Luan Van Thac Sy Hoa Hoc

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I HC THI NGUYN

TRNG I HC S PHM ------------------------------

NGUYN THANH TUN

XY DNG H THNG CU HI V BI TP

MT S CHNG TRONG PHN TCH L HO

Chuyn ngnh: Ho phn tch

M s: 60 44 29

LUN VN THC S KHOA HC HO HC

NG I HNG DN KHOA HC:TS. NG XUN TH

Thi N guyn, nm 2008


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S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn

MC L C

M u 1

I. L do chn ti 1

II. Ni dung chnh ca ti 2

III. Nhim v ca ti 2Ch- ng I Tng quan 3

I.1 ngha ca h thng bi tp 3I.1.1 Tng hp v n luyn kin thc 3I.1.2 Phn loi bi tp v cu hi ho hc 5I.1.3 Tc dng ca bi tp ho hc 6I.1.4 Vn dng kin thc gii bi tp 7I.2 Dy hc ch trng phng php t hc 7I.2.1 Dy hc ch trng rn luyn ph ng php t hc 8I.2.2 Hc thng qua t chc cc hot ng ca sinh vin 8

I.2.3 Tng c ng hc tp c th, phi hp vi hc tp hp tc 9I.3 Xu h- ng pht trin ca bi tp Ho hc hin nay 9I.4 C s phn loi cu hi v bi tp cn c vo mc

nhn thc v t- duy

10

Ch- ng II Ph- ng php phn tch quang ph hp th nguyn

t

13

II.1 C s l thuyt 13II.1.1 c im chung ca ph ng php o quang ph hp

th nguyn t 13

II.1.2 iu kin to thnh ph hp th nguyn t 13II.1.2.1 Qu trnh nguyn t ho mu 13


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II.1.2.2 Cc ph ng php nguyn t ho 14II.1.2.3 S hp th bc x cng h ng 15II.1.2.4 Ph ng php phn tch ph hp th nguyn t 15

II.1.3 Cch loi tr sai s do cc nguyn t i km v sai s phng 17

II.2 Cu hi t lun 18

II.3 Bi tp ch ng II 37Ch ng III Ph- ng php phn tch quang ph pht x nguyn t 60III.1 C s l thuyt 60

III.1.1 c im chung c a phng phpquang ph pht xnguyn t 60

III.1.2 S to thnh ph AES 60III.1.3 Bn cht ca ph ng php ph ph t x nguyn t 61III.1.4 S kch thch, s pht x v c ng vch ph t x

nguyn t 62

III.2 Cu hi t lun 63

III.3 Bi tp ch ng III 79Ch ng IV Cc phng php tch, chit v phn chia 97IV.1 C s l thuyt ca phng php chit 97IV.1.1 nh ngha v h s phn b 97

IV.1.2 Hng s chit 97

IV.1.3 Cc yu t nh h ng n qu trnh chit ho hc 98

IV.1.3.1 nh h ng ca H+ trong pha n c

98IV.1.3.2 nh h ng ca hiu ng mui 99

IV.1.3.3 nh h ng ca tc nhn chit 100

IV.1.3.4 iu kin chit 101


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IV.2 C s l thuyt ca phng php sc k 101IV.2.1 Thi gian l u 102

IV.2.2. Th tch l u 104

IV.2.3 Sc k kh (GC) 105IV.2.3.1 Sc k kh - rn (GSC) 105IV.2.3.2 Sc k kh - lng (GLC) 105

IV.3 C s l thuyt ca phng php tch 106IV.3.1 Tch cht bng ph ng php ch ng ct 106

IV.3.1.1 Cn bng lng hi ca h hai hay nhiu cu t 106

IV.3.1.2 Xc nh s a l thuyt v t s hi l u bng ph ng

php MC Cabe Thielo

106

IV.3.1.3 Xc nh s a l thuyt cc tiu v t s hi l u cc

tiu theo ph ng php MC Cabe Thielo

107

IV.3.1.4 Xc nh ng knh ca ct ch ng ct v chiu cao

ca ct ch ng ct cho yu cu tch cho

107

IV.4 Cu hi t lun 108

IV.5 Bi tp ch ng IV 131K T LU N 151

TI LIU THAM KH O


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S ha bi Trung tm Hc liu i hc Thi Nguyn http://www.lrc-tnu.edu.vn 1

PH N M UI L do chn ti:

Trong cng cuc cng nghip ho, hin i ho t nc, nn kinh tnc ta ang chuyn i t c ch k hoch ha tp trung sang c ch thtrng c s qun l ca Nh nc. Cng cuc i mi ny ra nhng yucu mi i vi h thng gio dc. Cng vi s pht trin ca khoa hc kthut, thng tin lin lc trn th gii rt pht trin kotheo s thay i vcng to l n vyu t con ng i trong x hi. Trong x hi m i, tri th c lyu tquyt nh, con ng i l yu t trung tm, l chthca ton x hi,

do gio dc con ng i ng vai tr quan trngtrong s pht trin ca tnc [26].

p ng yu cu con ng i - ngu n nhn lc, yu tquyt nh s pht trin ca t n c trong th i k cng nghip ha,hin i ha cn tora nhng chuy n bin c bn, ton din v gio dc v o to. Trong nh ngnm gn y, B gio dc v o to khuyn khch vic sdng a dng

cc phng php dy hc tch cc nhm hot ng ha ng i hc. Mun c nh th, ngun bi tp, cu hi cho n i dung ki n thc phi phong ph,a dng.

Tuy v y, v i nhng mn hc c mc t duy cao v mt khnngvn dng kin thc tng hp th vic chun b dng cu TNKQ l d ng nh cha y , cha c s sng to, nhy bn v spht trin tduy khoa h c

cao. Do v y, trong tr ng hp ny cn duy tr v pht trin hthng cu hiv bi t p t lun x l thng tin v l nh hi tri th c mn hc.[17]

V nhng l do trn y, chng ti mnh dn la chn ti Xyd ng h th ng cu hi v bi t p m t s chng trong phn tch l ha v sdng chng theo h ng dy v hc tch cc pht trin nng lc t duy, c lp, sng to ca ng i hc.


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II Ni dung chnh ca ti:H thng cu hi v bi tp t lun mn phn tch l ho ca 3

chng:

- Chng I: Phng php phn tch quang ph hp th nguyn t. - Chng II: Phng php phn tch quang ph pht x nguyn t. - Chng III: Cc phng php tch, chit v phn chia.

III Nhi m v ca ti:- Nghin cu c s l lun ca ti. - Nghin cu a ra hthng cu hi v bi t p trong n i dung ti.

- Nghin cu h ng dn cch gii.Phn loi thnh cc nhm bi t p theo ch , t hthng ha kin thcv bao qut c ni dung mn hc ca 3 chng ny.

T nng cao cht l ng ging dy hc phn Phn tch l hatrong cc tr ng HSP, C SP v HKHTN, c sdng hc phn phn tch ha l.


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Ch- ng I

TNG QUAN

I.1 ngh a ca h thng bi tp:UNESCO tng nhn mnh rng: tri vi thng l c truyn vic

ging dy phi thch nghi vi ngi hc, ch khng phi buc ngi hctun theo cc quy nh sn c t trc trong vic dy hc. Ngi hc vcng chng ni chung cn c ting ni nhiu hn trong vic quyt nh ccvn lin quan n gio dc.Tnhng n m 1980 tr li y, ni bt ln

mt h ng m i: vic ging dy phi m bo cho ng i hc tr thnh ng icng dn c trch nhim v hnh ng hiu qu. Nhvy mc ch ca vichc t p pht trin t hc hiu n hc hnh ri n hc thnhng i - mt con ng i t ch, nng ng v sng to. V th vic hc tpgii quyt vn trong hc tp, trong thc tin i hi con ngi phi c ckin thc v phng php t duy.[19]

I.1.1 T ng h p v n luyn ki n th c:Vic dy hc, c bit dy hc i hc khng ththiu bi t p, s

dng bi tp l mt bin php ht sc quan tr ng nng cao cht l ng dyhc. Hthng cu hi v bi tp c nhng ngha, tc dng to l n vnhiumt:

- Lm chnh xc ha cc khi nim ha hc, cng c, o su v m

r ng kin thc mt cch sinh ng, phong ph, h p dn. Khi ng i hc vndng kin thc vo vic gii bi t p, hm i nm kin thc mt cch su sc.

- n t p, hthng ha kin thc mt cch tch cc nht. Khi n t p,ng i hc skhng t p trung n u gi n t p ch yu cu hnhc li cckin thc c hc. Thc t cho th y ng i hc (hc sinh, sinh vin) ch thch trli cc cu hi suy l un v gii bi t p trong gi n t p.

- Rn luyn cc k nng khoa h c ha hc ca mn hc nh cn bng phng trnh phn ng, ph ng trnh ion, tnh ton theo cng thc ha hc,


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p dng cc nh lut, phng php xc nh nh lng cc cht Nu l bi t p thc nghi m s rn cc knng thc hnh, gp phn vo vic giodc k thut tng h p cho ng i hc.

- Rn luyn khnng vn dng kin thc vo thc tin i sng, laong sn xut v bo v mi tr ng.

- Rn luyn k nng sdng ngn ng ha hc v cc thao tc tduy.Pht trin hc sinh cc nng lc t duy logic, bi n chng, khi qut, clp, thng minh, sng to.

- Rn luyn c tnh chnh xc, kin nhn, trung th c v lng say m

khoa h c.C khnng t ng h p v khi qut ha ki n th c:Thng qua vic tr li cc cu hi v gii cc bi t p trong n i dung

hc phn hc t ng ng, ng i sinh vin st khi qut ha kin thc mtcch tt nht d i scvn, ch o ca ng i thy.

hnh thnh cho sinh vin nhng khi qut ng n, tiu biu cn

m bo cc iu kin sau:- Lm bin thin hoc m nht nhng duhiu kh ng bn cht ca vt

hay hi n t ng kho st, ng th i gi khng i du hiu bn cht.- Chn nhng dng bi t p a ra c s bin thin hp l nht

nu bt c du hiu bn cht v tru tng ha du hiu thyu.- C ths dng nhng cch bin thin khc nhau c cng mt

ngh a tm l hc, nhng li hiu nghi m. Qua thhin c smm doca tduy.

- Phi cho ng i hc t mnh pht biu c thnh li nguyn tc binthin v nu c tnh ca nhng du hiu khng bn cht. iu cngchng tr ng sinh vin nhn thc c du hiu bn cht. Ngoi vic bom nhng iu kin trn y, gio vin cn t p luyn cho ng i hc phttrin t duy khi qut bng nhng hnh thc quen thu c nh lp dn , xydng nhng k t lun v tm tt ni dung cc bi, cc ch ng.[26]


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I.1.2 P hn loi bi tp v cu hi ha hc:Da vo ni dung v hnh thc thhin c th phn loi bi tp ha

hc thnh 2 loi:- Bi t p nh tnh. - Bi t p nh l ng.

Cu hi c th phn loi thnh: - Cu hi ti hin kin thc- Cu hi vn dng kin thc- Cu hi suy l, chng minh.

* Bi tp nh tnh: L cc dng bi tp c lin h vi s quan st m t, gii thch cc hin tng ha hc. Cc bi tp nh tnh cng c rtnhiu cc bi tp thc tin gip hc sinh gii quyt cc vn thc tin sinhng.

* Bi tp nh lng (bi ton ha hc): L loi bi tp cn vn dngk nng ton hc kt hp vi k nng ha hc (nh lut, nguyn l, quy tc,

) gii.* Cu hi ti hin kin tc l dng cu hi ngi hc ch cn ti hin

trnh by li nhng ni dung m mnh tip thu c.* Cu hi vn dng kin thc l cu hi m ngi hc cn phi

nghin cu k phn kin thc c s l thuyt, t vn dng linh hot trongkhi nghin cu gii quyt bi ton.

* Cu hi suy l, chng minh l cu hi m ngi hc phi nm vngl thuyt, bit vn dng cc ni dung kin thc c lin quan gii quyt biton.[9] I.1.3 Tc dng c a bi tp ha hc:

* Tc dng tr dc:- Bi tp ho hc c tc dnglm chnh xc, cng nh hiu su h n

cc khi nim v nh lut hc.


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- Gip cho sinh vin nng ng sng to trong h c tp, pht huy nnglc nhn thc v tduy, t ng tr thng minh v l ph ng tin ng i hcv n t i nh cao tri th c.

- L con ng ni lin gia kin thc thc t v l thuyt to ra m tth hon chnh thng nht bin chng trong c qu trnh nghin cu. osu, m r ng shiu bit mt cch sinh ng, phong ph khng lm nng n thm khi l ng kin thc cho ng i hc. Ch c vn dng kin thc vovic gii bi tp, sinh vin m i nm kin thc su sc.

- L ph ng tin n t p, cng c, hthng ha, kim tra nh gi

vic nm bt kin thc mt cch tt nht (ch ng, sng to).- To iu kin pht trin tduy cho ng i hc: khi gi i bi t p bt

buc ngihc phi suy lun, quy n p, din d ch hoc cc thao tc tduyu c vn dng.

Trong th c thc tp, c nhng vn buc ng i hc phi o susuy ngh m i hiu c tr n vn. Thng th ng khi gi i mt bi ton nn

yu cu hoc khuy n khch ngihc gii bng nhi u cch- tm ra cch giingn nht, hay nh t.

* Tc dng gio dc:- Bi tp ha hc c tc dng gio dc t t ng cho h c sinh, sinh

vin v thng qua gii bi tp rn luyn cho HS, SV tnh kin nhn, trungthc trong h c tp, tnh sng to khi x l v vn dng trong cc vn hc

t p. M t khc, qua vic gii bi tp rn luyn cho cc em tnh chnh xc khoahc v nng cao hng th hc b mn.

- Cc bi tp ha hc cn c s dng nhmt ph ng tin nghincu ti liu mi, ngoi ra cc bi c ni dung th c nghi m c tc dng rnluyn tnh tch cc, t lc l nh hi tri th c v tnh cn thn, tun thtrit quy nh khoa h c, chng tc phong lum thu m, vi ph m nhng nguyn tckhoa h c.[25]

I.1.4 V n d ng ki n th c gii bi tp:


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gii bi t p ng i hc phi bit vn dng l thuyt hc nidung cc chng cc bi, qu trnh ny thc cht i hi ng i hc phi cmt k nng nhn thc v tduy nh t nh. Ho t ng nhn thc v pht

trin tduy ca sinh vin trong qu trnh dy hc ha hc. Nhn thc l mt trong ba m t c bn ca i sng tm l con ng i

(nhn thc, tnh cm, l tr). N l tin ca hai m t kia v ng thi cmi lin hcht chvi chng v cc hin tng tm l khc.

T duy l mt qu trnh tm l phn nh nhng thu c tnh bn cht,nhng mi lin h bn trong c tnh quy lut ca s vt hin t ng trong

hin thc khch quan m tr c ta cha bit.Theo M.N.Sac acop: T duy l snhn thc khi qut gin tip cc svtv hin t ng trong nh ng du hiu, nhng thu c tnh chung v bn cht cachng. Tduy c ng l snhn thc sng to nhng s vt hin t ng m i,ring lca hin thc trn c s nhng kin thc khi qut ha thu nhnc.[9]

I.2 D y hc ch trng ph ng php t hc:Tnh tch cc l mt phm cht vn c ca con ng i trong i sng

x hi, khc v i ng vt con ng i khng ch tiu thnhng g c sn trongthin nhin m ch ng to ra c s vt cht cn thit cho s tn ti v phttrin ca x hi.

I.2.1 D y hc ch trng rn luyn ph ng php t hc:Phng php tch cc xem vi c rn luyn phng php hc t p cho

hc sinh khng ch l mt bin php nng cao hiu qudy hc m cn lmt mc tiu dy hc. Trong m t x hi hin i ang bi n i nhanh cngv i s bng n thng tin, khoa hc, k thut, cng ngh pht trin v i tc cc k nhanh chng, chng ta thc s cn phi quan tm dy cho sinh vin

phng php hc t p, ph ng php nhn thc.


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Nu rn luyn cho ng i hc c c phng php, knng, thiquen, ch thc th sto cho h lng ham hc, kh i dy ni lc vn ctrong m i ngi th kt quhc t p s c nng ln g p bi. V vy, v i

phng php dy hc ngy nay rt nhn mnh hot ng t hc, n lc tora schuyn bin thc t p th ng sang t hc ch ng c sh ng dnca ng i thy. Bn cnh vic hc tp l sut i, ngi hc khng chhc trong nh trng, m cn hc mi lc, mi ni, sau khi ri gh nhtrng cn phi tip tc hc tp b sung cp nht kin thc.[10] I.2.2 H c thng qua tchc cc hot ng h c tp ca h c sinh:

Trong ph ng php dy hc tch cc, ng i hc l i t ng ca hotng dy, nh ng ng th i cng l chthca hot ng hc c cun vocc hot ng hc tp do gio vin tchc v ch o, do m sinh vin t khm ph nhng iu mnh cha r ch khng phi ti p thu nh ng tri th c c gio vin s p t sn. Khi t vo nhng tnh hung thc t, ng ihc tr c tip quan st, tho lun, gii quyt vn t ra theo nh ng suy

ngh ring ca mnh, t sinh vin va nm kin thc, va nm ph ng php lm ra kin thc m khng theo nhng khun mu sn c. Vi cchdy hc ny, Thy khng ch n gin l truyn t tri th c m sdy hc

phi gip cho tng sinh vin bit hnh ng v tch cc tham gia cc ch ngtrnh hnh ng cng t p th.[10]I.2.3 T ng c ng h c tp c th, ph i h p v i hc tp h p tc:

Khi trong m t l p hc m trnh kin thc, t duy ca sinh vinkhng ng u mt cch tuyt i th khi p dng phng php dy hctch cc buc phi ch p nhn s phn ha vc ng , tin hon thnhnhim vhc t p, nht l cc bi hc c thit k thnh mt chui cc hotng c lp. Khi p dng phng php tch cc trnh cao th s phnha cng l n. Vi c sdng cc cng ngh thng tin trong nh tr ng s png nhu c u c th ha hot ng hc t p theo nhu c u v khnng ca mihc sinh, sinh vin.


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Tuy nhin, trong h c tp khng phi mi tri th c, k nng, thi uc hnh thnh bng nhng hot ng c l p ca mi c nhn. L p hc lmt mi tr ng giao ti p thy - tr, to nn mi quan h hp tc gia cc c

nhn trn con ng chi m l nh nhng ni dung h c tp. Thng qua tholun, tranh lu n trong t p th, kin c nhn c bc l, khng nh hay bc b, qua ng i hc c nng mnh ln mt trnh m i. Ng i hcsvn dng vn hiu bit, kinh ngi m sng ca mi c nhn v ca c t pthl p ch khng phi ch da trn vn hiu bit v kinh nghim sng cagio vin.

Hnh thc hc tp theo nhm khng lm hn chsc mnh, hng thhc t p ca mi c nhn m tri li to mnh m tnh tch cc, ch ngca mi c nhn, gip hchuyn t i tng gio dc thnh chth giodc.[25]

I.3. Xu h ng pht trin ca bi tp ha hc hi n nay:Bi tp ha hc va l mc tiu, va l mc ch, va l ni dung v a

l phng php dy hc hu hiu do v y cn c quan tm, ch trng trongcc bi hc. N cung c p cho h c sinh khng nhng kin thc, nim say m

b mn m cn gip ng i hc con ng ginh ly kin thc, l b c mcho qu trnh nghin cu khoa h c, hnh thnh v pht trin c hiu qu trong ho t ng nhn thc ca hc sinh.

Bng h thng bi t p s thc y s hiu bit ca sinh vin, s vn

dng nhng hiu bit vo thc tin, s l yu t c bn ca qu trnh phttrin x hi, tng tr ng kinh t nhanh v bnvng.

Xu hng pht trin ca bi tp ho hc hin nay hng n rn luynkh nng vn dng kin thc, pht trin t duy ho hc. Nhng bi tpc tnh cht hc thuc trong cc cu hi l thuyt s gim dn m c thay bng cc cu hi i hi s t duy, tm ti.

Dy hc ch trng rn luyn phng php t hc trng i hcc xem l rt quan trng v c nhiu trng coi trng p dng. Ngoi


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ra, trong thi gian gn y, mt s chin lc i mi phng php dy hcc th nghim l dy hc hng vo ngi hc, hot ng hangi hc, tip cn kin to trong dy hc [6]

I.4. C s phn loi cu hi v bi tp cn c vo mc nhn thc vt duy:

Vn dng cc quan im v vic phn loi mc nhn thc v tduy ca GS.Bloom vGS. Nguyn Ngc Quang, cn c vo thc tin dyhc Vit Nam, chng ti thy vic kt hp vn dng hai quan im trn

cho ph hp l cn thit. Vic phn loi sp xp cc cu hi v bi tp hc phn phn tch l ha

cn c vo cc mc nhn thc v t duy ca qu trnh lnh hi kin thck nng k xo chng ti thy c th sp xp thnh 4 dng sau:


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Dng bi

Nng lc nhn thc Nng lc t duy K nng

I

Bit (nh li nhngkin thc hc mtcch my mc vnhc li)

T duy c th Bt chc theo mu

II

Hiu (ti hin kinthc, din gii kin

thc, m t kin thc)

T duy logic (suylun, phn tch,

so snh, nhnxt)

Pht huy sng kin(hon thnh k nng

theo ch dn, khngcn bt chc mymc)

III

Vn dng T duy h thng(suy lun tngt, tng hp, so

snh, khi qutha)

i mi (lp li knng no mt cchchnh xc, nhp

nhng)

IV

Vn dng sng to(phn tch, tng hp,nh gi)

T duy trutng (suy lunmt cch sngto)

Sng to (hon thnhk nng mt cch ddng c sng to, tti trnh cao)

Vic s dng cu hi v bi tp trong dy hc c bit l dy hc ihc c tm quan trng c bit. i vi sinh vin y l phng php hctp tch cc, hiu qu v khng c g thayth c, gip cho sinh vin nmvng kin thc mn hc, pht trin t duy, hnh thnh khi nim, kh nngng dng ha hc vo thc tin, lm gim nh s nng n cng thng cakhi lng kin thc v gy hng th cho sinh vin trong hc tp.


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Tuy nhin hiu qu ca vic s dng h thng cu hi v bi tp hahc cn ph thuc vo nhiu yu t nh: tnh t gic, tnh va sc v hngth hc tp ca sinh vin. Cng nh vn hc tp, nu nh cu hi v bi

tp d qu hoc kh qu u khng c sc li cun hc sinh, sinh vin. Vvy trong qu trnh dy hc, tt c cc kiu bi ln lp khc nhau, ngigio vin phi bit s dng cu hi v bi tp c s phn ha ph hp vitng i tng tc l gp phn rn luyn v pht trin t duy cho ngi hc.

Ty theo mc ch dy hc, tnh phc tp v quy m ca tng loi bi,gio vin c th s dng h thng cu hi v bi tp theo 4 bc ca qu trnh

nhn thc v t duy nh trn.[9]

CHNG IIPHNG PHP PHN TCH QUANG PH HP TH NGUYN T

II.1. C S L THUYT:


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I.1.1. c dim chung ca phng php phn tch quang ph hp thnguyn t (AAS):[7] - Phng php AAS (Atomic Absorption Spectroscopy) da trn kh nng

hp th chn lc cc bc x cng hng ca nguyn t trng thi t do.Cc nguyn t t do c to ra do tc dng ca ngun nhit bin cc chtt trng thi tp hp bt k thnh trng thi nguyn t, l qu trnh nguynt ho.- Cc nguyn t t do s hp th bc x in t tun theo nh lut hp th

bc x (nh lut Bouguer Lambert Beer): A = lg I I 0 = C l ..

Trong : A: mt quang. I0, I: cng nh sng trc v sau khi b nguyn t t do hp

th. : h s hp th mol phn t, ph thuc bc sng .

l: dy lp hi nguyn t.- chnh xc ca php o ph hp th nguyn t rt cao. - Phng php phn tch n gin, nhanh, i khi khng cn sinh ch mu,p dng rng ri trong nhiu lnh vc khoa hc. II.1.2. iu kin to thnh ph hp th nguyn t: [11] II.1.2.1. Qu trnh nguyn t ho: - Bin mt cht trng thi tp hp bt k thnh trng thi nguyn t t dogi l qu trnh nguyn t ho.- Gi thit kim loi nghin cu Me nm trong dung dch dng mui MeX.Dung dch MeX c phun vo ngn la n kh dng aeroson. Trongngn la n kh s xy ra qu trnh nhit phn: MeXMe + X- Trong ngn la c th xy ra qu trnh ion ho nguyn t to thnh lmgim nhy ca php phn tch. hn ch s ion ho ny, phi a vodung dch phn tch cc cht d b ion ho tng nn electron trong bu

kh.


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II.1.2.2. Cc phng php nguyn t ho: * Nguyn t ho mu bng ngn la: - Ngi ta dng nng lng nhit ca ngn la n kh ha hi v nguyn

t ho mu phn tch. V th mi qu trnh xy ra trong khi nguyn t homu ph thuc vo cc c trng v tnh cht ca ngn la n kh, nhngch yu l nhit ca ngn la. l yu t quyt nh hiu sut nguynt ho mu phn tch v mi yu t nh hng n nhit ca ngn la nkh u nh hng n kt qu ca phng php phn tch.- Trong php o ph hp th nguyn t, ngn la l mi trng hp th. N

c nhim v ho hi nguyn t ho mu phn tch, to ra m hi ca ccnguyn t t do c kh nng hp th bc x n sc to ra ph hp thnguyn t, v th ngn la n kh c dng vo mc ch ho hi vnguyn t ho mu phn tch th:

+ Ngn la n kh phi lm nng u c mu phn tch. + Ngn la n kh phi ln v n nh theo thi gian.

+ Ngn la n kh phi thun khit. + Ngn la n kh phi c b dy ln.

* Nguyn t ho mu khng dng ngn la: - K thut nguyn t ho mu khng dng ngn la hin nay ang c ngdng rt rng ri v c nhy rt cao. Do vy, khi phn tch lng vt cckim loi trong nhiu trng hp khng cn thit phi lm giu s b cc

nguyn t cn xc nh. c bit l khi xc nh cc nguyn t vi lngtrong cc loi mu ca y hc, sinh hc . - Tuy nhin, n nh ca php o khng dng ngn la km hn php odng ngn la. nh hng ca ph nn thng rt ln. Khoa hc pht trin,th vic khc phc nhng nhc im ny l khng kh khn. - Php o khng ngn la i hi mt lng mu tng i nh (mi ln okhong 20- 50 L).


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- Nguyn t ho mu khng dng ngn la l qu trnh nguyn t ho tckhc trong thi gian rt ngn nh nng lng ca dng in cng sut ln vtrong m i trng kh tr.

II.1.2.3. S hp th bc x cng hng: - Khi hng vo lp hi nguyn t t do Me chm bc x in t (chnh lcc tia pht x pht ra t n catot rng lm bng chnh kim loi cn xcnh) c tn s ng bng tn s cng hng ca nguyn t kim loi Me, sxy ra hin tng hp th cng hng chuyn ln cc mc nng lngkch thch gn nhau: Me + h Me *

Qu trnh ny tun theo nh lut Bouguer Lambert Beer.- Trong ph hp th nguyn t, ch mt s nguyn t c kh nng hp th bc x cng hng thc t bng s nguyn t chung ca nguyn t cn xcnh v t b nh hng ca nhit ngun nhit nn phng php ny c nhy cao. II.1.2.4. Phng php phn tch quang ph hp th nguyn t:

Hnh 2.1: S khi ca ph k hp th nguyn t dng ngn la

1. Ngun bc x (n catot) 4. Detector quang2. n 5. Cu trc ghi ph 3. My lc nh sng n sc

- Cu to ca n catot rng:


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Hnh 2.2: n catot rng 1. Bng thu tinh 2. Anot3. Catot 4. Ca s thch anh

n catot rng l mt bng thu tinh hnh tr, ng knh 3- 5cm, cca s bng thu tinh hay thch anh. Anot c ch to t thanh kim loi.C hai cc c t trong bng thu tinh c cha kh tr (agon hay neon)vi p sut khng ln (0,2 2MPa). n catot rng c ni vi dng in300 500V, n nh v phi c n nh cao. Dng phng ca n thng

l vi miliampe. Khi n lm vic, mt dng mt bn trong catot caohn mt ngoi. V vy ti l m ca catot s pht sng. Catot ca n cch to t kim loi, hp kim kh nng chy c cha nguyn t cn xc nh. - Vic gim cng ca bc x cng hng do hin tng hp th ca cnguyn t t do ca nguyn t nghin cu tun theo nh lut hp th bc xnguyn t Bouguer - Lambert - Beer:

A = lg I I 0

= C l ..

Mt quang t l thun vi nng cht nghin cu trong mu. Ta xc nh nh lng theo 2 phng php:

+ Phng php ng chun: Ngi ta o mt quang ca vi dungdch chun ri xy dng th A = f(CTC). Sau o mt quang ca dungdch nghin cu vi cng iu kin o dung dch chun ri da vo th

xc nh CX ca dung dch nghin cu.


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+ Phng php thm: Tin hnh nh trong php o ph pht xnguyn t, ch khc l thay choS ( en), ta xy dng s ph thuc mt quang ti vch ph cng hng ( ): A = f(C CT) v ngoi suy n A =

0.

Nu dng mt dung dch chun, u tin o X A ca dung dch nghin cu,

sau thm vo dung dch nghin cu mt nng ca dung dch tiuchun nguyn t nghin cu, o mt quang ca dung dch thm cAX + Ch .

T cc kt qu ta c:

AX = X C l ..

AX + Ch = )(.. ch X C C l CX = C ch

X ch X

X

A A A

II.1.3. Cc loi tr sai s do cc nguyn t i km v sai s phng:[11] Trong php o ph hp th nguyn t, ngi ta khc phc sai s

phng bng cch dng 2 phng php hiu chnh sai s ca phng nn nh

sau:- Hiu chnh sai s phng dng n teri (hiu chnh nh sng c bc snglin tc). - Phng php t o: y dng in nh sng c cng thp, ta oc tng tn hiu ca hp th nguyn t v tn hiu phng. Cn khi dngdng in nh sng c cng cao xut hin tn hiu hp th ca phng.

Sau my t ng tr i tn hiu tng phn tn hiu phng v nh vy tach o c tn hiu hp th nguyn t ca nguyn t nghin cu.

II.2. CU HI: Cu 1: Hy cho bit nguyn tc ca php o AAS? HDTL:

Phng php phn tch da trn c s o ph hp th nguyn t camt nguyn t c gi l php o ph hp th nguyn t (php o AAS).


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C s l thuyt ca php o ny l s hp th nng lng (bc x n sc)ca nguyn t t do trng thi hi (kh) khi chiu chm tia bc x qua mhi ca nguyn t y trong mi trng hp th. thc hin php o AAS

phi thc hin 3 nguyn tc sau: + Chn cc iu kin v mt loi trang b ph hp chuyn mu phn tcht trng thi ban u (rn hay dung dch) thnh trng thi hi ca cc nguynt t do. l qu trnh ho hi v nguyn t ho mu.+ Chiu chm tia sng bc x c trng ca nguyn t cn phn tch quam hi nguyn t va iu ch c trn. Cc nguyn t ca nguyn t

cn xc nh trong m hi s hp th chn lc nhng bc x nht nhv to ra ph hp th ca n. y, phn cng ca chm tia sng bnguyn t hp th ph thuc vo nng ca n trong mi trng hp th. Ngun cung cp chm tia sng pht x ca nguyn t cn nghin cu cgi l ngun pht bc x n sc hay bc x cng hng. + Tip , nh mt h thng my quang ph, ngi ta thu ton b chm

sng, phn li v chn mt vch ph hp th ca nguyn t cn nghin cu o cng ca n. Cng chnh l tn hiu hp th ca vch ph hpth nguyn t. Trong mt gii hn nht nh ca nng C, gi tr cng ny l ph thuc tuyn tnh vo nng C ca nguyn t trong mu phntch theo phng trnh:

A = a.C b a = K.K a : hng s thc

nghim b: hng s bn cht, 0 < b1

Thng thng, ngi ta thng s dng trong khong nhit b = 1(khng ph thuc tuyn tnh) ng dng phn tch.

Cu 2: thc hin c php o ph hp th nguyn t, h thng my o ph hp th nguyn t phi bao gm cc phn c bn no? HDTL:


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Mun thc hin c php o ph hp th nguyn t, h thng myo ph hp th nguyn t phi bao gm 4 phn c bn sau: + Phn 1: Ngun pht tia pht x cng hng (vch ph pht x c trng

ca nguyn t cn phn tch), chiu vo mi trng hp th cha ccnguyn t t do ca nguyn t. chnh l cc n catot rng (HCL), ccn phng in khng cc (EDL), hay ngun pht bc x lin tc c bin iu.+ Phn 2: H thng nguyn t ho mu phn tch. H thng ny c ch totheo hai loi k thut nguyn t ho mu. l k thut nguyn t ho bng

ngn la n kh (lc ny ta c php o F- AAS) v k thut nguyn t hokhng ngn la (lc ny c php o ETA- AAS).+ Phn 3: L my quang ph, n l b n sc, c nhim v thu, phn li vchn tia sng (vch ph) cn o hng vo nhn quang in pht tn hiuhp th AAS ca vch ph. + Phn 4: L h thng ch tn hiu hp th ca vch ph (tc l cng c

vch ph hp th hay nng nguyn t phn tch). H thng ny c th lcc trang b:

- n gin nht l mt in k ch nng lng hp th (E) ca vch ph.

- mt my t ghi pic ca vch ph. - hoc b hin s digiltal.

-

hay b my in (printer).- hoc my tch phn.

Cu 3: Hy nu nhng u, nhc im ca php o AAS? HDTL:

* u im: - Php o ph hp th nguyn t c nhy v chn

lc cao. Gn 60 nguyn t ho hc c th c xc nh bng phng phpny vi nhy t 1.10-4 n 1.10-5 %. c bit, nu s dng k thut


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nguyn t ho khng ngn la th c th t n nhy n.10-7%. Chnh v nhy cao, nn phng php phn tch ny c s dng rng ri trongnhiu lnh vc xc nh lng vt cc kim loi. c bit l trong phn

tch cc nguyn t vi lng trong cc i tng mu y hc, sinh hc, nngnghip, kim tra cc ho cht c tinh khit cao.

- ng thi cng do c nhy cao nn trong nhiutrng hp khng phi lm giu nguyn t cn xc nh trc khi phn tch.Do tn t nguyn liu mu, tn t thi gian khng cn phi dng nhiu hacht tinh khit cao khi lm giu mu. Mt khc cng trnh c s nhim

bn mu khi x l qua cc giai on phc tp. cng l mt u im lnca php o ph hp th nguyn t.

- Cc ng tc thc hin nh nhng. Cc kt qu phntch li c th ghi li trn bng giy hay gin lu gi li sau ny. ngthi, vi cc trang thit b hin nay, ngi ta c th xc nh ng thi haylin tip nhiu nguyn t trong

mt mu. Cc kt qu phn tch li rt n nh, sai s nh. * Nhc im:

- Php o AAS ch cho ta bit thnh phn nguyn t cacht trong mu m khng ch ra trng thi lin kt ca nguyn t trongmu. V th n ch l phng php phn tch thnh phn ho hc ca nguynt.

- Mun thc hin php o AAS cn phi c mt h thngmy AAS tng i t tin. Do nhiu c s khng iu kin xydng phng th nghim v mua sm my mc.

- Php o c nhy cao nn s nhim bn rt c ngha i vi kt qu phn tch hm lng vt. V th mi trng khng kh phng th nghim phi khng c bi, cc dng c ho cht dng trong phpo phi c tinh khit cao.


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Cu 4: Nu nhng yu cu v nhim v ca ngn la trong k thut nguynt ho mu? HDTL:

Trong php o ph hp th nguyn t, ngn la l mi trng hp th. N c nhim v ho hi v nguyn t ho mu phn tch to ra m hi cacc nguyn t t do c kh nng hp th bc x n sc to ra ph hpth nguyn t, v th ngn la n kh mun dng vo mc ch ho hiv nguyn t ho mu phn tch n cn phi tho mn mt s yu cu sau:

+ Ngn la n kh phi lm nng u c mu

phn tch. Ho hi v nguyn t ho mu phn tch vi hiu sut cao, m bo cho php phn tch t chnh xc v nhy cao.

+ Nng lng (nhit ) ca ngn la phi lnv c th iu chnh c tu theo tng mc ch phn tch mi nguyn t.ng thi li phi n nh theo thi gian v lp li c trong cc ln phntch khc nhau m bo cho php phn tch t kt qu ng n.

+ Ngn la phi thun khit. + Ngn la phi c b dy ln c c lp

hp th dy lm tng nhy ca php o.


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Cu 5: Hy trnh by cu to ca ngn la n kh? HDTL:

Hnh 2.3 : Cu to ca ngn la n kh.

Ngn la n kh gm 3 phn chnh l: + Phn ti ca ngn la. Trong phn ny hn hp kh

c trn u v t nng cng vi cc ht sol kh (th aerosol) ca mu phn tch. Phn ny c nhit thp (700- 1200 oC). Dung mi ho tan mus bay hi trong phn ny v mu c sy nng.

+ Vng trung tm ca ngn la. Phn ny c nhit cao v thng khng c mu hoc c mu xanh rt nht. Trong phn ny hnhp kh c t chy tt nht v khng c phn ng th cp. V th trong

php o ph hp th nguyn t ngi ta phi a mu vo phn ny nguyn t ho v thc hin php o. Ngha l ngun n sc phi chiu qua phn ny ca ngn la.

+ Phn v v ui ca ngn la: vng ny c nhit thp, ngn la c mu vng v thng xy ra nhiu phn ng th cp khngc li cho php o ph hp th nguyn t.

Cu 6: Hy trnh by nhng qu trnh xy ra trong ngn la?


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HDTL:

Ngn la l mi trng nguyn t ho mu ca php o ph hp thnguyn t (F-AAS). Trong ngn la c nhiu qu trnh xy ra ng thi: c

qu trnh chnh v cng c qu trnh ph. Trong nhit ca ngn la lyu t quyt nh mi din bin ca cc qu trnh.

Trc ht, khi mu th sol kh c dn ln n nguyn t ho, di tcdng nhit ca ngn la, ming n, l s bay hi ca dung mi ho tanmu v cc cht hu c trong th sol kh. Nh vy, mu cn li l cc ht rnrt nh mn trong ngn la v n c dn tip vo vng trung tm ngn la.

Tip l qu trnh ho hi v nguyn t ho ca cc ht mu bt kh . y cc cht s c cc qu trnh chnh sinh ra ph v qu trnh ph khngsinh ra ph din bin theo tnh cht nhit ho ca cht mu.

* Yu t quyt nh cc qu trnh chnh l: - Nhit ca ngn la. - Bn cht ca cht mu v thnh phn ca cht mu.

- Tc dng nh hng ca cht ph gia thm vo mu. * Cc qu trnh ph tuy c mc khc nhau, nhng trong mt mitng quan nht nh trong ngn la, c bit l nhit ngn la, th ttc cc qu trnh u c th xy ra cng vi cc qu trnh chnh ca php o F-AAS. Do phi chn cc iu kin ph hp hn ch nmc nh nht cc qu trnh ph v gi cho n khng i sut trong mt

php o xc nh mt nguyn t. Cu 7: Hy cho bit c ch ca cc qu trnh xy ra trong ngn la? HDTL:

Cc qu trnh xy ra trong ngn la thng theo 2 c ch chnh: + Nu Eh < E a tc l nng lng ho hi (Eh) ca cc hp

phn c trong mu nh hn nng lng nguyn t ho (Ea) ca n, th trcht cc hp phn ny s ho hi dng phn t. Sau cc phn t kh nymi b phn li (nguyn t ho) thnh cc nguyn t t do (c ch I).


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+ Nu Eh > E a tc l nng lng phn li Ea ca cc hp phn ca mu l nh hn nng lng ho hi Eh ca chnh n, th trc htcc hp phn ny s b phn li thnh cc nguyn t t do, ri sau mi ho

hi (c ch II). C ch I:

MnAm (l) MnAm (k) n.M (k) + m.A (k)

M (k) + n (h ) ph AAS C ch II:

MnAm (l) n.M (r,l) + m.A (l,r) n.M (k)

M (k) + n (h ) ph AAS Cu 8: Hy tm tt cc qu trnh nguyn t ho mu? HDTL : Cc qu trnh nguyn t ho mu:

1. Dn mu vo bung Aersol ho. 2. Qu trnh aerosol ho mu. 3. Ho hi, nguyn t ho: MeA (r) MeA (l) MeA (k)

4. S phn li, kch thch, hp th, ion ho, pht x: Me o + A . (phn li) Me o + h (hp th bc x)

MeA Me o + E (kch thch) Me o e (ion ho) Me o + .

5. S kh oxi ca oxit bi cacbon:

MeO + C Me + CO

6. Cc phn ng ho hc khc (hp cht bn nhit monoxit): Me + O MeO

Me + xC MeC x

Cu 9: Hy nu nhng yu cu i vi h thng nguyn t ho mu? HDTL : H thng nguyn t ho mu c 5 yu cu:


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1. H thng nguyn t ho hi v nguyn t ho mu phn tch vihiu sut cao v n nh, m bo cho php o c nhy caov lp li tt.

2. Phi cung cp c nng lng ln v c th iu chnh cd dng theo yu cu c th nguyn t ho c nhiu loi muv phn tch c nhiu nguyn t.

3. Cuvet cha mu nguyn t ho phi c tinh khit cao. Khnglm nhim bn mu, khng c ph ph gy kh khn cho php oca nguyn t cn phn tch.

4. Hn ch, c t hay khng c cc qu trnh ph trong qu trnhnguyn t ho mu thc hin php o.

5. Tiu tn t mu phn tch. Cu 10: Nu nhng yu cu ti thiu m ngun pht tia bc x n sctrong php o ph AAS phi tho mn? HDTL : C 4 yu cu:

- Ngun pht tia bc x n sc phi to ra c cc tia pht x nhy(cc vch pht x nhy, c trng) ca nguyn t cn phn tch.Chm tia pht x phi c cng n nh, phi lp li trongcc ln o khc nhau trong cng iu kin, phi iu chnh cvi cng mong mun cho mi php o.

- Ngun pht tia bc x phi cung cp c mt chm tia pht x

thun khit ch bao gm mt s vch nhy c trng ca nguyn t phn tch. Ph nn ca n phi khng ng k. C nh th mi hnch c nhng nh hng v vt l v v ph cho php o AAS.

- Chm tia sng n sc do ngun cung cp phi c cng cao. Nhng li phi bn vng theo thi gian v phi khng b ccyu t vt l khc nhiu lon, t b nh hng bi cc dao ng caiu kin lm vic.


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- Ngun pht tia bc x n sc phi bn lu, khng qu t tin vkhng qu phc tp cho ngi s dng.

Cu 11: Nu nhng yu t c th nh hng n kt qu phn tch trong

php o ph hp th nguyn t? HDTL:

Nhm 1: L cc thng s ca h my o ph. Cc thng s ny cn ckho st v chn cho tng trng hp c th. Thc hin cng vicny chnh l qu trnh ti u ho cc thng s ca my o cho mti tng phn tch.

Nhm 2: L cc iu kin nguyn t ho mu. Cc yu t ny th hin rtkhc nhau tu thuc vo k thut c chn thc hin qu trnhnguyn t ho mu.

Nhm 3: L k thut v phng php c chn x l mu. Trong cngvic ny nu lm khng cn thn s c th lm mt hay lm nhim bn thm nguyn t phn tch vo mu.

Nhm 4: Cc nh hng v ph. Nhm 5: Cc yu t nh hng vt l. Nhm 6: Cc yu t nh hng ho hc.


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Cu 12: Cc yu t v ph c nh hng nh th no trong php o AAS? HDTL: Cc yu t v ph gm:

- S hp th nn: yu t ny c trng hp xut hin r rng nhng

cng c th khng xut hin. N ph thuc vo vch ph c chn onm trong vng ph no. Ni chung trong vng kh kin th yu t ny thhin r rng, cn trong vng t ngoi th nh hng ny t xut hin do phnn trong vng t ngoi l yu. Hn na s hp th nn cn ph thuc vothnh phn nn ca mu phn tch.

V d: Khi xc nh Pb trong mu sinh hc bng php o ngn la th

s hp th nn l khng ng k. Khi xc nh Pb trong nc bin th nhhng ny li l v cng ln.

- S chen ln ca vch ph: yu t ny xut hin khi cc nguyn t th ba trong mu phn tch c nng ln v thng l nguyn t c s ca mu. Tuy nguyn t ny c cc vch ph khng nhy nhng do nng ln nn cc vch ny vn xut hin vi rng ln.

- S hp th ca cc ht rn: trong mi trng hp th, c bit l trongngn la n kh, nhiu khi cn c cha c cc ht rn rt nh li ti ca vtcht mu cha b ho hi v nguyn t ho, hay cc ht mui cacbon canhin liu cha c t chy hon ton. Cc loi ht ny thng c th c lp v ca ngn la. Yu t ny gi l hp th gi. N th hin rt r rngkhi chn khng ng chiu cao ca n nguyn t ho mu v khi hn hp

kh chy khng c t chy tt. Cu 13: nh hng ca yu t vt l trong php o AAS? Bin php loitr?

HDTL:

- nht v sc cng b mt ca dung dch mu: yu t ny nh hngnhiu n tc dn mu vo bung aerosol ho ca mu, t nh hngn kt qu phn tch. Tc dn mu t l nghch vi nht ca dungdch mu.


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Hnh 2.4 : nh hng ca nht n tc dn mu (1cp 1g/m.s) Trong mi qu trnh phn tch mt nguyn t, nht thit phi m bo

cho mu phn tch v cc mu u lp ng chun phi c cng nng axit, loi axit v thnh phn ho hc, vt l ca cc nguyn t khc, nht lcht nn ca mu. loi tr nh hng ny, ngi ta dng cc bin phpsau:

+ o v xc nh theo phng php thm tiu chun. + Pha long mu bng mt dung mi hay mt nn ph hp.

+ Thm vo mu mt cht m c nng ln. + Dng bm y mu vi mt tc xc nh m ta mong mun.

- Hiu ng lu: yu t ny th hin r rng trong phpo ph khng ngn la. N ph thuc vo bn cht cc nguyn t. Mun loitr nh hng ny, ngi ta lm nh sau:

+ Lm sch cuvet sau mi ln nguyn t ho mu.

+ Dng cc cuvet c ch to t cc loi graphit c hot hoton phn, c b mt chc v mn.

+ Khi phn tch nn o cc mu c nng nh trc.+ Thm vo mu nhng cht m c nng ph hp. + Trng b mt trong ca cuvet graphit bng mt lp cc hp cht bn

nhit.


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- S ion ho: qu trnh ion ho thng lm gim snguyn t t do ca nguyn t phn tch trong mi trng hp th to ra ph. loi tr yu t ny, ngi ta lm nh sau:

+ Chn iu kin nguyn t ho c nhit thp. + Thm vo mu phn tch mt cht m cho s ion ho.

- S kch thch ph pht x: yu t ny lm gim nng ca cc nguyn t trung ho c kh nng hp th bc x trong mi trnghp th. Do cng lm gim cng ca vch ph hp th. loi trn, ngi ta lm nh sau:

+ Chn nhit nguyn t ho mu thp ph hp m ti nhit s kch ph pht x l khng ng k hoc khng xy ra i vinguyn t phn tch.

+ Thm vo mu cc cht m hn ch s pht x ca nguyn t phn tch.

Cu 14: Cc yu t ho hc c nh hng nh th no trong php o AAS? HDTL:

Trong php o AAS, cc nh hng ho hc cngrt a dng v phc tp. N xut hin cng rt khc nhau trong mi trnghp c th v cng c nhiu trng hp khng xut hin. Cc nh hng caho hc c th lm kt qu xy ra theo cc hng sau:

- Lm gim cng ca vch ph ca nguyn t phntch, do s to thnh cc hp cht bn nhit, kh ho hi v kh nguyn tho. V d, nh hng ca cc ion silicat, sunfat, photphat, florua.

- Lm gim cng ca vch ph, do s to thnhcc hp cht d ho hi v d nguyn t ho hay do hn ch c nh hngca s ion ho v s kch thch ph pht x ca nguyn t phn tch. chnh l tc dng ca mt s hp cht, ch yu l mui halogen ca kim loikim v kim th hay lantan clorua.


35/155


- S tng cng vch ph khi nguyn t phn tchtn ti trong nn ca mu l nhng hp cht d ho hi. Lc cc cht nnny c tc dng nh l mt cht mang cho s ho hi ca nguyn t phn

tch v lm n c ho hi vi hiu sut cao hn. - S gim cng vch ph khi nguyn t phn tch

tn ti trong nn ca mu l nhng hp cht bn nhit, kh ho hi. Lc nycc nguyn t nn km hm s ho hi ca nguyn t phn tch. Cc cht nnny thng l nhng hp cht bn vi nhit ca cc nguyn t nh Al, thim ....

Cu 15: Phng trnh c bn ca php o AAS? HDTL:

Trong php o ph hp th nguyn t, mi quan h gia cng camt vch ph hp th ca mt nguyn t phn tch v nng ca n trongmi trng hp th tun theo nh lut hp th quang Lambert- Beer. Nghal nu chiu chm tia sng n sc cng I0 i qua mt mi trng cha

mt loi nguyn t t do nng N v c b dy L (cm), th mi quan hgia I0 v phn cng sng I i qua mi trng :

lg I I 0 = K . N . L

lg I I 0 : nng lng ca tia sng b mt i do s hp th ca cc

nguyn t t do trong mi trng , n chnh l cng ca

vch hp th A.Ta c: A = K . N . L

Gi C l nng ca nguyn t trong mu phn tch: N = k . C b

k: hng s thc nghim, ph thuc tt c cc iu kin ho hi vnguyn t ho mu.(k = const)

b: hng s bn cht, ph thuc C (0 < b1)


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A = K . L . C b ; K = K.k gi l hng s thc nghim ca php oAAS.

Tng qut: A= a . C b l phng trnh c s ca phng php phn

tch nh lng da theo vic o ph hp th ca mt nguyn t xc nhnng (hm lng) ca n. Khong nng 0C C0 (C0 l nng gii hn khi b bt u khc 1)gi l khong tuyn tnh.

Cu 16: Trong php o AAS, nng ca nguyn t phn tch thng c

biu din nh th no? HDTL:

Trong php o AAS, nng ca nguyn t phn tch c biu dintheo 4 cch sau:

* Nng phn trm (%): c biu th bng s gam ca cht phn tch c

trong 100 gam mu em phn tch:C% =m

m X .100

mX: s gam cht phn tch c trong mu ly phntch m: s gam mu phn tch.

* Nng microgam/mL hay microgam/L: c s dng ph bin trong phn tch lng vt v c biu th theo hai cch:

+ S microgam ca cht phn tch c trong mt lit dung dch mu( g/L) hay s microgam cht phn tch c trong mt mililit dung

dch mu ( g/L).

V d: CPb = 1,2 g/L ngha l trong 1 lit dung dch mu c 1,2 g ca

Pb

+ S g ca cht phn tch c trong 1 gam mu (hoc 1kg mu).

V d: CPb = 2,5 g/L th c ngha l trong 1 gam mu c 2,5

g Pb.


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* Nng ppm v c s ca n: + Vi mu rn: nu 1 gam mu c cha 10-6 gam cht phn tch th gi

tr ny (10-6g) c gi l ppm. Cc c s ca n v ppm l ppb,

ppp, ppa (cc n v km nhau 1000 ln). + Vi mu lng: nu 1 ml mu c cha 10-6 gam cht phn tch v khi

t trng ca dung dch mu bng 1 th i lng 10-6g/ml mu cngc nh ngha l 1 ppm.

* Nng mol/l (M): biu th s phn t gam cht phn tch c trong 1 lit

dung dch mu: CM =V

n

Cu 17: nhy l g? nhy ph thuc vo nhng yu t no? HDTL:

* nhy l mt i lng ch ra kh nng ca mt phng php phn tch theo mt k thut o no c p dng cho phng php phn tch . Phng php phn tch c nhy cao tc lnng gii hn di c th phn tch c l nh.

* nhy ph thuc vo nhiu yu t khc nhau nh: + Cc c trng, tnh cht ca h thng my o nh h

quang hc, dng c thu nhn v pht tn hiu o. chnh l cc c trngk thut ca h thng my quang ph.

+ Cc iu kin v k thut thc hin nguyn t ho mu o ph. V th phng php nguyn t ho trong ngn la c nhykm hn phng php nguyn t khng ngn la.

+ Kh nng v tnh cht hp th bc x ca mi vch ph,ca mi nguyn t. Vch ph no c kh nng hp th cng mnh th phpo theo vch c nhy cng cao. i vi mt nguyn t cc vch phkhc nhau cng s c nhy khc nhau.


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Ngoi ra cn mt s yu t khc cng nh hng n nhy ca mt phng php phn tch nh: axit ca mu, nguyn t cn tr c trongmu.

Cu 18: Xc nh nng ca mt nguyn t trong mu phn tch theo php o ph hp th nguyn t, ngi ta thc hin theo nhng phng php no?

HDTL:

Da theo phng trnh nh lng c bn ca php o AAS qua vico cng ca vch hp th ca nguyn t phn tch v xc nh nng

ca cht phn tch trong mu o ph theo mt trong hai phng php chunho sau:

* Phng php ng chun: cn gi l phng php 3mu u.

Nguyn tc: da vo phng trnh c bn ca php o: A =K.C v mt dy mu u xy dng mt ng chun v t ng chun

ny v gi tr AX xc nh nng CX ca nguyn t cn phn tch trongmu o ph, ri t tnh c nng ca n trong mu phn tch.

Cch x l: Trc ht ngi ta phi chun b mt dy muu, dy mu chun (thng thng l 5 mu u) v cc mu phn tch trongcng mt iu kin. Sau chn cc iu kin ph hp v o cng camt vch ph hp th ca nguyn t phn tch trong tt c cc mu u v

mu phn tch c chun b trn. Ta dng c mt ng biu thmi quan h gia cc gi tr cng ng vi cc nng tng ng. lng chun ca phng php.

* Phng php thm tiu chun: Nguyn tc: dng ngay mu phn tch lm nn chun b

mt dy mu u, bng cch ly mt lng mu phn tch nht nh v giathm vo nhng lng nht nh ca nguyn t cn xc nh theo tng bc nng (theo cp s cng).


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Cch x l: Trc ht thm lng vo mu u. Tip ,chn cc iu kin th nghim ph hp v mt vch ph ca nguyn t phntch, tin hnh ghi vch ph theo tt c dy mu u. T cc gi tr cng

ng vi cc nng thm vo ca nguyn t phn tch, ta dng c mtng, chnh l ng chun ca phng php thm. Cu 19: Cc thng s ca h my o ph AAS thng bao gm cc yu t

no? HDTL: Cc thng s ca h my o ph AAS thng bao gm cc yu t:

- Vch ph: vch ph c chn xc nh nguyn t m

mnh mong mun. Vch ph ny phi tho mn iu kin: + Khng b cc vch khc chen ln, gy nhiu hay trng lp. + Phi r rng v nt, c di sng chnh xc. + Cng vch ph hp th A phi ph thuc tuyn tnh vo nng

C trong mt vng nng nht nh. + Vch ph phi nm trn mt nn khng qu ti.

+ Tt nhin khi cn xc nh cc nng rt nh th phi chn vch ph c nhy cao v ngc li.

- Cng n n catot rng (HCL): cng dng inlm vic ca HCL nn chn cng dng nm trong vng t 60%- 85%so vi cng cc i ghi trn HCL. Khi cn nhy cao th chn cndi, khi cn n nh th chn cn trn.

- Khe o ca my quang ph: nn chn mt gi tr ph hpnht cho php o nh lng mt nguyn t theo vch ph c chn,lm sao ch cho va vch ph cn o vo khe o l tt nht.

- Chiu cao n nguyn t ho mu: Burner head Hight. - Thi gian o: ph thuc c trng k thut tng my o

ph. - Lng mu: tc dn mu, lng mu bm vo. - Vng tuyn tnh: c chn nh lng.


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- Phng tin ch th kt qu o. - B chnh nn khi o: ph thuc vo vch ph khi o. - Nhn quang nhn tn hiu AAS: l thang nng lng hp

th ca php o AAS. Cu 20: Cc phng php xc nh trc tip theo AAS gm my giai on?

cc giai on xy ra nh th no? HDTL:

V nguyn tc th tt c cc nguyn t v cc cht c ph hp thnguyn t chng ta u c th xc nh n mt cch trc tip theo ph hp

th nguyn t ca n t dung dch mu phn tch. Ngha l cc phng phpxc nh trc tip ch ph hp cho vic xc nh cc kim loi c vch phhp th nguyn t. V cc kim loi u c ph hp th nguyn t ca trongnhng iu kin nht nh.

V d: cc mu v c l qung, t, , khong liu, mui, oxit, kimloi, xi mng, hp kim, nc, khng kh v cc mu hu c l cc mu thc

phm, ng .... Khi phn tch cc loi mu ny th nguyn tc chung gm 2giai on: * Giai on 1: X l mu a nguyn t kim loi cn xc nh c trongmu v trng thi dung dch ca cc cation theo mt k thut ph hp chuyn c hon ton nguyn t cn xc nh vo dung dch o ph. * Giai on 2: Phn tch nguyn t cn thit theo ph hp th nguyn t ca

n theo nhng iu kin nht nh ph hp c nghin cu v chn ra. y, giai on 1 l cc k quan trng. V nu x l mu khng tt

th c th lm mt nguyn t cn phn tch hay lm nhim bn thm vo. Ngha l vic x l mu khng ng s ngun sai s rt ln cho kt qu phntch, mc d phng php phn tch chn ph hp. Cu 21: Ti sao n catot rng l ngun c s dng ph bin nht? HDTL:


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- n pht tia bc x n sc c dng sm nht v ph bin nht trong php o AAS l n catot rng v n ny ch pht ra nhngtia pht x nhy ca nguyn t kim loi lm catot rng. Cc vch pht x

nhy ca mt nguyn t thng l cc vch cng hng. V th n catotrng cng c gi l ngun pht tia bc x cng hng. N l ph pht xca cc nguyn t trong mi trng kh km. - Cu to ca n catot rng:

Hnh 2.5 n catot rng 1. Bng thu tinh. 2. Anot 3.Catot 4 .Ca s thch anh n catot rng l mt bng thu tinh hnh tr, ng knh 3- 5cm, c

ca s bng thu tinh hay thch anh. Anot c ch to t thanh kim loi.

C hai cc c t trong bng thu tinh c cha kh tr (agon hay neon)vi p sut khng ln(0,2 - 2MPa). n catot rng c ni vi dng in300 - 500V, n nh v phi c n nh cao. Dng phng ca n thngl vi miliampe. Khi n lm vic, mt dng mt bn trong catot caohn mt ngoi. V vy ti l m ca catot s pht sng. Catot ca n cch to ca kim loi, hp kim kh nng chy c cha nguyn t cn xc

nh.

Cu 22: Hy nu nhng thnh phn chnh ca n catot rng? HDTL:

n catot rng gm 3 phn chnh: (xem hnh 2.5) Phn 1 l thn n v ca s.

Phn 2 l cc in cc cato v anot.


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Phn 3 l kh cha trong n. l kh tr He, Ar hay Nit. + Thn v v: Thn n gm c v n, ca s v b

cc in cc anot v catot. B bng nha PVC. Thn v v n bng thu

tinh hay thch anh. Ca s S ca n c th l thu tinh hay thch anh trongsut trong vng UV hay VIS l tu thuc vo loi n ca tng nguyn t pht ra chm tia pht x nm trong vng ph no.

+ in cc: in cc ca n l catot v anot. Anot cch to bng kim loi tr v bn nhit nh W hay Pt. Catot c ch to cdng hnh xylanh hay hnh ng rng c ng knh t 3- 5mm, di 5- 6mm

v chnh bng kim loi cn phn tch vi tinh khit cao (t nht 99,9%).Dy dn ca catot cng l kim loi W hay Pt. Ngun nui l ngun 1 chiuc th 220 - 240V.

+ Kh trong n: Trong n phi ht ht khng kh v npthay vo l mt kh tr vi p sut t 5- 15mmHg. Kh tr l agon,heli hay nit nhng phi c sch cao hn 99,99%. Kh np vo n phi

khng pht ra ph lm nh hng n chm tia pht x ca n v khi lmvic trong iu kin nht nh th t s tu thuc vo tng loi n v tngnguyn t kim loi lm catot rng. Cng lm vic ca n catot rngthng l t 3- 50mA.II.3. BI TP: Dng 1: Xc nh nng ca kim loi trong mu, khi bit hm lng kim

loi trong dung dch chun v hp th.


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Cch gii: p dng phng php hi quy tuyn tnh theo nguyn l bnh phng

ti thiu, ta xy dng ng chun: A = aC + b.

Tnh a, b?

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 ; b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .

Bi 1: Ngy nay, xc nh s nhim thu ngn ca cc dung dch nc bng phng php hp th nguyn t ngi ta dng phng php khng

ngn la mi ca s phun m. Thit b gm mt bnh kh thu ngn nivi mt cuvet hp th. 10ml mu nc vo bnh kh thu ngn v phalong n 100ml, sau thm vo 25ml axit sunfuric c v 10ml dung dchthic (II) sunfat 10%, axit sunfuric 0,25M (dung dch cui ny dng lm chtkh). Thu ngn b kh n trng thi nguyn t (nguyn t) v cchuyn vo cuvet hp th bi dng khng kh, ngi ta cho dng khng kh

ny i qua dung dch trong bnh kh thu ngn. Cui cng, dng ncatot rng lm ngun, ngi ta o s hp th ca cc nguyn t thu ngn bc sng 2537A0, s hp th t c mc cc i gn 3 pht. Ngi ta nhn c cc gi tr sau ca hp th i vi dy cc dung dchchun ca thu ngn (II): Hm lng Hg trong dung dch chun, g hp th

0,00 0,002

0,30 0,090

0,60 0,175

1,00 0,268

2,00 0,440


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Cc gi tr ca hp th ca hai mu nc bng 0,040 v 0,305 tng ng.Vy hm lng ca thu ngn trong tng mu bng bao nhiu? Nng ( ml g / ) ca thu ngn trong tng mu bng bao nhiu? [28]

Gii:Tin hnh xy dng ng chun, hi qui tuyn tnh theo nguyn l bnh phng ti thiu A = aC + b :

STT Hm lng Hg (C) hp th (A) A.C C 2

1 0,00 0,002 0,000 0,000

2 0,30 0,090 0,027 0,090

3 0,60 0,175 0,105 0,360

4 1,00 0,268 0,268 1,000

5 2,00 0,440 0,880 4,000

n=5 C = 3,9 A = 0,975 AC =1,28 2C = 5,45

n n n

i i i ii 1 i 1 i 1

2 2n n2i i

i 1 i 1

n A C A C5.1, 28 0,975.3,9

a5.5, 45 3.9

n C C

= 0,2157

n n n n2

i i i i ii 1 i 1 i 1 i 1

2 2n n2i i

i 1 i 1

A C A C C0,975.5, 45 1, 28.3,9

b5.5, 45 3.9

n C C

= 0,02672

Phng trnh hi quy thu c l: A = 0,2157C + 0,0267


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* Vi A = 0,040 thay vo phng trnh thu c hm lng thu ngn:0,040 = 0,2157C + 0,02672

Suy ra C = 0,0617 g ng vi nng 0,00617g/ml.

* Vi A = 0,305 thay vo phng trnh thu c hm lng thu ngn:0,305 = 0,2157C + 0,02672

Suy ra C = 1,2902 g ng vi nng 0,12902g/ml.

Bi 2: C th dng php o ph hp th nguyn t xc nh cc vt cckim loi nng trong du mazut. phn tch 5,000 gam mu ca loi du

mazut dng, ngi ta t vo mt bnh nh mc c dung tch 25,00 ml,ho tan vo 2- metyl - 4 - pentanol v bng dung mi ny a th tch trong bnh n vch. Sau phun m dung dch nhn c trong ngn la khngkh - axetylen. xc nh ng v ch cn dng cc n catot rng vi ccvch pht x 324,7 v 283,3 nm tng ng. nhn c cc th chuncn mt dy cc dung dch chun cha nhng lng bit ca ng v ch

trong hn hp tng ng vi du mazut cha dng v 2-metyl-4-pentanol.Xy dng phng trnh hi quy ca ng v ch trong 5,000 gam mu dumazut dng theo cc s liu sau:

Dung dch hp th

283,3 nm (Pb) 324,7 nm (Cu)

a, Chun: 19,5 ml g / Pb 5,25 ml g / Cu 0,356 0,514 b, Chun: 4,00 ml g / Pb 4,00 ml g / Cu 0,073 0,392

c, Chun: 12,1 ml g / Pb 6,27 ml g / Cu 0,220 0,612

d, Chun: 8,50 ml g / Pb 0,155 0,101

, Chun: 15,2 ml g / Pb 2,40 ml g / Cu 0,277 0,232

e, Cha bit 0,247 0,371

Gii:


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Tin hnh xy dng ng chun, hi qui tuyn tnh theo nguyn l bnh phng ti thiu A = aC + b :

Stt Hm lng ch (C) hp th (A) A.C C 2

1 4,00 0,073 0,292 16,000

2 8,50 0,155 1,318 72,250

3 12,1 0,220 2,662 146,410

4 0,00 0,247 0,000 0,000

5 15,2 0,277 4,210 231,040

6 19,5 0,356 6,942 380,250

C = 59,3 A = 1,382 AC = 15,424 C2 = 845,950

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 23,59950,845.63,59.328,1424,15.6 = 0,1099

b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 2

3,59950,845.6

3,59.424,1595,845.328,1 = 0,1339

Phng trnh hi quy thu c l: A = 0,1099C+ 0,1339Stt Hm lng ng (C) hp th (A) A.C C 2

1 0,00 0,101 0,000 0,000

2 2,40 0,232 0,557 5,760

3 0,00 0,371 0,000 0,000

4 4,00 0,392 1,568 16,000

5 5,25 0,514 2,699 27,563

6 6,27 0,612 3,837 39,313

C = 17,92 A = 2,222 AC = 8,661 C2 = 88,635

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 292,17635,88.692,17.222,2661,8.6 = 0,0557


47/155


b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 292,17635,88.6

92,17.661,8635,88.222,2 = 0,1981

Phng trnh hi quy thu c l: A = 0,0557C + 0,1981 Bi 3: xc nh hm lng ng trong mu FFDT (mu cht bo t dokh thu c t dch chit khp chn), ngi ta dng 11,23 mg FFDT phatrong bnh 5,00 ml bng dung dch HNO3 0,75 M n vch. Tin hnh o ph hp th nguyn t dng ngn la khng kh- axetylen thu c hpth 0,023 bc sng 324,8 nm. Tin hnh o cc dung dch chun trong

iu kin tng t, kt qu thu c nh sau : ppm Cu 0,000 0,100 0,200 0,300 0,400 0,500 0,600 0,700 1,000

hp th0,000 0,006 0,013 0,020 0,026 0,033 0,039 0,046 0,066 Nng ca ng (g/gam FFDT) l bao nhiu?

Gii:Tin hnh xy dng ng chun, hi qui tuyn tnh theo nguyn l bnh phng ti thiu A = aC + b :

Stt Hm lng ng (C) hp th (A) A.C C 2

1 0,000 0,000 0,000 0,000

2 0,100 0,006 0,001 0,010

3 0,200 0,013 0,003 0,040

4 0,300 0,020 0,006 0,090

5 0,400 0,026 0,010 0,160

6 0,500 0,033 0,017 0,250

7 0,600 0,039 0,023 0,360

8 0,700 0,046 0,032 0,490

9 1,000 0,066 0,066 1,000

n = 9 C = 3,800 A = 0,249 AC = 0,158 C2 =

2,400


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a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 2800,3400,2.9800,3.249,0158,0.9 = 0,0665

b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 2800,3400,2.9

800,3.158,0400.2.249,0 = - 0,00039

Phng trnh hi quy thu c l: A = 0,0665C- 0,00039(C l nng ppm Cu) thay A = 0,023

C = 0,352 ppm = 0,352 g/ml

Hm lng Cu : gam

ml ml g 01123,0

5./352,0 = 157 gCu/gam FFDT.

Bi 4: Mt nh phn tch a ra kt qu d liu xy dng ng chun phn tch hm lng photpho bng ph hp th nguyn t ngn la : ppm P 2130 4260 6400 8530

hp th 0,048 0,110 0,173 0,230

xc nh tinh khit ca mu Na2HPO 4, ngi ta ho tan 2,469 gammu vo nc, chuyn vo bnh nh mc 100,00 ml v nh mc n vch. hp th nguyn t ca mu thu c l 0,135. Xc nh tinh khit camu Na2HPO 4.

Gii: Tin hnh xy dng ng chun, hi qui tuyn tnh theo nguyn l bnh phng ti thiu A = aC + b:

Stt Hm lng phot pho (C)

hp th(A)

A.C C 2

1 2130 0,048 102,2400 4536900

2 4260 0,110 468,6000 18147600


49/155


3 6400 0,173 1107,2000 40960000

4 8530 0,230 1961,9000 72760900

n =4 C = 21320 A = 0,561 AC = 3639,940 C2 = 136405400

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 221320136405400.421320.561,0940,3639.4 = 2,854.10 -5

b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 221320136405400.4

21320.940,3639136405400.561,0 = -0,0119

Phng trnh hi quy thu c l: A = 2,854.10-5C - 0,0119(C l nng ppm Cu) thay A = 0,135

C = 4896,7 ppm = 4896,7 g/ml

Hm lng P : 4896,7g/ml . 100,00ml = 489670 g 0,490g

Khi lng Na2HPO 4 tinh khit l:31490,0 .142 = 2 ,243gam, tinh khit

91%.

Bi 5: Ch c xc nh bng phng php hp th nguyn t ngn la. gim nh hng nhiu, phng php thm chun c s dng. Tin hnhchun b ba mu, mi mu cha 5,00 ml dung dch mu cn xc nh hmlng ch, trong mu th nht thm vo 5,00 ml nc ct. Mu th hai cthm 1,00 ml dung dch ch nitrat 10,0M v 4,00 ml nc ct. Cn mu th

ba c thm 2,00 ml dung dch ch nitrat 10,0M v 4,00 ml nc ct. Bamu em o ph hp th nguyn t, hp th thu c ln lt l: 0,13;0,30 v 0,47. Hy xc nh hm lng ch trong mu ban u? Gii: - Gi nng ch trong mu ban u l Co ( M), trong mu th nht l:

CX =5,00

5,00 5,00Co = 0,5C o


50/155


- Nng ch trong mu th 2 l CX + 1,00.10,05,00 1,00 4,00

= C X + 1,00 ( M)

- Nng ch trong mu th 3 l CX + 2,00.10,05,00 2,00 3,00

= C X + 2,00 ( M)

V hp th ph thuc tuyn tnh vo nng nn s dng phng php th theo bng s liu : hp th 0,13 0,30 0,47 Nng ch (M ) 0 1,00 2,00(Ngoi suy th, im ct trc honh l-Cx = 0,5C o;) hoc xy dng nghi qui tuyn tnh A = aC + b v cho A = 0 thu c CX = - C.

Stt Hm lng P (C) hp th (A) A.C C 2

1 0 0,13 0,0000 0,00

2 1,00 0,30 0,3000 1,00

3 2,00 0,47 0,9400 4,00

n =3 C = 3,00 A = 0,90 AC = 1,24 C2 = 5,00

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 200,300,5.300,3.90,024,1.3 = 0,17

b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 200,300,5.3

00,3.24,100,5.90,0 = 0,13

Phng trnh hi quy thu c l: A = 0,17C + 0,13


51/155


y = 0,17x + 0,13R 2 = 1

0

0,05

0,1

0,15

0,2

0,25

0,3

0,35

0,4

0,45

0,5

-2 -1 0 1 2 3

Hnh 2.6: ng thm chun xc nh hm lng ch Kt qu thu c A = 0,17C + 0,13- C = C X = 0,7647 = 0,5C o.Vy nng ch trong mu ban u l: Co = 1,5294 M.

Bi 6: Canxi trong dung dch mu c xc nh bng phng php quang ph hp th nguyn t. Dung dch chun c chun b bng cch ho tan

1,834 gam CaCl 2.2H 2O trong nc, pha long thnh 1,00 lt, dung dch nyc pha long 1: 10. Cc dung dch chun o c chun b bng cch pha long dung dch th hai theo t l 1:20; 1:10 v 1:5. Mu c pha longtheo t l 1:25. Dung dch SrCl2 c thm vo cc dung dch trc khi phalong t 1% (gam/100ml) trnh nh hng photphat. Mu trng cchun b l dung dch SrCl2 1%. Tn hiu hp th c ghi khi cc dung

dch c phun vo ngn la khng kh- axetylen cho kt qu sau : Mu Trng Mu 1:20 Mu 1:10 Mu 1:5 Mu phn tch hp th 1,5 10,6 20,1 38,5 29,6Xc nh nng canxi trong mu di dng ppm? Gii: - Nng Ca trong dung dch chun:

+ Dung dch thu c khi pha 1,834 gam/L l:


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1,834.40

147gam Ca/L

+ Dung dch thu c khi pha t l 1:10 l:

1,834 .4147 = 0.050 gam Ca/L = 50 mg Ca/L.

hay nng Canxi theo ppm l: 50 ppm. - Gi nng Ca trong mu phn tch l Co (ppm). Nng Ca trong cc dung dch o v tn hiu hp th tng ng l: Mu Trng Mu 1:20 Mu 1:10 Mu 1:5 Mu phn tch Nng (ppm) 0 2,5 5,0 10,0 C o/25 hp th 1,5 10,6 20,1 38,5 29,6- Xy dng s ph thuc hp th A vo nng da trn nguyn l bnh phng ti thiu A = aC + b:

Mu Nng (C)

hp th (A) A.C C 2

Trng 0 1,5 0 0

Mu 1:20 2,5 10,6 26,5 6,25Mu 1:10 5,0 20,1 100,5 25,0Mu 1:5 10,0 38,5 385,0 100,0n =4 C = 17,5 A = 70,7 AC = 512,0 C2 = 131,25

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 25,1725,131.45,17.7,700,512.4 = 3,7063

b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 25,1725,131.4

5,17.0,51225,131.7,70 = 1,46

- Phng trnh hi qui thu c: A = 3,7063.C + 1,46

- Hm lng Ca trong mu:oC A 1, 46 29, 6 1, 46

25 3, 7063 3, 7063

Co = 189,8 ppm.


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Bi 7: Clorua trong mu nc c xc nh gin tip bng phng phpquang ph hp th nguyn t bng cch lm kt ta hon ton n di dngAgCl vi lng d chnh xc AgNO3, lc b kt ta o nng cn li ca

Ag trong dung dch nc lc. 10,00 ml ca cc dung dch mu v dung dchchun 100 ppm clorua c cho vo bnh eclen 100,0 ml kh, thm vo mi bnh 25,0 ml dung dch AgNO3, sau mt thi gian kt ta hon ton, lcchuyn cc hn hp sang cc ng li tm kh v em li tm. Mi dung dchnc lc (sau li tm) c phun vo bung nguyn t ho xc nh hp th. Mu trng c chun b tng t khi dng 10,0 ml nc ct. K

qu thu c trn bng sau: Mu Mu trng Mu chun Mu phn tch hp th (chiu caovch hp th (cm))

12,8 5,7 6,8

Nng ca clorua trong mu phn tch ban u l bao nhiu ?

Gii: - Gi nng ca AgNO3 trong dung dch chun gc l C01 mM.- Phn ng xy ra trong qu trnh chun b mu l :

Ag + + Cl - AgCl

- Nng Ag+ trong mu trng :25,0V

C01 = C 1 (mM)

- Nng Ag+ tr ong mu chun :o1 100

25, 0C 10 35,5V

= C 2 (mM)

- Nng Ag+ trong mu phn tch :o1 o

X25,0C 10,0CV

= C 3 (mM)

- Nng v tn hiu ca cc dung dch o: Mu Mu trng Mu chun Mu phn tch Nng C1 C2 C3


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hp th (chiu caovch hp th (cm))

12,8 5,7 6,8

- Ta c : A1 = KC 1 = 12,8 (1)A2 = KC 2 = 5,7 (2)

A3 = KC 3 = 6,8 (3)

o1

1

o12

C 12,8 25C100C 5,7 25C 1035,5

C01 = 2,0313 mM.

1

3

C 12,8C 6,8

=o1

o1 oX

25,0C

25,0C 10,0C CX

o = 2,3804 mM.

Hm lng clorua trong mu l: 2,3804.35,5 = 84,5 ppm.

Bi 8: Bonert and Pohl bo co kt qu phn tch ph hp th nguyn t ivi mt s kim loi trong cc sn phm ph ca qu trnh sn xut soda bng phng php ammonia-soda.a, Nng ca Cu c xc nh bng cch axit ho 200 ml mu bng 20,0ml dung dch HNO3, thm vo 1,0 ml dung dch H2O2 27% (khi lng/thtch) v un si trong 30 pht. Dung dch thu c c pha long thnh 500 ml v em phn tch hp th nguyn t. Kt qu phn tch cho trn bng: Mu Nng Cu (ppm) hp th trng 0 0.007chun 1 0.200 0.014

chun 2 0.500 0.036chun 3 1.000 0.072chun 4 2.000 0.146 phn tch 0.027Hy xc nh nng ca Cu trong mu phn tch? b, Nng ca Cr c xc nh bng cch axit ho 200 ml mu bng 20,0

ml dung dch HNO3, thm vo 0,20 gam Na2SO 3 v un si trong 30 pht.Lng Cr c tch t mu bng cch thm 20 ml NH3 to ra kt ta cha


55/155


Cr v mt s oxit khc. Kt ta c tch ra, ra chuyn qua vi nc ravo cc. Sau khi c axit ho bng 10 ml dung dch HNO3, c lm bayhi n kh. Cn rn c ho tan li bng dung dch HNO3 v HCl, lm

bay hi n kh. Cui cng cn rn c ho tan trong 5,00 ml dung dchHCl v pha long thnh 50,00 ml. Dung dch thu c em phn tch bng phng php thm chun. Kt qu hp th nguyn t c tng hp trn bng sau: Mu Nng Cr thm (ppm) hp th trng 0 0.001

mu 0 0.045thm chun 1 0.200 0.083thm chun 2 0.500 0.118thm chun 3 1.000 0.192Hy xc nh nng ca Cu trong mu phn tch? Gii: a) - Xy dng s ph thuc hp th A vo nng da trn nguyn l bnh phng ti thiu A = aC + b:

Mu Nng (C) hp th (A) A.C C 2

Trng 0 0,007 0 0Chun 1 0,200 0,014 0,003 0,040Chun 2 0,500 0,036 0,018 0,250Chun 3 1,000 0,072 0,072 1,000Chun 4 2,00 0,146 0,292 4,000

n =5 C = 3,700 A = 0,275 AC = 0,385 C2 = 5,290

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 2700,3290,5.5700,3.275,0385,0.5 = 0,071


56/155


b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 2700,3290,5.5

700,3.385,0290,5.275,0 = 0,0024

- Phng trnh hi qui thu c: A = 0,071C + 0,0024C = 0,3465 ppm

- Hm lng Cu trong mu :o 500,00.CC 0,866ppm200,00

.

b) Xy dng th thm chun vi s hiu chnh mu trng:V hp th ph thuc tuyn tnh vo nng nn s dng phng php th theo bng s liu:


57/155



Mu 0 0,045 0,001 0 0

Chun 1 0,200 0,083 0,001 0,016 0,040Chun 2 0,500 0,118 0,001 0,059 0,250Chun 3 1,000 0,192 0,001 0,191 1,000

n = 4 C = 1,700 A = 0,434 AC = 0,266 C2 = 1,290

(Ngoi suy th, im ct trc honh l-Cx = 0,5C o;) hoc xy dng ng

hi qui tuyn tnh A = aC + b v cho A = 0 thu c CX = - C.

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 2700,1290,1.4700,1.434,0266,0.4 = 0,1435

b =

n

i

n

i ii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 2700,1290,1.4

7,1.268,0290,1.438,0 = 0,0475

- Phng trnh hi qui thu c: A = 0,1435C + 0,0475

y = 0.1435x + 0.0475R 2 = 0.9957

0

0.05

0.1

0.15

0.2

0.25

-1 -0.5 0 0.5 1 1.5

Hnh 2.7: ng thm chun xc nh hm lng ng Kt qu thu c A = 0,1435C + 0,0475- C = C X = 0,331.


58/155


Vy nng ch trong mu ban u l: Co = 50,00.0,331200,00

= 0,0828 ppm.

Bi 9: Quigley v Vernon bo co kt qu xc nh lng vt kim loi trongnc bin bng quang ph hp th nguyn t dng l graphit s dng phng php thm chun. Lng vt cc kim loi u tin c tch t cc phc ca chng, lng mui ln c ng kt ta vi Fe3+. Theo phng php ny 5,0 ml dung dch Fe3+ 2000 ppm c thm vo 1,00 lt nc bin.Dng NH3 iu chnh pH dung dch n 9,0 v kt ta Fe(OH)3 c qua

m. Sau khi tch, ra kt ta, Fe(OH)3 v lng cc kim loi ng kt tac ho tan trong 2,0 ml HNO3 c v pha long thnh 50,00 ml. phntch Mn2+, 1,00 ml mu thu c c pha long thnh 100,0 ml v bmmu vo l graphit v phn tch :

Mu hp th

2,5 l mu + 2,5l dung dch 0 ppb Mn2+ chun 0.223

2,5 l mu + 2,5l dung dch 2,5 ppb Mn2+

chun 0.2942,5 l mu + 2,5l dung dch 5,0 ppb Mn2+ chun 0.361

Hy tnh kt qu theo part per billion (ppb) ca Mn2+ trong mu nc bin? Gii: - Nng Mn2+ trong 100,0 ml l CX (ppb)

- Nng Mn2+

trong mu o 1 l 0,5CX - Nng Mn2+ trong mu o 2 l 0,5CX + 1,25 (ppb)- Nng Mn2+ trong mu o 3 l 0,5CX + 2,50 (ppb)


59/155



Chun 1 0 0,223 0 0

Chun 2 2,5/2 0,294 0,368 1,563Chun 3 5,0/2 0,361 0,903 6,250

n = 3 3,75 0,878 1,270 7,813

a =

n

i

n

iii

n

i

n

iii

n

iii

C C n

C AC An

1

2

1

2

1 11 = 275,3813,7.375,3.878,0270,1.3 = 0,0552

b =

n

i

n

iii

n

i

n

iiii

n

i

n

iii

C C n

C C AC A

1

2

1

2

1 11 1

2 .= 275,3813,7.3

75,3.270,1813,7.878,0 = 0,2237

y = 0.0552x + 0.2237R2 = 0.9997

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

-8 -6 -4 -2 0 2 4 6

Hnh 2.8: ng thm chun xc nh hm lng mangan Da trn th hay phng trnh ng thm chun A = 0,0552C + 0,2237ta tnh c: 0,5CX = -C = 4,0525 ppb CX = 8,105 ppb.

Hm lng Mn2+ trong nc bin l :100,00.8,105 50,00.1, 00 1000

= 40,525 ppb.


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Dng 2: Xc nh hp th ca nguyn t kim loi. Cch gii:

p dng nh lut Bouguer Lambert Beer: A = lg I

I 0 = C l ..

Trong : A: mt quang. I0, I: cng nh sng trc v sau khi b nguyn t t do hp

th. : h s hp th mol phn t, ph thuc bc sng .l: dy lp hi nguyn t.

Bi 10: Dung dch 12 ppm ca ch cho tn hiu hp th nguyn t l 8%, nhy hp th nguyn t ca ch l bao nhiu? (Bit nhy hp th nguynt llng ti thiu ca cht hp th hp th l 1%).

Gii:

Ta c : A = I I 0

lg = thuhap I I I

0

0

lg = K.C

K =C 1 lg

8100100 =

121 lg

8100100 = 3,0177.10 -3

Mt khc: khi hp th l 1% th gi tr ca nng l nhy hp thnguyn t :

o3

I1 1 100C lg lg

K I 100 13,0177.10

1,45 ppm.

Bi 11: nhy hp th nguyn t ca bc l 0,050 ppm trong mt iukin xc nh. hp th thu c s l bao nhiu i vi dung dch cha0,70 ppm bc?

Gii: oI 100A lg KC K.0,050 lgI 100 1

K = 8,7296.10 -2


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oIA lg KC 0,0611I

oht

I 100A lg lg 0,0611

I 100 I

Iht = 13,13%.

Bi 12: xc nh hm lng ch trong nc tiu nh php o ph hp thnguyn t c th dng phng php thm chun. Tng 50,00ml nc tiuc chuyn vo mi phu di c dung tch 100ml, thm vo mt phu300 l dung dch chun cha 50,0 mg/l ch. Sau pH ca mi hn hp

c a n 2,8 bng cch thm tng git dung dch HCl. Trong mi phungi ta a vo 500l dung dch amoni pyroliinithiocacbaminat mi

chun b 4% trong metyl- n - amylxeton, trn cn thn cc pha nc v phahu c chit ch. Hm lng ca ch trong pha hu c c xc nh bng phng php hp th nguyn t, mt khc ngi ta dng n catot rngvi ng hp th 283,3 nm. Nng ch bng bao nhiu mg/l trong munc tiu ban u, nu s hp th ca dch chit ca mu khng thm ch l0,325, cn dch chit ca mu pha thm lng bit ca ch l 0,670? Gii:Gi s nng ch trong mu nc tiu l Co g/ml, khi chit vo 0,5 mlmetyl-n- amylxeton th nng trong pha hu c l : CX = 100C o.Khi thm vo dung dch 0,3 ml dung dch ch 50,0g/ml, khi chit vo 0,5ml metyl-n- amylxeton th nng trong pha hu c l:

C = 100C o + 30,0

hp th trong vng tuyn tnh nn: o

x xo

t t

A C 100C 0,325A C 0,670100C 30

= 0,02826 g/ml

Dng 3: Xc nh bc sng (m); tn s (s-1) hay (hec); s sng (cm-1); nng

lng (J.mol-1), khi bit mt trong cc s liu trn.


62/155


Cch gii:

* p dng cng thc:

hc E

v h E c

s sng = 1 (cm -1)

Trong : c: tc nh sng = 3.108 m/s : bc sng ca bc x.

: tn s ca bc x.

* Qui i n v: 1cm = 10 -2m ; 1 m = 10 -6m ; 1nm = 10 -9m ; 1

o

A = 10 -10m ;

Bi 13: Hy tnh tn s (trong giy-1) tng ng vi mi mt trong cc bc sng sau ca bc x n sc?

a, 222nm b, 530nm c, 17o

A

Gii:a, Ta c: 222nm = 222.10-9m

p dng cng thc:

c = 9

8

10.22210.3 = 1,35.10 15 giy-1

b, Ta c: 530nm = 530.10-9

m

p dng cng thc:

c = 9

8

10.53010.3 = 5,66.10 14 giy-1

c, Ta c: 17o

A = 17.10 -10m

p dng cng thc:

c = 10

8

10.1710.3 = 1,76.10 17 giy-1

Bi 14: Hy tnh tn s (trong hec) tng ng vi mi mt trong cc bcsng sau ca bc x n sc?


63/155


a, 0,030cm b, 1,3.10 -7cm c, 6,1 m

Gii:a, Ta c: 0,030cm = 3.10-4m

p dng cng thc:

c = 4

8

10.310.3 = 1.10 12 hec

b, Ta c: 1,3.10-7cm = 1,3.10 -9m

p dng cng thc:

c = 9

8

10.3,110.3 = 2,31.10 17 hec

c, Ta c: 6,1 m = 6,1.10 -6m

p dng cng thc: c

= 68

10.1,6

10.3= 4,92.10 13 hec

Bi 15: Hy tnh s sng (trong cm -1) cc bc sng v xc nh vng ph tng ng? a, 1,3.10 -7cm b, 5,6 m c, 530nm

Gii:

p dng cng thc:c

1 v da vo s phn b cc dng bc x theo

cc bc sng v s sng: a, Ta c: 1,3.10-7cm

= 710.3,1

1 = 7,69.10 6 cm -1 vng Rnghen

b, Ta c: 5,6 m = 5,6.10 -4 cm

= 410.6,51 = 1,79.10 5 cm -1 vng hng ngoi gn

c, Ta c: 530nm = 5,30.10-5cm

= 510.3,51 = 1,89.10 4 cm -1 vng nhn thy


64/155


Bi 16: Hy tnh di sng (trong cm v nm) tng ng vi cc tn s sauca bc x in t? a, 1,97.10 9 hec

b, 4,86.10 15 hec

Gii:

p dng cng thc:

c

c

a, = 98

10.97,110.3 = 1,52.10 -1 m = 15,2cm = 1,52.10 8 nm

b, = 158

10.86,410.3

= 6,17.10-6

cm = 61,7 nm

Bi 17: Hy tnh s sng (trong cm-1) trong cc trng hp sau v xc nhvng ph ca n?a, 7,32.10 19 hec

b, 2,31.10 17 hec

Gii:

p dng cng thc:c v da vo s phn b cc dng bc x theo cc

bc sng v s sng:

a, = 819

10.310.32,7 = 2,44.10 9 cm -1 vng

b, = 817

10.310.31,2 = 7,7.10 7 cm -1 vng Rnghen

Bi 18: B sung nhng gi tr cn thiu trong bng sau : Bc sng (nm) Tn s s-1 S sng (cm-1) Nng lng (J.mol-1)

1,01.10 -8

1,33.10 15

3215

7,20.10 -19

Gii:


65/155


Bc sng (m) Tn s s-1 S sng (cm-1) Nng lng (J.mol-1)4,50.10 -9 6,67.10 16 2,22.10 6 4,42.10 -17

2,26.10 -7 1,33.10 15 4,42.10 4 8,81.10 -19

3,11.10 -6 9,65.10 13 3215 6,39.10 -20

2,76.10 -7 1,09.10 15 3,62.10 4 7,20.10 -19

Bi 19: B sung nhng gi tr cn thiu trong bng sau : Bc sng (nm) Tn s (hec) S sng (cm-1) Nng lng (J.mol-1)

1,97.10 -24

421,18.10 10

3,02.10 6

Gii:Bc sng (nm) Tn s (hec) S sng (cm-1) Nng lng (J.mol-1)

1,01.10 8 2,97.10 9 99.10 -3 1,97.10 -24

2,38.10 5 1,26.10 12 42 8,35.10 -22

2,55.10 7 1,18.10 10 39,2.10 -2 7,82.10 -14

3,02.10 6 0,99.10 11 33.10 -1 6,56.10 -23

Kt lun chng II:Chng Phng php phn tch Quang ph hpth nguyn t : xy dng h thng cu hi t lun v h thng bi tpc phn loi theo cc dng bi tp v xut cch gii: Xc nh nng ca kim loi trong mu, khi bit hm lng kim loi trong dung dch chunv hp th. Xc nh hp th ca nguyn t kim loi. Xc nh bcsng (m); tn s (s-1) hay (hec); s sng (cm -1); nng lng (J.mol-1).


66/155


CHNG III: PHNG PHP PHN TCH QUANG PH PHT X NGUYN T

III.1 C S L THUYT: III.1.1 c im chung ca phng php:[7] - Phng php phn tch quang ph pht x nguyn t (AES- AtomicEmission Spectroscopy) c s dng nh lng hu ht cc kim loi vnhiu phi kim nh: P, Si, As, C, B vi nhy nh hn hoc bng 0,001%. - Phng php AES c u im l mt ln phn tch c th xc nh ng

thi c nhiu nguyn t v phn tch c cc i tng phn tch xada vo nh sng pht x t cc i tng .

III.1.2 S to thnh ph AES:[7] - Phng php AES da vo vic o bc sng, cng v cc c trngkhc ca cc bc x in t do cc nguyn t hay ion trng thi hi pht

ra. Vic pht ra cc bc x in t trong min nh sng quang hc ca ccnguyn t l do s thay i trng thi nng lng ca nguyn t. - Theo thuyt cu to nguyn t, cc nguyn t ch c th c mt s mcnng lng gin on E0, E1, E 2 m khng c cc trng thi nng lng trung gian.

Trong iu kin bnh thng, cc nguyn t trng thi nng lng c

bn. Khi ta cp nng lng cho nguyn t mt cch no , cc nguyn t cth chuyn sang mc nng lng cao hn. Khi nguyn t b kch thch. N tn ti trng thi kch thch khong 10-8 giy. Sau n lun lun cxu hng tr v trng thi c bn ban u bn vng. Ngha l gii phngnng lng m chng hp th c trong qu trnh trn di dng ca cc bc x quang hc. Bc x ny chnh l ph pht x nguyn t, n c tn sc tnh theo cng thc:


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E = (E n E0) = h hay E =

hc

Trong : En, E 0 l nng lng ca nguyn t trng thi c bn (E0) v

trng thi kch thch(En)h l hng s Planck (6,626.10-27erk.s) hay h = 4,1.10 -15eV.sc: tc nh sng (3.108m/s); : tn s ca bc x .

: bc sng ca bc x .

III.1.3 Bn cht ca phng php quang ph pht x nguyn t:[11]- Trong ngn la nhit cao (hay l graphit, tia la in, in h quang ...),t cc phn t cht nghin cu to cc nguyn t t do, cc nguyn t t dony do hp th nng lng bn ngoi chuyn t trng thi c bn ln ccmc nng lng kch thch, lc tr v km theo s pht ph pht x nguynt c trng cho nguyn t cn xc nh. - Ph pht x nguyn t c dng phn tch nh tnh, bn nh lng vnh lng.

Hnh 3.1 S ph k pht x nguyn t dng ngn la. III.1.4 S kch thch, s pht x v cng vch pht x nguyn

t:[11]


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* nhn c nhy cao, trong php o ph pht x nguyn t cn cungcp ngn la c nhit cao. Phng trnh Boltzman:

Nm* = N m

)(T B

Gu . e -Eu/KT

Trong : N m

* l s nguyn t b kch thch trong ngn la. N m l s cc nguyn t t do trong ngn la. Gu l trng lng thng k (tch ra) ca trng thi nguyn t b kch

thch.

B(T) l hm phn b ca cc nguyn t theo tt c cc trng thi. Eu l nng lng ca trng thi kch thch. K l hng s Botlzman. T l nhit tuyt i.

T phng trnh Boltzman: khi tng nhit (T) th s cc nguyn t b kchthch N* tng, nhy tng.

Cng sut ca ph pht x nguyn t (PT) c tnh theo cng thc:

PT = (h 0 ).(N*, A T)

trong : h 0 l nng lng E ca mi ln chuyn.

h: hng s Planck (6,624.10-34J.s)

0 : tn s (giy-1) ca pic quan st c.

AT: h s Anhxtanh (s chuyn trong mt giy m nguyn t b

kch thch phi c). * Cng vch ph pht x:

+ Cng vch quang ph c c trng bng chi sng cavch quang ph v cng vch quang ph, k hiu l I.

+ Cng I ca vch pht x nguyn t ph thuc vo: - iu kin kch thch ph.

- Trng thi vt l ca mu nghin cu. - Ph thuc vo nng ca nguyn t nghin cu trong mu.


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+ S ph thuc ca cng vch pht x nguyn t vi nng c biu din bng phng trnh Lomakin: I = a.c b Trong : a, b: hng s ph thuc iu kin kch thch v trng thi vt

l ca mu nghin cu. Ta c: lgI = lga + blgclgI l hm tuyn tnh ca lgc: lgI = f(lgc). y l biu thc c s cho phng php phn tch ph nh lng.

III.2 CU HI:

Cu 1: Nu nguyn tc ca php o AES? HDTL:

+ Mu phn tch cn c chuyn thnh hi (kh) ca nguyn t t do trongmi trng kch thch. l qu trnh ho hi v nguyn t ho mu. Sau dng ngun nng lng ph hp kch thch m hi nguyn t chng pht x. l qu trnh kch thch ph mu.

+ Thu, phn li v ghi ton b ph pht x ca vt mu nh my quang ph.Trc y, ph c ghi ln knh nh hay phim nh. Chnh my quang phs lm nhim v ny. Nhng nhng trang b hin i ngy nay c th thu vghi trc tip cc tn hiu cng pht x ca mt vch ph di dng c pic trn bng giy hay ch ra cc s o cng vch ph trn my in hocghi li vo a t ca my tnh.

+ nh gi ph ghi v mt nh tnh v nh lng theo nhng yu cu t ra. y l cng vic cui cng ca php o.

Cu 2: Hy nu nhng u im v nhc im ca php o AES? HDTL:

* u im: + Php o AES c nhy cao, c dng kim tra, nh gi tinh khit ca nhiu ho cht v nguyn liu tinh khit cao, phn tch


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lng vt cc kim loi nng c hi trong i tng thc phm, ncgii kht, mi trng. + C th phn tch ng thi nhiu nguyn t trong mt mu, m

khng cn tch ring chng ra khi nhau. Mt khc, li khng tnnhiu thi gian, c bit l ng dng phn tch nh tnh v bn nhlng. + L php o chnh xc tng i cao. + L phng php phn tch tiu tn t mu, ch cn t 1 n vi chcmiligam mu l .

+ C th kim tra c ng nht v thnh phn ca vt mu nhng v tr khc nhau. V th cng c ng dng kim tra ng nht ca b mt vt liu. + Ph ca mu nghin cu thng c ghi li trn phim nh, knhnh hay trn bng giy. N l ti liu lu tr v khi cn thit c thnh gi hay xem xt m khng cn phi c mu phn tch.

* Nhc im:+ AE S ch cho bit thnh phn nguyn t ca mu phn tch, mkhng ch ra c trng thi lin kt ca n trong mu. + chnh xc ca php phn tch ph thuc vo nng chnh xcca thnh phn ca dy mu u v cc kt qu nh lng u phida theo cc ng chun ca dy mu u c ch to sn

trc.

Cu 3: Nu nhng yu cu ca ngun kch thch ph AES ? HDTL:

Mt ngun sng mun dng c lm ngun kch thch ph pht xn cn cn phi tho mn mt s yu cu sau:


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+ Ngun sng phi m bo cho php phn tch c nhy cao v cng ca vch ph phi nhy vi s bin thin nng ca nguyn t phn tch,nhng li khng nhy vi s dao ng ca iu kin lm vic.

+ Ngun nng lng phi n nh v bn vng theo thi gian, m bocho phng php c lp li v n nh cao. Ngha l cc thng s cangun sng chn nht thit phi duy tr v lp li c. + Ngun nng lng phi khng a thm ph ph vo lm ln vi ph camu nghin cu. Nu khng s lm kh khn thm cng vic nh gi nhtnh v nh lng v c khi lm sai lc c kt qu phn tch.

+ Ngun kch thch phi c s cu to khng qu phc tp, nhng li ckh nng thay i c nhiu thng s c th chn c nhng iu kin ph hp theo tng i tng phn tch. + Ngun nng lng kch thch phi lm tiu hao t mu phn tch.

Cu 4: Hy nu hai qu trnh chnh ca qu trnh kch thch ph trong ngn

la trong ph AES? HDTL:

Trong ph AES, nu dng ngn la lm ngun kch thch th mu phn tch phi chun b dng dung dch. Hai qu trnh chnh ca qu trnhkch thch ph trong ngn la: * Nu Eh < E nt th mu phn tch s ho hi, nguyn t ho to ra cc nguyn

t t do, chng b kch thch v pht x c ph pht x. Me nXm (r) Me nXm (k) mMe (k) + nX (k)

Me (k) + E Me (k) * Me 0 + n(h ) Nng lng kch thch Chm tia pht x

C ch 1 ny cho nhy v n nh cao ca s pht x. Cc hp chtmui halogen (tr F), axetat, mt s nitrat, sunfat ca kim loi thngtheo c ch ny.


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* Nu Eh > E nt th mu s b nguyn t ho trc, sau ho hi thnhnguyn t, ri b kch ph sinh ra ph pht x.

Me nXm (r) Me n (r) + X m (r) mMe (k) + nX (k)

Me (k) + E Me (k) * Me 0 + n(h ) Nng lng kch thch Chm tia pht x

C ch 2 ny cho nhy v n nh km hn c ch 1. Cc hp chtmui SiO23 , PO 34 , F -, mt s nitrat, sunfat ca kim loi thng theo c

ch ny.

Cu 5: Hy cho bit c im ca h quang in c dng lm ngunnng lng kch thch cho phng php ph AES?

HDTL:

- H quang l ngun kch thch c nng lng trung bnh v cng l ngunkch thch vn nng. N c kh nng kch thch c c mu dn in v

khng dn in. Nhit ca h quang ph thuc rt nhiu vo bn cht cnguyn liu lm in cc. Cng dng in trong mch h quang l yut quyt nh nhit ca h quang. - H quang l ngun kch thch cho nhy tng i cao. - H quang l ngun kch thch c s mch tng i n gin, nhng lic nhiu thng s c th iu chnh c chn c mt nhit tng

i ph hp cho php phn tch mi cht. - H quang l s phng in gia hai in cc c th thp (di 260V) vdng cao (t 8- 20A). Th v dng lin h vi nhau qua biu thc:

U = A +m I

B v Im =21 R R

U

A, B: cc hng s. R 1 l in tr plasma h quang.


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Cng thc ny gii thch ti sao h quang km bn vng. V th phi avo mch h quang mt bin tr ph R 2 tht ln (R 2 >> R 1) hn ch sthay i cng dng in trong mch v gi cho h quang n nh.

Cu 6: Yu t no quyt nh nhit ca plasma h quang? HDTL:

- Yu t th nht quyt nh nhit ca h quang l cng dng introng mch h quang. Trong mt khong nht nh ca nhit , khi tngcng ca dng in th nhit ca h quang cng tng theo:

T = k.I k: h s t l T: nhit ca h quang (oC)

- Yu t th hai quyt nh nhit ca h quang l bn cht ca nguynliu lm in cc. - Yu t th ba l th ion ho ca cc cht c trong mu phn tch. Nguynt no c th ion ho cng nh th lm gim nhit ca h quang nhiu.

- Trong plasma ca h quang thng lun xy ra cc phn ng ho hc vcc qu trnh vt l. l qu trnh phn li, ion ho, t hp phn t v skch thch ph. Nhng qu trnh ny thng l cc qu trnh nhit ng hc.V m hi trong plasma cng c mt nhit nht nh ng vi nhngthng s xc nh ca my pht h quang c chn. Nhit m bo cho plasma c mt p lc in t xc nh duy tr s kch thch ph.

Cu 7: Hy cho bit nguyn tc v cch chn ngun kch thch ph? HDTL:

Cc iu kin kch thch ph l nhng yu t c nh hng trc tipn cng ca vch ph v nhy ca phng php phn tch. chnc nhng iu kin, ngun kch thch ph ph hp cho mt i tng phn tch no , cn da vo mt s nguyn tc sau:


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+ Phi xut pht t i tng phn tch thuc dng mu no (bt, rn, haydung dch), cng nh tnh cht c l ca tng loi mu. V d

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Luan Van Thac Sy Hoa Hoc