H v tn sinh vin:T Kim Cn 1014021M s dung dch kim tra: 07L Ngc Cng 1014029M s dung dch kim tra: 65M s nhm thc tp: CN N 2 Ch nht, ngy 11 thng 3 nm 2012BI TNG TRNH PHN TCH NH LNGPHNG PHP CHUN COMPLEXON1. Vi gii thiu ban u:1.1. Dung dch m:Phn ng chun lun c thc hin ti mt gi tr pH rt xc nh ca dung dch m thch hp.V d chun Mg2+ bng EDTA ti pH = 10:Mg2+ + H2Y2- MgY + 2H+Phn ng cho thy vi mi mol Mg2+ tham gia phn ng li gii phng 2 mol H+ i vo trong dung dch. Vy phi c dung dch m duy tr pH n nh, trnh s thay i ca cc hng s iu kin lm sai lch kt qu chun .1.2. Cc lu khc:Chun complexom l phng php phn tch nh lng dng xc nh nng ca cc hu ht cc ion kim loi ni chung, tr kim loi kim. EDTA phn ng to phc bn vi hu ht cc kim loi theo t l 1:1 nn rt thun li trong vic tnh ton kt qu chun .Phng php chun complexom c nhiu u im trong vic xc nh hm lng cc ion kim loi trong dung dch. Tuy nhin phng php ny hn ch trong vic xc nh hn hp cc kim loi khng tinh khit (v EDTA phn ng hu ht vi cc kim loi nn tnh chon lc khng cao). Do ngi ta c th nng cao tnh chon lc bng vic s dng cht che.C nhiu phng php chun complexom nh: chun trc tip, chun thay th, chun ngc... mi phng php c p dng trong tng trng hp ring bit ty thuc vo i tng cn xc nh, iu kin hin c ca phng th nghim...2. Pha cc dung dch v ch th mu kim loi:2.1. Gii thiu:Dung dch ETDA ban u do phng th nghim chun b c nng c 1.000000N, ta pha long thnh cc dung dch EDTA c nng thp hn s dng cho cc th nghim t y tr v sau.2.2. Pha dung dch ETDA nng 0.01N:Ly ng 5ml dung dch ETDA ban u (do phng th nghim cung cp) nh mc trong bnh 500ml:( )( )( )( )0.95, 0.95nh 500ml0.95, nh 500ml 0.95nh 500ml1.000000 0.0000051.000000 0.0000050.030* 1.96* 0.0591 15.000 0.0590.050* 1.96* 0.0981 1500.000 0.098EDTAEDTAburetVburetburet EDTAbVbbC MN NzV V mlzV t t t t ml 1N1N nh 500ml 1N1NEDTA 0.01Nnh 500ml2 2EDTA 0.01N EDTA 0.01N EDTA 0.01N0.95,0.01N 1N1N nh 500ml* 5.000*1.000000N = 0.01500.000N N N* * *EDTA EDTA bEDTA EDTAbNEDTA V N VEDTA EDTA bV NVV N V _ _ _ + +

, , ,( )22 2 20.01 0.01 0.01*0.059 *0.000005 *0.098 0.000125.000 1.000000 500.0000.01000 0.00012EDTAN N

_ _ _ + + , , , t2.3. Pha dung dch ETDA nng 5.10-4N:Sau khi np dung dch EDTA 0.01N va mi pha long vo buret. M kha buret ly ng 25ml dung dch ny mang i nh mc trong bnh 500ml:( )0.95, 0.95 0.01N0.030* 1.96* 0.0591 125.000 0.059buretVburetburet EDTAzV V ml t( )-4-4-4nh 500ml0.95, nh 500ml 0.95nh 500mlEDTA 0.01N EDTA 0.01NEDTA 5.10 Nnh 500mlEDTA 5.10 N0.95,5.10 N 00.050* 1.96* 0.0981 1500.000 0.098* 25.000*0.01000N = 0.0005500.000NbVbbbNEDTAEDTAzV mlV NVV t -4 -4 0.01N0.01N nh 500ml2 2 2EDTA 5.10 N EDTA 5.10 N.01N0.01N nh 500ml2 2 2N N* * *0.0005 0.0005 0.0005*0.059 *0.00012 *0.098 0.025.000 0.01000 500.000EDTA EDTA bV N VEDTA bN V _ _ _+ + , , , _ _ _ + + , , ,000061( ) 0.0005000 0.0000061EDTAN N tEDTA phng th nghim cung cp c s dng lm cht chun gc, khng cn tin hnh chun li. y cng l mt thun li ca phng php complexon so vi acid baz.3. Xc nh Mg2+:3.1. Gii thiu:Chun mt th tch chnh xc mui Mg2+bng dung dch complexon III trong m amoniac pH = 10 vi ch th NET n khi mu chuyn t nho sang chm xanh chm.3.2. Phn ng chun :Mg Y MgY + (tt c u khng mu)Phn ng ch th:Xy ra khi cho d mt git complexon III:{{MgIn Y MgY In + + nho xanh chmC bn l mu ca MIn khc vi mu ca In- 3.3. Gii thch l thuyt:Do / /MgY MgInpK pK > nn phc MgIn c xu hng b phn hy to thnh phc MgY bn hn.Vi nng C0 gn bng 0.01, khong bc nhy pMg tnh gn ng bng4.30 6.24 . Ta c pMg cui 1 = pMgIn pKa + pH m = 7 11.6 + 10 = 5.4, suy ra pMg cui 2 = 5.4 + 1 = 6.4. Ta thy ch c pMg cui 1 l nm trong khong bc nhy, vy chn pMg cui 1 kt thc chun . Tc l mu trung gian (chm xanh chm).3.4. Tin hnh chun :Dng pipet ht 10.00ml dung dch kim tra cho vo erlen 250ml. Thm 10ml m pH = 10, 20mg ch th NET v chun bng EDTA 0.01M n khi dung dch chuyn t nho sang chm xanh chm (mu trung gian). Lm thao tc tng t nh trn 3 ln v ghi li kt qu VEDTA.3.5. S liu thc nghim:Dung dch EDTA 0.01000NChun Mg2+Ch th NET, m pH = 10VETDA(ml) V1V2V3Vtb9.80 9.85 9.80 9.8173.6. Kt qu v tho lun:( )( )( )2220.95, 0.9520.950.070* 1.96* 0.00793 310.0000 0.00790.01000 0.000120.02886750.95, * 4.30* 0.0723 39.817 0.072* 0.01000*9.8170.010.0000pipetV pipetMgEDTAEDTAETDA ETDAMgMgzV mlN NsV EDTA tV mlN VNV+++ t t t 098172 2 2222222 20.95,2 2 2* * *0.009817 0.009817 0.009817*0.00012 *0.072 *0.0079 0.000140.01000 9.817 10.00000.009ETDA ETDAMgMg Mg MgN V VNMgETDA ETDAMgMgN N NN V VN + + +++++ _ _ _ + + , , , _ _ _ + + , , , ( ) 82 0.00014 N tNguyn tc c bn quyt nh chn kt thc chun mu trung gian hay r rt c trnh by phn gii thch l thuyt bn trn. Tc l im i mu phi nm trong khong bc nhy, hoc ch t phi cng gn khong bc nhy cng tt. Nu pM cui 1 v c pM cui 2 u nm lt vo trong khong bc nhy, ta u tin chn pM cui 2 hn (do mu r rt, d nhn). T y tr v sau nu khng c gii thch g thm th vic chn mu trung gian hay r rt kt thc chun u tun theo nguyn tc nh vy.Khi ta cho ch th NET, mu c th m nht khc nhau ty thuc vo lng ch th s dng. Mu sc t cam (hay cam) cho n nho m. Nu s dng lng ch th nhiu, mu dung dch rt m th s kh khn trong vic nhn ra s i mu. V vy lm nhiu php chun s gip ta nhn bit c lng cht ch th c s dng bao nhiu l hp l nht.4. Xc nh cng chung ca nc my:4.1. Gii thiu: cng chung ca nc l do s c mt ca ca mui Mg2+v Ca2+tan trong nc. c biu din bng s mili ng lng gam kim loi (m g ) c trong mt lt nc cng. Khng phn bit t l Mg2+: Ca2+ l bao nhiu.Thng th nng Ca2+ c trong nc my kh thp (nh hn 0.001M). Thc nghim cho thy php xc nh ny ch cho kt qu ng vi iu kin trong mu nc phi hin din Mg2+. Tc l khng th xc nh ng cng ca nc my nu trong ch c mi Ca2+. Ti sao vy? Ion Mg2+ ng vai tr g trong php chun ny?4.2. Phn ng chun :CaIn Y CaY InMgIn Y MgY In+ ++ +Phn ng thay th nh lng:CaIn MgY CaY MgIn + +Phn ng ch th xy ra khi cho thm mt git complexon III:MgIn Y MgY In + +In xanh chm.4.3. Gii thch l thuyt:Ta chun mt th tch chnh xc ca nc my trong mi trng m pH = 10 bng dung dch ETDA 5.10-4N vi ch th NET. Trong iu kin cc cation Ca2+ v Mg2+ ng thi c chun v tha iu kin / /8.24 5.4MgY MgInpK pK > v / /10.24 3.8CaY CaInpK pK > . Ta phi chothmKCNlinkt kimloi nngctrongnc(nht lvt Cu2+), chothm NH2OH.HCl ngn nga s to thnh Mn (IV) trong mi trng km ph hy ch th NET.4.4. Tin hnh chun :Ly Vmu= 50ml nc my cho vo erlen 250ml. Thm 10ml m pH=10, ri thm 10 git KCN 10%, 10 git dung dch NH2OH.HCl 1%, lc u, thm khong 40mg ch th NET. Tin hnh chun bng dung dch EDTA 0,01M n khi chuyn mu t nho sang xanh chm r rt. Lm thao tc tng t nh trn 3 ln v ghi li kt qu VEDTA.4.5. S liu thc nghim:Dung dch EDTA 5.10-4NVnc my = 50mlCh th NETVETDA(ml) V1V2V3Vtb23.90 24.00 23.90 23.934.6. Kt qu v tho lun:Ta xem nh Vrng = 0( )( )( )nh500.95, 0.9520.95, 0.950.017* 1.96* 0.0193 350.000 0.0190.0577350* 4.30* 0.143 323.93 0.140.0005000 0.0000061b mlVmaumauV ETDAEDTAETDAzV mlstV mlN N t t t10002 2 22 2* *2* 23.93*0.0005000*2*10000.478650* * *0.4786 0.4786 0.4786*0.14 *0.0000061 *0.0123.93 0.0005000 50.000EDTA ETDA mauEDTA ETDAmauH V N VEDTA ETDA mauV N VHVH H HV N V _ _ _ + + , , , _ _ + + , ,( ) ( )29 0.0110.479 0.011 H m g _ , t Vi phc CaY bn hn phc MgY vic xc nh iu kin chun ch cn xt da trn iu kin ca pMgY. (pKCaY = 10.7, pKMgY = 8.7). Nh vy im cui ca chun l s chuyn mu ca phc MgIn (do phc MgY km bn hn CaY nn s xy ra CaIn MgY CaY MgIn + +). D thy s c mt ca Mg2+ s lm cho php chun chnh xc (s lm r vai tr ca Mg2+ hn th nghim sau).Do nng Mg2+ v Ca2+ trong nc kh thp nn khi chun ta s dng mt lng ln mu (50ml) ng thi s dng dung dch EDTA c nng rt thp 5.10-4N tng th tch s dng, lm cho phn ng chun t chnh xc cao.V trong thnh phn ca nc cng ngoi 2 ion chnh l Mg2+ v Ca2+ cn nhiu ion kim loi nng khc cho nn ta cn s dng cht che v cht bo v . Ion Cu2+ s to phc rt bn vi ch th NET, lm cho khng c phn ng ch th ti im tng ng (gii phng In- nhn bit kt thc qu trnh chun ), bng cch s dng CN- to phc bn vi ion Cu2+ s lm mt kh nng to phc ca ion ny vi ch th. Ion Mn2+trong mi trng baz (pH = 10), s chuyn thnh Mn4+ph hy ch th NET. Nn ta cho NH2OH.HCl vo nhm ngn nga s to thnh Mn4+.5. Xc nh cng chung ca nc cng:5.1. Gii thiu:Tng t nh vic xc nh cng ca nc my, nhng ta tin hnh vi loi nc cng cng hn do phng th nghim cung cp.5.2. Phng trnh chun :CaIn Y CaY InMgIn Y MgY In+ ++ +Phn ng thay th nh lng:CaIn MgY CaY MgIn + +Phn ng ch th xy ra khi cho thm mt git complexon III:MgIn Y MgY In + +In xanh chm.5.3. Cch thc tin hnh:Ly Vmu= 5ml nc cng cho vo erlen 250ml. Thm 10ml m pH=10, 20mg ch th NET. Tin hnh chun bng dung dch EDTA 5.10-4M n khi chuyn mu t nho sang xanh chm r rt. Lm thao tc tng t nh trn 3 ln v ghi li kt qu VEDTA.5.4. S liu thc nghim:Dung dch EDTA 5.10-4NChun cng ca nc cng (5.00ml)Ch th NETVEDTA(ml) V1V2V3Vtb26.35 26.45 26.40 26.405.5. Kt qu v tho lun:Ta xem nh Vrng = 0.( )( )( )0.95, 0.9520.95, 0.950.0070* 1.96* 0.00793 35.0000 0.00790.05* 4.30* 0.123 326.40 0.120.0005000 0.0000061pipetVmaumauV ETDAEDTAETDAzV mlstV mlN N t t t10002 2 22 2* *2* 26.40*0.0005000*2*10005.285.0000* * *5.28 5.28 5.28*0.12 *0.0000061 *0.007926.40 0.0005000 5.0000EDTA ETDA mauEDTA ETDAmauH V N VEDTA ETDA mauV N VHVH H HV N V _ _ _ + + , , , _ _ + + , , ( ) ( )20.0695.280 0.069 H m g_ , t Tng t nh xc nh cng chung ca nc my, vic xc nh cng ca nc cng cng da vo s c mt ca ion Mg2+.Mt s thi im quan trong ca chun :F = 0.99: pCa = pC0 + pDF + 2 = 5.3F = 1.00: pCa = 0.5(pC0 + pDF+ pKCaY) = 6.77F = 1.01: pCa = pKCaY p(F1) =8.24.Khong bc nhy pCa = 8.24 5.3 = 2.94.Khi s dng ch th NET :Mu trung gian: pMg = pKMgIn = 5.4 pCacui = 7.4 + p([CaY]/[MgY])Mu r rt: pMg = pKMgIn + 1 = 6.4 pCacui = 8.4 + p([CaY]/[MgY])Ta nhn thy pCacuity thuc vo t l nng CaY v MgY. Nu [CaY]/[MgY] =10/1 pCa cui= 6.4 i vi im cui mu trung gian v pCa cui= 7.4 i vi im cui mu r rt. C hai trng hp ny u c pCacuinm trong khong bc nhy 5.3 8.24 nn ta chun cho ti khi dung dch chuyn t mu nho sang xanh chm r rt.6. Xc nh Ca2+ nng thp khi c thm MgY:6.1. Gii thiu:Khi nng ban u C0 ca Ca2+ tng i ln ta c th chun Ca2+ bng complexon III hon ton tng t nh Mg2+ bng cch dng ch th NET vi m pH = 10. Nhng khi nng C0ban u ca Ca2+ qu thp (c nh hn 0.001N) liu iu ny c cn chnh xc na hay khng?6.2. Gii thch l thuyt:6.2.1. Kh khn khi nng ban u ca Ca2+ qu thp:T cng thc tnh khong bc nhy ti F = 0.99: pCa = pC0 + pDF + 2, ta nhn thy khi nng ban u C0 ca Ca2+ cng nh, im u ca khong bc nhy cng cao ln trong khi im cui hu nh khng thay i trong cng iu kin pH tng t. iu c ngha l khong bc nhy ngy cng b thu hp.Mt s thi im quan trong ca chun , tnh gn ng khi nng Ca2+vo c 0.0005 0.001N:F = 0.99: pCa = pC0 + pDF + 2 = 5.3F = 1.00: pCa = 0.5(pC0 + pDF+ pKCaY) = 6.77F = 1.01: pCa = pKCaY p(F1) = 8.24Tc l cn phi c mt cht ch th sao cho 5.30