 M Cd2+ with 0.0080 M EDTA in the presence of the auxiliary complexing agent NH3. The solution is buffered at a pH of 11.00 and the NH3 concentration is fixed at 0.100 M. The formation constant for the Cd2+−EDTA complex is Kf=2.9×1016.

Calculate pCd2+ at each of the given points in the titration of 55.00 mL of 0.0080 M Cd2+ with 0.0080 M EDTA in the presence of the auxiliary complexing agent NH,. The solution is buffered at a pH of 11.00 and the NH, concentration is fixed at 0.100 M. The formation constant for the Cd2+-EDTA complex is K = 2.9 x 1016. Oml PCd2+ = 5.7 5.00 mL pCd2+ = 13 50.00 mL pCa2+ = 5.7 55.00 mL pCd2+ = 14.8 60.00 mL 60.00 ml pcă+ = 6 pCa2+ =

Answer 1

8DL: the fraction Cadmium, in the forms of cat is af follows : Req2t 1+BI CL] + P2 [L]+B[]+By [LJU Here (L) = 0.1, B23162,5), B, 147, p.3=16**, B4 = 16096 4.4 6,56 10 Ca2+ = catet 1720" scous +10%-4760197653013 +16*5*10.04 1+324+295 +599 +36 3 = 7.81x104 the value of averat. PH 11,0 is 0.81. Now, using all these valus, Kf = a cost dys.kg .: +(7. $t xc04) (0-811(2.9 x006) = 1i8x103 a) . Bebore Equivalence point - OML the Equivalence point is 56m1the fraction of cont? remaining is SS :55 the diluhion factor is see thes, the conc. of Cadmium is not bound to EDTA is : Cca2+ = () Co-doko m) (SD 20.00 80 mg foithe Cone.of edzt is as followe

[cq24) = 2 cq2t. Cca24 = (4.81x104) (0.0080) = 6.24 106 pealt = -log Cca2+] -10g (6.24x276 ) = 5.204 Å 5.2 b) Before the equivalent point - 5mi the equivalence polut is ssmi, the traction of Cd2+ remaining som 60 the dilution factor is 55 itus , the conc. Of cadmium Dot bound to EDTA IS 9682. - 5.0 (0.008) (66) = 0.0066M efence calculate the cone of ca²t as follows (Cd2+] = d.cg27 C Ca2+ = (7i glx104) (0,0066) = 5.15X10CM -5p642+ = -log Cca2+] s log (5.15X106) = 5.288 5.29 © Before equivalence point - somi the equivalence point is ssmi, fraction of Ca2+ remaining is so the dilution factor is lor

ерте е thers, the conc. Of cadmium is not bound to Ceght = (sa) Co.oOeum) 20.00038MY for the conc. Of cart is as follows [code+] = &qet e copt (ti 81x104) (0.00038) 5.2.96x107 m of pcd²f = -log [ ca ²+ ] - Flog.C.296X16*m) - 6,528 Ž 6.53 d) At Equivalence point -55m At Ezuivertence point, the diction factor is for fer [cay-]. = 55.60.0060) llo 0.0040M Construct ICE table Fedzt. + EPTA I(M) 0 0 = Cay?? 0.004m 0.004-2 Esex ke"= [cayer] [[cart] [epta] 1, Axio!¿ 0.004.-2

C1.8x4o'3) +2 -0.004 50 = 1,49x107 so the conc. of Ca 2+ is as follows [Ca 2+] = Xcq2t Ccalt = (1.814164). (1.49x15%) 511163x1012 Mi pea2t = -log [cart] = -log C1163x10-12M) 510-934 = 10.93 @ After the equivalence point -com Almost all cadmium is in the form caly 2 with a dilution factor or so for cadmium [C1423 = C0-0090) = 0.0038 mg we also know the conc. of excess EDTA, whose dilution factor is 5 Eota) = f (00oso) = 0.00034 M Now, use the Equilibrium constant to find (ca2+] [aya) = 444 kg [cazt[ EDTA]

(0.0038) t(0.41(2.9x1016) [Ca2+] (000034) (Ca2+] = (0.0038) (0:1) (219x1016) (0.00034) 4.45761076 mg & pcd2f = log C cazt} E-log C 4.45841566) = 15.322 1532