Bao Cao Full

Bo co thc tp phn tch mi trng

Bi 1: Xc nh dung lng trao i ca cationit v h s phn b ca mt s ion kim loi trn cationitI. II.1.

Mc ch th nghim Nguyn tc xc nh

- Xc nh dung lng trao i ca cationit - Xc nh h s phn b ca mt s kim loi trn cationit nh dung lng trao i ca cation Cho mt lng ionit xc nh tip xc vi mt th tch dung dch cht in ly cho trc n khi thit lp cn bng trao i ion: R-H + NaOH(dd) R-Na + H2O Hoc 2 R-H + CaCl2(dd) R2-Ca + 2 HCl Sau chun li lng NaOH cn li trong dung dch bng HCl hay xc nh lng HCl gii phng ra bng NaOH ta s x nh c dung lng trao i ca cationit R-H. - Xc nh h s phn b Lc mt th tch dung dch bit bngV(mL) cha ion cn trao i M n+ vi mt lng cn chnh xc cationnit kh l gR n khi thit lp cn bng trao i ion sau: nR-H + Mn+(dd) Rn-M + nH+(dd) + H s phn b khi lng DmDm = QR QS

QR lng ion Mn+ (mmol) i vo pha ionit QS -lng ion Mn+ (mmol) cn li trong dung dch + H s phn b trao i ion DgDg = V QR / g R = Dm * dd QS / Vdd gR

+ H s phn b trao i nng D c nu bit khi lng ring d(g/mL) ca ionit tc l s gam ionit c th tch Vionit (mL)DC = D g * d = Dm * Vdd g * R g R Vionit

+ H s phn b th tch DVDV = D g * = D m V dd gR VM = Dm g R VM + VS VM + VS

- nh gi th tch lu ca cht cn tch C th tnh th tch lu bng phng trnh sau: VR = V M + D c VS Hoc VR = VM(Dm + 1)M M Ngoi ra nu k hiu V= VM + VS l th tch ct, cn a a = V + V = V l phn M S th tch pha ng trong ct (hoc phn th tch lm vic ca ct), c th chng minh rng:

V

V

V R = V ( DV + a )

Thng thng a 0.4-1-


- nh gi chn lc ca mt php tch sc k trao i ion nh gi tnh chn lc ca php tch sc k hai ion A v B c th dng 2 i lng sau: c,B H s tch = D ( > 1) c, A

D

T s tch = D + a V ,A III. Kt qu tnh ton: 1. Xc nh dung lng trao i ca cationit - Xc inh m ca cationit Cc 1 (cha c cationit) Lc cha sy: m1 = 61.6252 g Lc sy ri : m2 = 61.4174 g Cc 2 (c cationit) Lc cha sy: m3 = 62.4133 g Lc sy ri : m4 = 62.4133 g m X% ca cationit l:X = 2.0 g *100 2.0

DV , B + a

g (khi lng cationit kh) = m4-m2 = 62.4133-61.4174 = 0.9959 g => X = (2-0.9959)/2*100 = 50.2% - Xc inh dung lng trao i Cho vo mi bnh nn I v II 1,0000 g cationit bit m (cn chnh xc bng cn phn tch). Thm vo bnh I 100 mL dung dch NaOH 0.1 N ; cho vo bnh nn II 100 mL dung dch CaCl2. y cht np bnh ri lc lin tc trong 3.0 h. Sau ht t mi bnh 25,00ml dung dch v chun , dng ch th phenolphtalein . Chun 2-3 ln ly kt qu trung bnh. Bnh I : chun NaOH d bng HCL 0.1 N. Bnh II : chun HCl sinh ra trong qu trnh trao i bng NaOH 0.1 N. K hiu : N1 nng ng lng ca dung dch chun NaOH; N2- nng ng lng ca dung dch chun HCl ; V1- th tch dung dch chun HCl tiu tn,mL ; V2 th tch dung dch HCl chun tiu tn,mL ; m- khi lng cationit ly ; X- m ca cationit, %.Q NaOH = 100 N 1 4V2 N 2 100 X (1) m( ) 100

Chng minh : S milli ng lng Na+ d trong 10 mL chun : (CNV)NaOH = (CNV)HCl = V2N2 meq => s mili ng lng ca Na+ d trong 100 mL dd u : V2N2*100/25= 4V2N2 meq s mili ng lng ca Na+ ban u : 100 N1 meq => s mili ng lng ca Na+ i vo cationit (100N1 - 4V2N2 ) meq

-2-

Bo co thc tp phn tch mi trng 100 N 1 4V 2 N 2 Q NaOH = 100 X => Dung lng trao i l : m( ) 100QCaCl 2 = 4V1 N 1 100 X (2) m( ) 100

Chng minh : s mili ng lng ca HCl trong 25 mL chun : (CNV)HCl = (CNV)NaOH = V1N1 meq => s mili ng lng ca HCl trong 100 mL dd u : V1N1*100/25=4V1N1 meq => s mili ng lng ca Ca2+ i vo cationit cng bng 4V1N1 meq => Dung lng trao i l :QCaCl 2 = 4V1 N 1 100 X m( ) 100

Kt qu th nghim : Bnh I: Chun NaOH d bng HCl 0.1 N V1 22.20 mLVH l C

V2 22.00 mL

V3 22.50 mL

= (22.20+22.00+22.50)/3 = 22.23 mL100 * 0.1 4 * 22 .23 * 0.1 100 50 .2 =2.22 meq/g 1( ) 100

Q NaOH =

Bnh II : Chun HCl sinh ra bng NaOH : V1 2.00 mLVNaOH

V2 1.80 mL

V3 1.80 mL

= 1.87 mL4 * 1.87 * 0.1 100 50 .2 = 1.50 meq/g 1( ) 100

QCaCl 2 =

2. Xc nh h s phn b ca mt s kim loi trn cationitKt qu th nghim :-3-

Bo co thc tp phn tch mi trngTh tch EDTA tiu tn khi chun Zn2+(mL) B1( 0.05N HCl) B2 ( 0.20N HCl) B3 ( 0.50N HCl) V1 V2 V3 Vtb 1.20 1.10 1.00 1.10 4.00 4.10 4.10 4.07 7.40 7.30 7.20 7.30

B4 ( 1.00N HCl) 9.10 9.10 9.10 9.10

V1 V2 V3 Vtb

Th tch EDTA tiu tn khi chun Cd2+(mL) B1( 0.05 NHCl) B2 ( 0.20 NHCl) B3 ( 0.50N HCl) 0.90 3.80 6.10 0.80 0.80 0.83 3.80 3.80 3.80 6.20 6.20 6.17

B4 ( 1.00N HCl) 7.40 7.20 7.40 7.33

CN(EDTA) = CM(EDTA) * 2= 0.01*2=0.02 N + S mmol ban u ca : Zn2+ : 20*1/65=0.31 mmol Cd2+ : 20*1/112,41= 0.18 mmol + Vdd = VM = 50 mL + gR= 1*(100-50.2)/100 = 0.498 g + Vionit = VS = 0.498/d= 0.498/0.64= 0.73 ml Me2+( Zn2+ v Cd2+) Xt trong 10.00 mL dd chun :C N (M e2+

)=

(CV ) EDTA a = V Me 2 + Me 2 + * V Me 2 +

=> a = (CV ) EDTA *Me2+ => n =(C N V ) EDTA * Me 2 + (C N V ) EDTA = M Me 2 + 2

=> suy ra s mmol ca Me2+ d trong 50 mL dd (QS)(C NV ) EDTA (C V ) *50/10= N EDTA *5 mmol 2 2

=> suy ra s mmol ca Me2+ i vo cationit (QR)S mmol ban u - QS p dng cng thc, ta c bng sau:

V 1.10

QS 0.055

QR 0.255

Dm 4.636

Tnh vi Zn2+ Dg Dc 465.498 317.559

Dv 4.570

VR 281.818

a 0.010 0.986-4-


4.07 7.30 9.10

0.204 0.107 0.365 -0.055 0.455 -0.145

0.523 -0.151 -0.319

52.544 -15.129 -31.996

35.845 -10.321 -21.827

0.516 -0.149 -0.314

76.167 42.466 34.066

V 0.83 3.80 6.17 7.33

QS QR 0.042 0.139 0.190 -0.010 0.309 -0.129 0.367 -0.187

Dm 3.337 -0.053 -0.417 -0.509 0.5 0.362 1.456

[HCl] 0.05 0.2 1.389 -9.943 1.299 1.608 -

Tnh vi Cd2+ Dg Dc Dv 335.075 228.586 3.289 -5.284 -3.605 -0.052 -41.820 -28.530 -0.411 -51.091 -34.854 -0.502 H s tch v t s tch 1 0.626 1.387

VR 216.867 -2.632 -20.827 -25.443

0.010

a 0.99

th biu din cc h s phn b theo HClZn2+ 5.00 4.00 3.00 2.00 1.00 0.00 -1.000.00

Dm th biu din Dm theo HCl

Zn2+ Cd2+ [HCl] (N) 0.50 1.00 1.50

Dg

-5-

Bo co thc tp phn tch mi trngZn2+ 500.00 400.00 300.00 200.00 100.00 0.00 -100.000.00 0.50 1.00 1.50 Zn2+ Cd2+ [HCl] (N)

th biu din Dg theo HCl

DcZn2+ 400.00 300.00 200.00 100.00 0.00 -100.000.00 0.50 1.00 1.50 [HCl] (N)

th biu din Dc theo HClZn2+ Cd2+

DVZn2+ 5.00 4.00 3.00 2.00 1.000.00 0.00 -1.00 [HCl] (N) 0.50 1.00 1.50

th biu din Dv theo HClZn2+ Cd2+

- th biu din h s tch theo HCl-6-

Bo co thc tp phn tch mi trngTy so tach 2 1.5 1 0.5 0 0.00

th biu din t s tch theo [HCl]

0.20

0.40

0.60

0.80

1.00

1.20

[HCl] (N)

IV. Nhn xt1. Xc nh dung lng trao i QNaOH = 2.22 meq/g > QCaCl2 = 1.50 meq/gV theo s trao i ion ca nha din ra theo t l 1RH :1NaOH v 2RH :1CaCl2 2. Xc nh cc h s phn b - Vi Zn2+ : Bnh 3,4 c QR < 0 ( v l) - Vi Cd2+ : bnh 2,3,4 c QR g R = * V = 0.45*1.26x106= 567000g= 567 Kg VM + VS V 567 100 X = 1138.55 Kg (nha kh) => mcationit = 100

Ta c =

b. Tng dung lng trao i c th t c ca ct ny Dung lng trao i ca R-SO3-H+ l 5meq/g => Tng dung lng trao i c th t c l 567*5=2835meq.

Bi 2: Tch Cu2+ v Ni2+ trn ct sc k trao i cationI.

Mc ch.-8-


- Tch Cu2+ v Ni2+ trn ct sc k trao i cation. - Nm r cch nhi ct v cc thao tc k thut ca phng php ng.

II. C s l thuyt.1. K thut sc k trao i.Sc k trao i ion c s dng rng ri ti cc ion v c v nhng cht hu c c th tn ti trng thi ion. Ct sc k trao i lm bng thy tinh hoc thy tinh hu c, y ct c kha iu chnh tc dng chy ca dung dch qua ct. Mt php tch sc k trao i ion gm cc giai on sau: Hp thu hn hp cn tch nn ct gi ton b cc ion trong dung dch nh phn, cn dng mt lng ionit ln hn lng tnh ton hn hp phn tch ch chim mt phn rt nh pha trn ct. Gii hp cc ion c hp thu trn ct: giai on ny, ln lt cc ion c gii hp ra khi ct bng cc dung dch thch hp nh cc acid c nng khc nhau hoc dung dch mt cht c th to phc bn vi cc ion cn tch, ty tng trng hp c th m bng mt dung dch c th gii hp lin tip mt s ion, c trng hp vi mi ion ngi ta dng mt dung dch gii hp ring ( nh bi thc hnh th nghim ny), cng c khi nng cht gii hp c lm gia tng dn dn trong qu trng gii hp ( gii hp gradient). Ra ionit sau khi gii hp Dng dung dch thch hp ui ht dung dch gii hp cn lu li trong ct trong cc khong trng gia cc ht ionit. Ti sinh ionit a iont v dng ban u. Ra ionit sau khi ti sinh Ra ct bng nc ct loi ht dung dch ti sinh cn trong ct, sau giai on ny c th dng ct cho php phn tch mi.2.

Tch Cu2+ v Ni2+ trn ct trao i cation.

Dng cationit acid yu dng Na+ lm pha tnh. Do cation acid yu c cc nhm sinh ion l nhm cacboxylic, loi cationit ny ch dng c trong mi trng kim hoc trung tnh v nhm carboxylic l acid yu, khng phn ly trong mi trng acid. Cation acid yu dng H+ c k hiu l R-COOH, cation acid yu dng Na +c k hiu l: R-COONa trong , R l nn polymer tr ca cationit. Tch sc k hn hp Cu2+ v Ni2+ trn ct nhi cationit acid yu dng Na+ gm cc giai on sau: Thay i dng ion linh ng ca cationit t dng H+sang dng Na+, Nhi vo ct Cationit acid yu dng H+, sau cho qua ct mt dung dch hn hp NaOH NaCl phn ng trao i ion xy ra nh sau: R-COOH + Na+ (dd) +OH-(dd) RCOONa + H2O Sau ra ct bng nc ct ui ht dung dch NaOH-NaCl d, kim tra bng cch nhng giy qu vo nc chy ra y ct ( pH=7). Hp thu hn hp Cu2+ v Ni2+ nn ct Cho hn hp cn tch Cu2+ v Ni2+ qua ct khi din ra cc phn ng trao i ion nh sau: 2RCOONa + Cu2+(dd) (RCOO)2Cu +2Na+(dd) -9-


2RCOONa + Ni2+(dd) (RCOO)2Ni + 2Na+(dd) Gii hp - Gii hp Cu2+ bng cch di qua dung dch glyxerin - NaOH, Ion Cu2+ to phc mang in m mu xanh dng vi glyxerinat, khng b gi bi cationit, trong khi Ni2+ khng to phc vn c gi trn cationit, - Gii hp Ni2+ bng dung dch HCl 3M, (R-COO)2Ni + 2H+(dd) 2R-COOH + Ni2+(dd) Mi ion kho st c gii hp bng mt dung dch khc nhau, Cht gii hp Cu2+ l mt tc nhn to phc, Ni2+ c gii hp n gin bng cch lm chuyn dch cn bng trao i ion. Ra ct sau khi gii hp: Dng nc ct ra cationit ui ht HCl cn li trong ct. nh lng Cu2+ v Ni2+ trong cc giai on gii hp - Xc nh Cu2+ trong phn on gii hp u tin bng phng php Iod thiosulphat. - Xc nh Ni2+ trong phn on gii hp bng HCl c xc nh bng phng php Complexon.

III. Kt qu th nghim1. nh lng Cu 2+ :

Phn ng nh lng ng : CuI + I2 Cu2+ + I I3I2 + I 2I3 + S2O3 S4O62- + I(CNN)Cu2+ = (CNV)I2 = (CNV)(S203)2=> CN Cu =(C N V ) ( S 2 O 3 ) VCu

Cthiosulphat = 0,01N Cu2+ = 64/2 = 32 Th tch dung dch thiosulphat tiu tn khi chun Cu2+ l: V1 V2 V3 9.30 mL 9.20 mL 9.30 mL V = 9.27 mL CN Cu 2+ ( trong 10mL dd chun ) =(C N V ) ( S 2O 3) 10

=

0.01 * 9.27 = 9.27x10-3 N 10

CN Cu 2+ ( trong 100mL phn on gii hp th nht) = 9.27x10-3 N => khi lng ng trong 100 ml ung dich ny l: m(Cu2+) = CNCu2+V= 9.27x10-3*32*100=29.66 mg 2. nh lng Ni 2+ V1 V2 V3 7.30 mL 7.20 mL 7.20 mL V = 7.23 mL CN EDTA= CM EDTA * 2= 0.01*2=0.02 N Ni2+ = 58.71/2 = 29.36 Phn ng chun :- 10 -

Ni + H3Ind- NiH2Ind- + H+ (tm) (mu vng) 2+ 2Ni + H2Y NiY2- +2H+ (CNN)Ni2+ = (CNV)H2Y22+


CN Ni2+ ( trong 10 mL dd chun) l: CN Ni2+ =

(C N V ) H 2Y 2 0.02 * 7.23 = = 0.0145 N V Ni 2 + 10

CN Ni2+ ( trong 100 ml dd phn on gii hp th hai) = 0.0145 => khi lng Niken trong 100 ml ung dich ny l: m(Ni2+) = CNNi2+V = 0.0145*29.36*100= 42.57 mg

Th tch dung dch EDTA tiu tn khi chun Ni2+ l: p dng quy lut ng lng, ta c: (CN*V)Ni = (CN*V)EDTA CEDTA = 0,01N; VNi = 10 mL; CN.Cu2+ = (0.01*8.4)/10 = 8.4*10^-3 N Ta c: CN =a/( *V) vi = 58.71/2 a = CN **V = 8.4.10^-3*(58.71/2)*0.01 = 2.46*10^-3 g

Bi 3: Xc nh cht ty ra trong nc bng phng php chit trc quang vi thuc th xanhmetylen.I. Tm tt l thuyt :

Nguyn tc ca phng php : cc cht ty ra dng c ,cha alkylbenzen sulfonat ( ABS) c mch cacbon phn nhnh cao , cc vi khun khng th lm thoi bin (phn hy) sinh hc n hon ton v cc vi khun mi ln phn hy hai cacbon trong mch v dng li ch phn nhnh.Do vy m mt loi cht ty ra mch thng LAS linear alkylbenzen sulphonat c kh nng bin thoi sinh hc c thay th cho cht ty ra dng c. Phng php chun xc nh cc cht ty ra trong nc l chit - trc quang sau khi ln mu vi thuc th xanhmetylen y l phng php nhy (xc nh di 0.1ppm cht ty ra ),ch cn dng c ,thit b n gin. Xanh metylen phn ng vi cc cht ty ra cha nhn benzen to thnh mt lin hp ion mu xanh tan trong chloroform .Cng mu tuyn tnh theo nh lutBeer trong khong hm lng 0.1 n 4 ppm . Cn tr trong php xc nh c cc sulphat ,sulphonat, carbocylat , phosphat hu c v cc phenol v chng lm tng cng mu, trong khi cc amin li lm gim mu do to phc cnh tranh vi xanhmetylen.Cn vi cc ion v c nh : Cl- , NO3- , SO42- , PO43- , SCN- , S2- khng cn tr. I.2 Tnh nng dung dch phn tch bng phng php ng chun:- 11 -


Quan h gia mt tnh cht vt l hoc ha l c o y no ca cht nh phn vi hm lng x ca n trong mu thng l quan h tuyn tnh: Y = a + bx Bng phng php chit- trc quang sau khi ln mu vi thuc th xanh metylen ta s xy dng c ng chun ,t phng trnh ng chun ta xc nh c hm lng cht ty ra trong mu nc phn tch. Trong phng php ng chun, ngi ta chun b mt lot n mu chun c hm lng cht nh phn bit x1 , x2 ,, xn .o tnh cht y tng ng vi cc mu ny c cc ga tr y1 , y2 ,,yn. Bng phng php bnh phng ti thiu, c th xc nh gi tr cc h s a v b trong quan h tuyn tnh: y = a + bx theo cng thc sau :

x . y x . x . ya=i =1 2 i i =1 n i i =1 i i =1 i

n

n

n

n

i

n n. x xi i =1 i =1 2 i n n n i =1 i =1 i =1 2

2

n. xi . y i xi . y i

b=

n n n. xi2 xi i =1 i =1 * Nu gi x - hm lng cht nh phn trong mu phn tch. Y* - gi tr y o c ng vi mu phn tch trong cng iu kin khi xy dng ng chun. Ta c th tnh x* t phng trnh ng chun xy dng. II. Kt qu v tnh ton. II.1 Xy dng ng chun:

VSDS ( mL) 2 4 0.2 0.4 CSDS (: g/mL) ABS 0.032 0.042 Vy phng trnh ng chun c dng : Y = 0.049x + 0.022 ng chun dng nh sau:0.07 0.06 0.05 ABS 0.04 0.03 0.02 0.01 0 0 0.2 0.4 C ppm 0.6 0.8 y = 0.049x + 0.022 R = 0.99422

6 0.6 0.050

8 0.8 0.062

1

- 12 -


II.2.Xc nh hm lng cht ty ra trong mu nc : V (mL) ABS 1.0 0.018

T phng trnh ng chun ta xc nh c hm lng cht ty ra trong m nc phn tch (theo n v mg/mL ). T phng trnh ng chun : y = 0.049x + 0.022 ,ta suy ra: Ta c nng ca cht ty ra trong mu nc gi nh: CN = (A-0.022)/0.049 = -0,082 Vy hm lng cht ty ra trong mu nc phn tch l : -0.082 g/mL. Nhn xt :Hm lng cht ty ra trong nc nh hn chun cho php.

Bi 4: Xc nh St Fe trong nc bng phng php trc quang vi thuc th Thyocyanat. I. Tin hnh th nghim v gii thch cc giai on trong phn tch trc quang st bng thuc th Thiocyanat. Chun b 10 bnh nh mc dung tch 100mL v cc dung tch 100mL. Cho ln lt vo mi cc t th hai n cc th mi dung dch chun st vi cc th tch ln lt l 1,0; 1,5; 2,0; 2,5; 3,0; 3,5; 4,0; 4,5 v 5,0 mL. Thm nc ct vo cc cc sao cho t th tch diu khong 50mL. Ring cc th nht khng cho dung dich chun st, ch cho khong 50mL nc ct lm mu trng (mu rng, dung dich so snh). Thm vo tng cc 1,5mL dung dch H2SO4; 1,5mL dung dch KMnO4; t cc ln bp in c li amian ngn cch, un si dung dch trong cc t 3-5 pht oxy ha Fe(II) ln Fe(III), ngoi ra nu dung dch c cha anion nitrit NO2- ( gy sai s ch yu n t cht cn khi xc nh st trong nc_v anion ny to hp cht mu NOSN vi thuc th, cng mu vi phc ([Fe(SCN)]2+[Fe(SCN)]3- ) th cng xy ra qu trnh oxy ha anion nitrit NO2 thnh NO3 phng trnh phn ng: 5Fe2+ + MnO4 + 8H+ = 5Fe3+ + Mn2+ + 4H2O 2MnO4- + 5NO2- + 6H+ = 2Mn2+ + 5NO3- + 3H2O Sau khi dng un, va thm tng git dung dch H2C2O4 va lc iu n khi mt mu TmLoi tr KMnO4 d bc trn: 2MnO4+ 5C2O42- + 16H+ = 2Mn2+ + 10CO2 + 8H2O Ngoi ra mt lng Fe(III) cng s b kh v Fe(II) bi dung dch H2C2O4: 2Fe3+ + C2O42- = 2Fe2+ + 2CO2 Li thm cn thn tng git dung dch KMnO4 ng thi lc iu cc n khi dung dch va c mu hng nhtiu ny gip chng chc chn Fe(II) chuyn hon ton thnh Fe(III)__(im tng ng ca qu trnh oxy ha Fe(II) ln Fe(III) ): 5Fe2+ + MnO4 + 8H+ = 5Fe3+ + Mn2+ + 4H2O ngui , chuyn dung dch vo bnh nh mc dung tch 100mL, trng cc bng mt t nc ct ri chuyn c vo bnh nh mc( Nu dung dch b c th phi lc, khi cn gom c nc lc v nc ra phiu lc vo bnh nh mc).

- 13 -


Thm 1,5mL dung dch HCL 1:1, lc iu( to mi trng H+). Tip tc cho vo bnh nh mc 2,5mL dung dch thiocyanat, nh mc bng nc ct n vch, y kn np v lc iu. o ngay mt quang ca dung dch bc sng 495nm, vi mu dung dch so snh l mu rng. V ta c c kt qu di y: Cc V(ml) CFe (mg/l) A 1 50 0.05 0.021 2 1.0 0.10 0.064 3 1.5 0.15 0.106 4 2.0 0.20 0.174 5 2.5 0.25 0.205 6 3.0 0.30 0.207 7 3.5 0.35 0.236 8 4.0 0.40 0.303 9 4.5 0.45 0.341 10 5.0 0.50 0.373

0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 0.00

A

y = 0.7433x + 0.0002 R2 = 0.9757

0.10

0.20

0.30 C (mg/l)

0.40

0.50

0.60

II. o mu nc. T phng trnh ng chun: y = 0,7433x + 0.0002 Ta c nng ca Fe trong mu nc thc t: X=(y-0.0002)/0.7433=(0.021- 0.0002)/0.7433= 0.028(mg/L). Bi 5: Xc nh nng carbon dioxid (CO2) trong khng khI.

Mc ch. Xc nh nng carbon dioxid trong khng kh, bit c cch lp t, s dng cc thit b ly mu, I. C s l thuyt. a. Phng php xc nh:- 14 -


Mu cha kh cn nh phn i qua dung dch hp thu c cha lng d mt dung dch base, Carbon dixid phn ng vi base to thnh bicarbonat hoc carbonat theo phn ng: HCO3-(aq) CO2(g) + OH-(aq) CO32-(aq) CO2(g) + 2OH-(aq) + H2O(L) Nu NaOH hoc KOH c s dng lm cht hp thu l Ba(OH)2, Ba2+(aq) + CO2(g) + 2OH-(aq) BaCO3(s) + H2O(L) Lng Ba(OH)2 d c chun bng H2C2O4, BaC2O4 + 2H2O Ba(OH)2 + H2C2O4 b. Carbo dioxid (CO2): Trong kh quyn ngoi kh Oxy v Nit c vai tr quan trng hn c cn c cc kh khc c trnh by trong bng sau: Tn Nit Oxy Argon CO2 Ne % v th tch 78,08 20,95 0,93 0,033 0,0018 Tn He CH4 Kr N2O H2 % v th tch 0,00052 0,00020 0,00011 0,00005 0,00005

CO2 l mt cht kh khng mu, khng mi, d khng c nhng CO2 khng h tr qu trnh h hp ca con ngi v ng vt, nng cao c xem l nguyn nhn dn n s ngp th v cht,CO2 mc nng cao l nguyn nhn gy hiu ng nh knh v kh nng m ln ton cu. II. Kt qu th nghim v gii thch. 1. Kt qu th nghim: - Th tch khng kh sc qua cc bnh hp th l: S o trc: A0 = 118590 S o sau : A1 = 190945 Vkk = (190945 118590).0,529 = 38,28 L - Th tch acid oxalic tiu tn khi chun mu trng: V = 4 ml - Th tch acid oxalic tiu tn khi chun mu: bnh VH2C2O4 1 4.1 2 3.9 3 3.8 Vtb 3.9 - Nhit trung bnh khi ly mu: t = 23.5 0C T=296,5 0 F - Xt phng trnh phn ng: Ba2+(aq) + CO2(g) + 2OH-(aq) BaCO3(s) + H2O(L) Ta thy 1 ng lng gam ca Ba2+ phn ng ht vi 1 ng lng gam ca CO2, hay s ng lng gam Ba2+ phn ng bng vi s ng lng gam CO2 b hp th. Mt khc: CN.Ba = 0,015 N Lng Ba2+ phn ng trong 10 mL l:- 15 -

Bo co thc tp phn tch mi trng C (V0 Vm ). N 2 = nCO2 nBa2+ = 10 0, 015 (4 3,9). -5 nCO2 = 2 = 7.5.10 mol 10

Lng CO2 c trong 600 mL mu l:-5 mCO2 = 7,5.10 .600.44 = 0.198 g = 198 mg 10

Th tch khng kh thu c quy v iu kin tiu chun l:P0 .V0 P.V P.V .T0 0,89.38, 28 .273 = V0 = = =31,37 L T0 T T .P 296,5.1 0

Nng CO2 c trong kh quyn l:CCO2 = 198 = 6,31mg/L 31,37

Kt qu s chu nhng nh hng dn n sai s: - Dng lu ca khng kh i vo my, s chu tc ng bi gi, lu lng xe qua li. - Nhit ca khng kh ni ly mu khng n nh. - V kt qu nh lng CO2 c th sai s t thao tc pha ha cht, chun . Nhng cht nhim khc c th gy cn tr n vic xc nh CO2: - Bi, tuy c thit b lc bi tuy nhin dng lu ca bi khi i vo my nh hng n kh nng lu ng ca CO2 vo bnh thu mu. nh hng ca CO2 trong mi trng ch yu v nng bng hin ny l n ng gp vai tr chnh trong hin tng bin i kh hu v m ln ton cu ca tri t

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