do khong dam bao do 1

VN PHNG CNG NHN CHT LNG Bureau of Accreditation (BoA) H THNG CNG NHN PHNG TH NGHIM VIT NAM Vietnam Laboratory of Accreditation Scheme (VILAS)

HNG DN CC V D C LNG KHNG M BO O TRONG PHN TCH HO HC NH LNGM s: AGL 19 Ln ban hnh: 1.04 Ngy ban hnh:

Bin son H tn K tnHong Thanh Dng

Xem xt V Xun Thy

Ph duytTS.H Tt Thng

THEO DI SA I TI LIU TT V tr Ni dung sa i Ngy sa i

Hng dn cc v d c lng KB o trong phn tch ho hc nh lng

Li m u: Hng dn ny c xy dng da trn nguyn tc ca cc ti liu sau: 1. EURACHEM: Quantifying Uncertainty in Analytical Measurement, Laboratory of the Government chemist, London, UK, 1995. ISBN 0-948926-08-02. 2. Guide to the Expression of Uncertainty in Measurement, ISO, Geneva, Switzerland 1993. ISBN 92-67-10188-9. 3. Protocol for uncertainty evaluation from validation data, Valid Analytical Measurement, report number LGC/VAM/1998/088, January 2000. 4. ISO 5725:86: Accuracy (trueness and precision) of measurement methods and results Part 1-6. 5. SAC-SINGLAS Technical Guide 2. A guide on measurement uncertainty in chemical analysis, First edition, April 2000

GII THIU CHUNGHng dn ny a ra cc v d c th v c lng khng m bo cho cc ch tiu th nghim. Trong hng dn s a ra cc v d c th cho cc ch tiu phn tch ho hc nh lng. T v d n gin n v d c lng KB cho cc ch tiu phn tch phc tp, nhiu bc trong qu trnh th nghim. Mc ch ca hng dn l h tr, cung cp cho cc phng th nghim ha hc c mt bc tranh tng qut v cc bc c lng KB cho cc ch tiu c th v c th p dng c lng KB cho cc ch tiu c th ca PTN. Ni dung hng dn gm 3 phn chnh nh sau: I. II. III. c lng khng m bo thnh phn t nhng d liu c sn c lng khng m bo cho mt s bc trong qu trnh phn tch c lng khng m bo cho mt s ch tiu th nghim c th

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I.

C LNG KB THNH PHN T NHNG D LIU C SN

1.

Qui nh k thut ca nh sn xut cung cp cho bnh nh mc 100mL cp A l 0.08mL. Tnh khng m bo chun ca th tch bnh? Qui v Phn b dng hnh ch nht - khng m bo chun l: 0.08/ 3 = 0.046 mL

2.

Qui nh k thut ca nh sn xut cho pipet 2mL l 0,01mL. Tnh khng m bo chun ca th tch cht lng chuyn qua pipet? Qui v Phn b dng hnh ch nht - khng m bo chun l: 0.01/ 3 = 0.0058 mL

3.

Chng ch hiu chun cho cn 4 s cho bit khng m bo o l 0,0004g vi mc tin cy khng t hn 95%. Tnh khng m bo chun? Mc tin cy 95% nn qui v phn b chun v KB chun bng KB m rng chia 2 0.0004/2 = 0.0002 g

4.

tinh khit ca mt hp cht ho hc c nh cung cp a ra l (99,9 0,1) %. Xc nh khng m bo chun ca tinh khit ca hp cht? Qui v Phn b dng hnh ch nht - khng m bo chun l: 0.1/ 3 = 0.058 %

5.

Mt qu cn hiu chun c chng nhn l 10.00000g 0.04mg vi mc tin cy t nht l 95%. Tnh khng m bo chun ca qu cn? Mc tin cy 95% nn qui v phn b chun v KB chun bng KB m rng chia 2 0.04/2 = 0.02 mg

6.

lch chun ca cc ln cn lp li ca qu cn 0,3g l 0,00021g. Tnh khng m bo chun? lch chun chnh l khng m bo chun nn KB chun l: 0.00021 g

7.

Chng ch hiu chun ca pipet 25mL cp A c ghi khng m bo l 0.03mL. khng m bo ny c da vo khng m bo chun nhn vi h s ph k=2, cho mt mc tin cy xp x l 95%. Tnh khng m bo chun ca th tch cht lng chuyn qua pipet? C h s ph k = 2 nn qui v phn b chun v KB chun l: 0.03/2 = 0.015mL

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8.

khng m bo ca th tch dung dch trong bnh nh mc dung mi hu c ho tan vo bnh nh mc 100ml n vch nh mc. Tnh khng m chun ca th tch cht lng trong bnh nh mc? Cho cc d liu sau: - Kt qu ca 10 ln thc hin (cn v) vo bnh nh mc cp A 100mL c lch chun l 0.01732mL - Qui nh k thut ca nh sn xut cho bnh l 0.08mL. H s n ca th tch ca dung mi hu c ho tan l 1x10-3 0C-1 - S khc nhau gia nhit phng th nghim v nhit hiu chun bnh nh mc c c lng l 3 0C vi mc tin cy l 95%. Tnh

khng m bo chun do s khc nhau gia nhng ln ong v cn chnh l lch chun: 0.01732mL T qui nh k thut ca nh sn xut qui v phn b hnh ch nht v tnh ra khng m bo chun ca th tch bnh l 0.08/ 3 = 0.046mL khng m bo do s khc nhau gia nhit phng th nghim v nhit hiu chun bnh c c lng l Vx3x1x10 -3mL vi V l th tch ca bnh, 3 l s bin thin nhit c th v 1x10 -3 l h s n th tch ca dung dch cht hu c. V th tch gin n ca cht lng ln hn nhiu th tch gin n ca bnh nn ch quan tm n th tch gin n ca dung dch. S khc nhau v th tch do nh hng nhit c tnh (da vo mc tin cy l 95%) l:

100 x 3 x 1 x 10-3 = 0.3mL V c chuyn thnh lch chun bng cc chia cho h s ph k=2 khng m bo do nh hng ca nhit l 0.3/2=0.15mL Tng hp 3 thnh phn khng m bo trn s c khng m bo ca th tch dung dch trong bnh nh mc l:

uv = 0.01732 + 0.0462 + 0.1502 = 0.16 mL9. khng m bo ca th tch cht lng chuyn qua pipet Mt pipet 2mL cp A c s dng pha dung mi hu c. Tnh khng m bo chun th tch cht lng c chuyn qua pipet? Cho cc d liu sau: - Cc th tch t 10 ln lp li vic chuyn cht lng t pipet 2mL cp A c lch chun l 0.0016mL - Qui nh k thut ca nh sn xut pipet l 0.01mL. H s n th tch ca dung mi hu c l 1x10-3 0C-1 - S khc nhau gia nhit phng th nghim v nhit hiu chun bnh nh mc c c lng l 30C vi mc tin cy l 95%.

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Tnh

khng m bo chun do vic ong lp li dung mi qua pipet chnh l lch chun: 0.0016mL khng m bo do hiu chun pipt t nh sn xut c qui theo phn b dng hnh ch nht. Vy khng m bo do hiu chun pipet l: 0.01/ 3 = 0.0058mL

khng m bo do s khc nhau gia nhit phng th nghim v nhit hiu chun pipet c lng l V x 3 x 1 x 10 -3 mL vi V l th tch ca pipet, 3 l s thay i nhit v 1x10-3 l h s gin n th tch ca dung mi hu c. V th tch gin n ca cht lng ln hn th tch gin n ca pipet nn ch cn tnh gin n ca dung mi. S bin thin v th tch do nh hng ca nhit c tnh l:

2 x 3 x 1 x 10-3 = 0.006 mL Sau chuyn thnh lch chun bng cch chia cho 2 nn khng m bo do nh hng ca nhit l: 0.006/2 = 0.0030 mL Tng hp 3 thnh phn khng m bo trn s cho khng m bo ca th tch dung mi chuyn qua pipet l:

uv = 0.0016 + 0.0058 + 0.0030 = 0.00672 2 2

mL

10.

khng m bo cn Mt phng php yu cu cn mt chun ni b 100mg trn cn 4 s. Tnh khng m bo chun ca vic cn? - Chng ch hiu chun ca cn c nu khng m bo o l 0.0004g vi mc tin cy khng di 95% - Cn lp li qu cn 100mg trn cn 4 s c lch chun l 0.000041g Tnh khng m bo t vic hiu chun cn c tnh ton t chng ch hiu chun. khng m bo c trch dn l 0.0004g vi mc tin cy l 95%. Bin i thnh lch chun bng cch chia khng m bo cho 2. khng m bo hiu chun l: 0.0004/2= 0.0002g = 0.200mg

khng m bo do s bin thin ca cc ln c cn l lch chun ca cc php cn lp li: 0.000041g = 0.041mg Tng hp khng m bo trn s cho khng m bo ca khi lng vt liu l:

u w = 0 .200 2 + 0 .041 2 = 0.208 mgAGL 18 Ln ban hnh : 1.04 Trang : 4


11.

khng m bo ca nng mt dung dch Dung dch chun ni b c chun b bng ho tan khong 100mg vt liu (cn bng cn 4 s) trong mt dung mi hu c ho tan v y vo bnh nh mc ti vch 100mL . Tnh nng ca dung dch theo mg/L. Tnh khng m bo chun v khng m bo m rng ca nng dung dch ? D liu: - 100.5mg vt liu c cn. khng m bo chun lin quan ti vic cn ny c tnh v d 10 trn. - tinh khit ca vt liu c trch dn t nh sn xut l (99.9 0.1)%. - khng m bo chun ca th tch cht lng trong bnh nh mc 100mL c tnh trong v d 9 trn. Tnh Cc ngun khng m bo gp phn vo ton b khng m bo ca nng dung dch l: Vic cn vt liu chun b dung dch; tinh khit ca vt liu; Th tch cui cng ca dung dch.

Cc khng m bo lin quan l: Cn vt liu (uw): 0.208mg-

tinh khit ca vt liu (up): 0.1/ 3 = 0.00058% Th tch cui cng: 0.16 mL W P 1000 V

Nng ca dung dch l C (mg/L) c tnh theo cng thc sau: C (mg / L) = Trong : W: P: V: khi lng ca vt liu s dng (mg) tinh sch ca vt liu s dng (% tinh sch chia 100) Th tch cui cng ca dung dch (mL)

Nng ca dung dch l:

100 ,5 99 ,9 1000 = 1004 , 0 mg / L 100

V cch tnh nng ch gm php nhn v chia, cc thnh phn khng m bo c tng hp nh cc lch chun tng i (relative standard deviations) T chng ta c th tnh khng m bo tng hp ca nng dung dch chun ni b l: u uc = w W C 2

u p uv + + P V 2

2

2

2

0,208 0,00058 0,16 u c = 1004 + + = 2,69 mg/L 100,5 0,999 100 2

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Gi tr ca khng m bo chun ca nng dung dch chun ni b l 2.69mg/L khng m bo m rng l 5.38mg/L c tnh ton bng cch s dng h s ph k=2 Nng ca dung dch c th c cng b trong bo co l: 1004 5 mg/L v ghi ch l khng m bo c tnh da vo khng m bo chun nhn vi h s ph k=2 v cho mt mc tin cy l xp x 95%.

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II.

C LNG KB CHO MT S BC TRONG QU TRNH PHN TCH

1. 1.1

CN Mc ch

1.1.1 Cn 500mg Cu bng cc phng php cn khc nhau 1.1.2 Cc bo co khi lng Wt bnh ng + Cu, g Wt bnh ng, g Wt Cu, g Ch thch: Wt l Khi lng 1.2 Xc nh ngun khng m bo ca php phn tch 1.2.1 th nguyn nhn v kt qu/nh hng PCu tinh khit Khi lng WCu lp li Khi lng m (b) Khi lng m (tng) Tuyn tnh nhy Hiu chun 1.3 lp li Khi lng m Tuyn tnh nhy Hiu chun 32,5829 32,0822 0,5007

nh lng cc khng m bo thnh phn

1.3.1 S tinh tinh khit ca kim loi ng Cc nh cung cp a ra thng bo v tinh khit ca Cu trong chng ch phn tch khi lng Cu l 99,99 0,01% m khng cp n tin cy ca n. V khng ai a ra gii hn tin cy ca tinh sch ny, chng ti a ra 1 php tnh v khng m bo theo phn b dng ch nht nn khng m bo chun u (PCu) l

0 , 0001 / 3 = 0 , 0000581.3.2 Qui trnh cn 1.3.2.1. Hiu chun tuyn tnh Hiu chun bn ngoi ca cn c s dng tuyn b rng s khc nhau gia trng lng tht trn a cn v s (trng lng) c trn thc chia trong khong 0,05mg vi tin cy 95%.AGL 18 Ln ban hnh : 1.04 Trang : 7


Vi s phn b bnh thng, tin cy 95% cho mt h s l 1,96. Bi vy khng m bo lin quan din t nh lch chun l 0,05/2 = 0,025mg Ch : khng m bo ca thnh phn ny tng ln gp i bi lin quan n 2 ln cn mt ln l trc khi thm kim loi Cu v 1 ln l sau khi thm Cu 1.3.2.2. lp li Lp li 10 ln php o c b v trng lng tng s c 1 lch chun ca cc sai khc gia cc ln cn l 0,06mg vi khong trng lng trong khong t 20mg n 100mg Ch : Chng ta tnh lp li ch duy nht 1 ln bi v n c tnh v s khc nhau ca trng lng a n mt lch chun ca cc ln cn khc nhau 1.3.2.3. nhy nhy ca cn c th khng c quan tm v nhng trng lng khc nhau c o trn cng 1 cn phm vi rt hp 1.3.2.4. Tnh khng m bo chun tng hp trong quy trnh cn u (mCu) = 1.4

2(0,025) + 0,062 = 0,07mg2

Tng kt cc gi tr ca khng m bo M t tinh khit ca kim loi Cu, P Wt ca kim loi Cu (mg) Gi tr x 0,9999 500,7 u(x) 0,000058 0,07 u (x)/x 0,000058 0,00014

1.5

Tnh khng m bo tng hp v khng m bo m rng Bi v khng m bo tng hp u (WCu) / WCu

=

0 , 000058

2

+ 0 , 00014

2

= 0,00015 vy u (wCu) = 0,00015 . 500,7 = 0,07

khng m bo m rng vi h s ph k = 2 l: U (WCu) = 0.07 x 2 = 0.14 khi lng ng 500.7mg bo co khng m bo l: 500,7 0,14 mg vi h s ph k=2 [mc tin cy xp x 95%] Ghi ch phn b khng m bo ca tinh khit ca ng l rt nh c th khng tnh n.

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2. 2.1

S CHUN B TH TCH Mc ch nh mc 500mL

2.1.1 chun b mt axt ng ho dung dch chun nitrat Cu t 500.7mg Cu pha trong bnh 2.1.2 Cc bc tin hnh l: a) b) Cn 500mg Cu tinh khit trong cc cn Dng 5mL axit nitrit c ho tan Cu

c) Khi phn ng ngng v Cu ho tan hon ton trong dung dch axit. Chuyn dung dch ny vo bnh nh mc 500mL. Thm nc ct n vch nh mc. 2.2 Xc nh ngun khng m bo U (W)/W

2.2.1 S nguyn nhn v nh hng

500mL dung dch Hiu chun th tch V lp li nh hng T0

2.3

Xc nh khng m bo thnh phn

2.3.1 S khng m bo khi cn l yu t hnh thnh u tin khi cn 500,7mg 0,14mg vi h s ph l 2 2.3.2 Hiu chun th tch ca nh sn xut Tuyn b ca nh sn xut l bnh nh mc 500mL c sai s 0,15mL nhit 20 0C. Khng c 1 tuyn b v tin cy no. Bi vy chng ta cho rng c mt phn b dng tam gic v th tch tht dao ng gn tm hn l khong gii hn xa. Do vy khng m bo trong hiu chun l 0,15/ 6 = 0,06mL. 2.3.3 Lp li cc phn tch th tch Lp li 10 ln y v cn bnh nh mc 500mL cho mt khng m bo chun di dng lch chun l 0,04mL s ny c dng tnh trc tip kt qu cui cng. 2.3.4 nh hng ca nhit Theo nh sn xut, bnh nh mc c hiu chun 200C, bi th trong phng th nghim gii hn thay i trong khong 40C. khng m bo sinh ra t nh hng ny c th c tnh t vic c lng khong nhit v h s n th tch. V n

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th tch ca cht lng ln hn n th tch bnh nh mc nn n ca cht lng cn thit c coi trng, h s n ca nc l 0,00021 0C-1 Do th tch n l: 500mL x 4 0C x 0, 000210C-1 = 0,420mL Tnh khng m bo chun i vi s thay i nhit bng s dng phn b theo dng ch nht: 0,420/ 3 = 0,25mL 2.3.5 Tnh khng m bo chun tng hp cho php o th tch u (V) u (V) = 2.4

0,062 + 0,042 + 0,252 = 0,26

Tm tt cc gi tr ca nhng khng m bo trong vic chun b th tch M t Trng lng Cu, mg Th tch V, mg Gi tr X 500,7 500 U (x) 0,07 0,26 U (x)/x 0,00014 0,0005

2.5

Tnh khng m bo tng hp v m rng khng m bo tng hp ca vic chun b 500,7mg Cu trong 500ml dung dch xem nh l khng m bo ca vic cn v ong th tch l: u (Conc) / Conc = = 0,00052 Cng nh nng ca dung dch Cu l 500,7mg/500mL = 1001,4mg/L U(conc) = 0,00052 x1.001,4mg/L = 0,52mg/L khng m bo m rng ca vic chun b 500,7 mg Cu trong 500ml dung dch hay nng 1001,4 mg/L l 0,52 x 2 = 1,04mg/L vi h s ph k=2 Bi th, nng ca dung dch Cu l 1001,4 1,0mg/L vi h s ph k=2

[u (W ) / W ]2 + [u (V ) / V ]2

=

[0,075 / 500,7]2 + [0,26 / 500]2

2.6

Nhn xt vi vic chun b th tch

2.6.1 Vi nhng iu trnh by trn, phn b ca khng m bo khi cn nh hn nhiu so

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3. 3.1

TNH TRNG LNG PHN T CA DUNG DCH Mc ch chun b 1 mol hoc dung dch thng, trng lng phn t ca dung dch cn bit v khng m bo trong nh gi khi lng phn t. VD, chng ti yu cu tnh khng m bo bng cch tnh khi lng phn t KMnO4

3.2

Hi ng IUPAC v khi lng nguyn t v s d tha ng v IUPAC xy dng mt danh sch nhng nguyn t vi khi lng nguyn t ring r v khng m bo lin kt trong bi Pure Appl.chem, vol 69,pp.2471-2473 (1997). Bng danh mc y ca tt c cc nguyn t v khng m bo ca chng c th tm thy trong trang web sau: http://www.chem.qmw.ac.uk/iupac/AtWt/

3.3 Tnh khi lng phn t ca KMnO4 v khng m bo 3.3.1 Khi lng nguyn t v khng m bo c lit k (t IUPAC) i vi tng (thnh phn) nguyn t cu to ca KMnO4 Nguyn t Khi lng nguyn t AW (e) 39,0983 54,938049 15,9994 khng m bo u (e) 0,0001 0,000009 0,0003 khng m bo chun u (e)/3 0,000058 0,0000052 0,00017

K Mn O Ch :

3 c s dng t bng IUPAC sau khi tnh ton khng m bo trong nh gi do s to nn cc lin kt bng phn b dng ch nht.

3.3.2 Khi lng phn t KMnO4 l: NWKMnO4 = 39,0983 + 54,938049 + 4x15,9994 = 158,0339 g.mol-1 u (NW KMnO4) = 0,000058 + 0,0000052 + (4 x 0,00017) = 0,0007 g.mol-12 2 2

3.4

S phn b ca cc nguyn t KMnO4 l s tp hp n gin nhng s phn b ca nguyn t n. Bi vy, khng m bo tng hp c tnh l bnh phng ca tng bnh phng ca phn b ca tng nguyn t.

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Danh mc mt s nguyn t thng thng

Nguyn t H2 C N2 O2 F Na Mg Al P S Cl2 K Ca Cr Mn Fe Co Ni Cu Zn as Br Ag Cd Ti Sb I Ba Hg Pb

Khi lng nguyn t 1.00794 12.0107 14.00674 15.9994 18.9984032 22.989770 24.3050 26.981538 30.973761 32.066 35.4527 39.0983 40.078 51.9961 54.938049 55.845 58.933200 58.6934 63.546 65.39 74.92160 79.904 107.8682 112.411 118.710 121.760 126.90447 137.327 200.59 207.2

KB lin quan 0.00007 0.0008 0.00007 0.0003 0.0000005 0.000002 0.0006 0.000002 0.000002 0.006 0.0009 0.0001 0.004 0.0006 0.000009 0.002 0.000009 0.0002 0.003 0.02 0.00002 0.001 0.0002 0.008 0.007 0.001 0.00003 0.007 0.02 0.1

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4. 4.1

NG CONG HIU CHUN Quan h tuyn tnh Mt phng php hoc mt dng c phn tch thng c hiu chun bng kho st nhng xu hng, y, tm nhng gi tr ca x. Trong phn ln cc trng hp quan h ny l mi quan h tuyn tnh v d: y = a + bx vi a,b bit th v c mt ng cong hiu chun. Trong trng hp ny nng xobs bit t phn tch 1 mu m quan st c s tr li yobs t c cng thc: xobs = (yobs - a)/b Trong vi trng hp, phng php phn tch yu cu quan h tuyn tnh i t 0 VD: a = 0, trong trng hp ny quan h tuyn tnh l y=bx v xobs = yobs/b Phng php thng thng ca quan h tuyn tnh da trn tng cp hiu chun (xi, yi) c s dng phng php hiu chun bnh phng tuyn tnh nh nht (cng hoc khng cng nh hng, tc ng = 0). .

4.2

Ngun khng m bo C 4 ngun khng m bo chnh ng quan tm khi nh gi khng m bo ca xobs a. S thay i ngu nhin ca php o y (gm yi v yobs) b. nh hng ngu nhin qui cho nhng gi tr ca xi c. Hng s khng bit gia xi v yi d. Gi nh khng c gi tr tuyn tnh Trong 4 ngun trn th mt ngun c ngha nht l (a). Phng php nh gi (a) thng qua s thay i cn li S, S c th tnh bng cng thc: S2 = (yi - yc)2/(n-2) Trong : yi: l s c im hiu chun th i yc: l s c tnh t y = a+bx n: s im hiu chun v u (xobs,y)= var(x) vi var (x)=S2/b2

4.3

V d Trong v d ny c 3 chun s dng hiu chun tp trung, xi 5 50 200 Thu c, yi 125 1,197 4,754

y = a + bx thch hp vi hiu chun trn a v b c tnh nh sauAGL 18 Ln ban hnh : 1.04 Trang : 13


xi+yi+ - nx y b= xi - nx2 2

a = y - bx

Trong v d ny x 5 50 200 Tng Trung bnh Trong : xiyi+ - nxy b= xi2 - nx2 = 42525 - 3 x 852 1011275 - 3 x 85 x 2025,333 = 23,732 255 85 125 1197 4754 6076 2025,333 y 625 59850 950800 1011275 xy 25 2500 40000 42525 x2

a = y - bx = 2025,333 23,7321343 x 85 = 8,102 V th y = a + bx = 8,102 + 23,732x Trong v d ny c th tnh ton thu c yc t gi tr bit ca x v bnh phng ca hiu (y - yc)2 x 5 50 200 y 125 1197 4754 yc c tnh 126,76259 1194,7086 4754,5288 (y - yc) 3,10672 5,25036 0,27961

V th S2 = (yi - yc)2/(n-2) = (3,10672 + 5,25036 + 0,27961)/(3 - 2) = 8,63669 var(x) = S2/b 2 = 8,63669 / 23,73213432 = 0,0153346 u(xobs,y) =

var(x) =

0.0153346= 0,124

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5. 5.1

NG DNG CA GC-MS SC K KHI PH Mc ch V d di ch ra khng m bo o trong k kh khi ph GC- MS (gas chromatographic - mass spectrometric).

5.2

Th t cc bc nh gi khng m bo lin quan Bc 1: Cc yu cu k thut Ho cht s dng cho k thut GC-MS phn tch nng biphenyl tinh khit trong benzen. Cht chun s dng hiu chun l 50ug/ml dung dch chun v dung dch trng (vd 0ug/ml). Nng (ug/mL) ca biphenyl trong benzen c th c tnh ton s dng phng php hiu chun hai im (Phng php Bracketing/ng hng) Cspl = Aspl x C50/ (A50 - A0) Trong : Aspl : Vng tr li/phn ng ca GC-MS cho biphenyl mu phn tch A50 : Vng tr li/phn ng ca GC-MS cho biphenyl ca chun A0 : Vng tr li/phn ng ca GC-MS cho biphenyl mu trng C50 : Nng biphenyl dung dch chun c nng 50ug/ml biphenyl Bc 2: Xc nh ngun khng m bo chm C50 Tuyn tnh Ho tan chm Tinh khit Cn chm Cspl

A0 chm

A50 chm

Aspl chm

Bc 3: Xc nh khng m bo tng hp C50: - Dung dch chun c chun b t cht rn biphenyl, u tin cn v sau ho tan v pha long trong benzen. Vi cc ln cn khc nhau tnh c khng m bo chun l 0.000206g i vi 0,052g biphenyl

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- tinh khit ca biphenyl c nh cung cp tuyn b l ln hn 99.0% . Do s tinh khit ca nguyn liu c tnh ton l 99.5% vi khng m bo lin quan ( 100% - 99%)/2 3 = 0,289% - Cht rn byphenyl sau c ho tan v pha long thnh 1,000mL vo bnh nh mc. Chi tit k thut ca bnh nh mc 1L s dng tuyn b cp chnh xc l 1000.22 0.20mL. Trong trng hp ny phn b hnh ch nht c s dng nh mt phn ca kim tra vic kim sot cht lng (QC) ch ra rng trung tm phn b ph hp hn cc vng gn phm vi. khng m bo chun t chng ch hiu chun ca bnh gm thu tinh l 0,20/ 6 = 0,0816mL Lp li vic v cn ch ra c khng m bo chun l 0,15mL. V nh hng ca nhit khc nhau i vi khi lng benzen s dng ho tan v pha long khng c d liu ca h s n ca benzen chng ta cho rng h s n ca n l gp i nh nc ti nhit thng (m c h s n l 2,1x10-4mL/0C). T kinh nghim chng ta bit rng c lng ny c th y . Nhit phng dao ng trong khong 40C vi mc tin cy xp x 95% khng m bo chun pht sinh t s thay i nhit 1000,22x (4,0/2) x 4,2 x 10-4 = 0,840mL khng m bo chun v ho tan v pha long l (0,816 + 0,15 + 0,840 ) = 0,857mL2 2 2

Bi vy, C50 v khng m bo chun ca n c tnh ton nh sau: m Gi tr KB m P V C50 0.052 99.5% 1000.22 51.72862 0.066529 u (C50) 0.257933 0.052 0.000206 0.052206 99.5% 1000.22 51.9335 0.20492 0.04199 P 99.5% 0.289% 0.052 99.789% 1000.22 51.8789 0.15025 0.02257 0.857 0.052 99.5% 1001.077 51.6843 -0.04428 0.00196 V 1000.22

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A0 , A50 , Aspl o lp li a ra kt qu sau: Vng phn ng ca GC - MS A0 A50 2 390 0 397 0 395 1 394 0 398 2 396 2 391 1 392 0 396 1 395 0.9 394.4 0.876 2.633 0.277 0.833

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Trung bnh lch chun lch chun trung bnh

Aspl 265 260 269 266 263 268 265 262 267 265 265 2.749 0.869

Cc lch chun ca cc gi tr trung bnh bng trn c s dng trc tip nh khng m bo chun lin quan v cc gi tr trung bnh m s c s dng trong tnh ton cui cng Tuyn tnh: Hiu chun hai im p dng tuyn tnh trong phm vi tp trung xc nh. Tuy nhin cc nghin cu ch ra rng bng cch phn tch dung dch biphenyl ti cc mc tp trung khc nhau, lch ln nht t kt qu thc l 1.0ug/mL. Phn b dng hnh ch nht c p dng v khng m bo chun thay v tuyn tnh l 1.0/ 3 = 0.577. Bc 4: Tnh ton khng m bo tng Nh s tuyn tnh l kt qu cui cng th s bao gm u tin cc khng m bo chun thay v C50, A0, A50 v Aspl l phi hp/t hp bi phng php chia bng nh trnh by bng di a ra tp trung trong mu l 34.836ug/mL vi khng m bo chun 0.266ug/mL. Nh vy tng khng m bo chun l: Gi tr KB C50 A0 A50 Aspl Cspl 51.72862 0.9 394.4 265 34.8363 0.04924 0.2219 C50 51.72862 0.257933 51.98655 0.9 394.4 265 35.01 0.173703 0.030173

(0 , 222

2

+ 0 , 577

2

) = 0 , 618 ug

/ mL

A0 0.9 0.277 51.72862 1.177 394.4 264 34.861 0.0245 0.0006

A50 394.4 0.833 51.72862 0.9 395.233 265 34.7627 -0.07359 0.00542

Aspl 265 0.869 51.72862 0.9 394.4 265.869 34.9505 0.11424 0.01305

u(Cspl)AGL 18

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Tnh khng m bo m rng vi mc tin cy 95% h s ph k = 2.26 c s dng nh ch xc nh 10 ln (bc t do =9). khng m bo m rng l U (Cspl) = 0.618 x 2.26 = 1.397 ug/mL Bi vy kt qu l: 34.8 1.4 (ug/mL)* khng m bo c bo co l khng m bo m rng tnh ton s dng h s ph k=2.26 vi bc t do l 9 m a ra mc tin cy tng ng l 95%. 5.3 Cch la chn cc khng m bo chun tng hp C50 = 1000000 x mP/V V th, Cspl = Aspl x C50 / (A50 - A0) = 1000000 x mPAspl/ [V (A50 - A0)]: m Gi tr KB 0.052 p 99.5% V 1000.22 0.857 A0 0.9 0.277 A50 394.4 0.833 Aspl 265 0.869

0.000206 0.289%

m P V A0 A50 Aspl

0.052 99.5% 1000.22 0.9 394.4 265

0.052206 0.052 99.5% 1000.22 0.9 394.4 265 99.789% 1000.22 0.9 394.4 265

0.052 99.5%

0.052 99.5%

0.052 99.5% 1000.22 0.9 395.233 265

0.052 99.5% 1000.22 0.9 394.4 265.869

1001.077 1000.22 0.9 394.4 265 1.177 394.4 265

Cspl

34.8363

34.9743 0.13801

34.9375 0.10118 0.01024

34.8065 -0.0298 0.00089

34.8608 0.02454 0.0006

34.7627 -0.0736 0.00542

34.9505 0.11424 0.01305

0.04924 u(Cspl) 0.221902

0.01905

Bng ny a ra kt qu tng t nh bng kt qu trc

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6. 6.1

SAI S RING/C TRNG CA CC PHNG PHP C BIT Gii thiu ca dung dch chun b c mi lin h n gin v khng phc tp xc nh. Cc c tnh phn tch ca vt cht m gi tr thc v bn cht ca thnh phn c xc nh l phc tp bi v trong tin trnh phn tch chng ta tin hnh khng ch php phn tch cui cng nhng hu ht cc trng hp mt th tc tin phn tch m c th t bn thn n l ngun gy sai khc ln hn l s sai khc c tm thy trong bn thn php phn tch.

6.1.1 khng m bo ca c tnh cht liu ca vt cht nh trng lng v khi lng

6.1.2 Cc th tc phn tch thng da vo cc c tnh ho hc ca vt cht v d dng ca hp cht cho php phn tch ngay hay tch ra qua tnh cht cht liu c s dng tch v d tnh tan, tnh bay hi/d bay hi...Cc tnh cht s dng cho cc th tc phn tch khng l cc tng chng l cc ch s mt hn ch nh hng n chnh xc v cp chnh xc ca cc kt qu phn tch bi nguyn nhn sai s c th ca khng m bo a ra trong phng php. V d: cc kt ta khng tan thng tan trong mt chng mc no , qui trnh khng tch c th lin quan nh s hon ho v ...v.v. 6.1.3 Khng th tho lun tt c cc nguyn nhn c th m c th ch dn cc sai s trong cc qui trnh phn tch ring v nh phn tch phi bit trc cc sai s h thng v hiu chnh chng khi khng m bo phn tch c tnh n. Hn na bt c phng php phn tch v bt c s thay i no trong phng php l i tng ca cc sai s c th ca bn thn phng php . y chng ta phi cp duy nht cc ngun chung ca sai s c th m tnh cht quan trng ca sai s c th c din t bng s. 6.2 Sai s trong vic ho tan kt ta

6.2.1 Ngun c th ca sai s trong php phn tch trng lng qua s kt ta l ho tan ca cc kt ta. S cn bng ABn A + Bn M c thit lp trong s khng ho tan ca mt tnh cht c th ho tan ca ABn. l cc c im bi sn phm ho tan Ksp = [A][B] n tp trung [A] v [B] s dng trong sn phm pha long phi thng c din t nh nng phn t gam. Cc gi tr Ksp ca mt vi vt cht quan trng c a ra bng di:

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Cc sn phm c th tan ca mt vi hp cht v c khng ho tan ti nhit (C) trong ngoc ( ) Hp cht Hp cht halogen AgCl AgBr Agl Hp cht hydro Fe (OH)3 Al (OH)3 Zn (OH)2 Hp cht sunphat CaSO4 BaSO4 PbSO4 Hp cht cacbon CaCO3 MgCO3 BaCO3 6.2.2 V d 1 tnh mi quan h sai s, nguyn nhn do ho tan ca phn ca cht kt ta trong qu trnh ra bng cch gn/cht: 0.1g ca AgCl c ra vi 250ml nc ti 250C y Ksp = [Ag+][Cl-] = 1.56 x 10-10 mol2.L-2 Nu mt mol ca AgCl ho tan trong mt lt nc th: [Ag+] = [Cl -] = a T vic ho tan ca mt mol AgCl, mt mol ion Ag v mt mol ion Cl c sinh ra Thay th vo s din t cho Ksp chng ta c: Ksp = a2 = 1.56 x 10-10 mol2.L-2 v do : a = 1.25 x 10-5 mol.L-1 Nng ca AgCl g/L l a ra bi cng thc nhn trng lng gram ca AgCl 1.25 x 10 -5 mol.L-1 x 143.3g.mol-1 = 1.8 x 10-3 g/L Bi vy ch 4.5 x 10 -4 g ca AgCl s ho tan trong 250ml nc ra s dng Sai s i km l: 4.5 x 10 -4 x 100 % = 0.45% 0.1 Ch thch: Sai s ny khng nh nn khng th b qua c. Sai s ny phi c hiu chnh trong vic c lng khng m bo. 6.2.3 V d 2 tnh nguyn nhn sai s trong phn tch trng lng ca 0.1g AgCl bng ra cht kt ta vi 250ml dung dch HCl 0.01M Trong trng hp ny, [Cl-] = 0.01M hoc 10-2mol.L-1 Ksp 1.56 x 10 -10 (25) 7.7 x 10-13 (25) 1.5 x 10-16 (25) 1.1 x 10-36 (18) 1.1 x 10-15 (18) 1.6 x 10-14 (18) 2.45 x 10-5 (25) 1.08 x 10-10 (25) 1.5 x 10 -8 (18) 0.87 x 10-8 (25) 2.6 x 10 -5 (12) 7.0 x 10 -9 (16)

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Trang : 20


Khng ch vic so snh nng ion Cl m c th l nguyn nhn bi s ho tan AgCl, chng ta thu c: [Ag+] = Ksp /[Cl-] = (1.56 x 10 -10mol2.L-2) / 1 x 10-2mol.L-1 = 1.56 x 10-8 mol.L-1 Tng ng vi: 1.56 x 10 -8 mol.L-1 x 143.3g.mol-1 = 2.24 x 10 -6 g/L cho s tan ca AgCl Bi vy 5.6 x 10-5 ca AgCl tan trong 250 ml ca axit s dng Trong trng hp ny sai s l: 5.6 x 10 -5 x 100 % = 0.056% 0.1 Ch thch: Gi tr ny c th khng tnh ng trong cc phn tch chnh xc. Hn na n c th cn nhc trong thc hin, tnh v t l khng hon ton c thit lp gia dung dch ra v cht kt ta nn sai s thc t l nh hn nhiu so vi mt tnh ton. 6.3 Sai s trong chun cht kt ta im cn bng c th c ch ra bi s to thnh ca ho tan t t cht kt ta qu mc chun v trong s phc tp chun m mt cht kt ta c ho tan li. 6.3.2 V d 3 Trong chun Mohr dung dch nitrat bc vo dung dch NaCl m thm cromat kali. AgCl kt ta hon ton trong khi chun nhng rt nhanh chng nng Cl t mt gi tr hin ti, dng clo bc nu ch ra mc cn bng hoc im kt thc. Tng cc ion Cl khng c chun tng ng ti nng ion bc dng clo bc. Tng s khng chun ny c th c xc nh t sn phm K(Ag2CrO4) = [Ag+]2[CrO42-] = 2 x10-12mol3.L-3 Bng tnh nng ion trong dung dch v thay th gi tr ny vo cng thc ton. Nu chng ta cn nhc trng hp m 1mL ca 5% w/v dung dch clo kali c thm vo ho 25mL ca 0.1M dung dch Clo v 0.1 nitrat bc, nng ion Clo l: 0.05 1000 [CrO4 ] = ---------- x -------- mol.L-1 = 5.15 x 10-3 mol.L-1 194.2 502-

6.3.1 Sn phm tan c cng c s dng tnh ton trong chun cht kt ta m

Khi nng ion Clo khng c chun l tng ng vi nng ion bc c a ra l: [Ag+] = 2 x10 12 1 mol.L = 2 x 10 -5mol.L-1 3 5.15 x10

Tng clo khng c chun l:

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50 --------- x 2 x 10 -5 mol = 10-6 mol 1000 Nh tng tnh l 25 -------- x 0.1mol 1000 10 -6 x 100 --------------- %, hoc 0.04% 2.5 x 10-3

V th sai s km theo l:

C th thy t tnh ton trn trong trng hp ny sai s chun ph thuc vo nng ca cht ch th thm vo; iu ny c din t s cn nhc rng cht ch th vn chuyn vai tr ca cht lm kt ta. Nu cho qu nhiu clo vo dung dch, clo bc c nh dng trc khi t c im kt thc; mt khc nu nng clo l qu nh, thm nitrat bc thm vo dung dch vt qu v bi vy m mu clo bc s kh nhn thy hn. V th sai s c th lm nh i bi vic gi iu kin chun (nh tng cht ch th nng v dung tch ca dung dch...) gn nh ging nhau qua s tiu chun ho ca chun bng chun ca clorua natri tinh khit s dng sau khi chun trong xc nh mu. 6.4 Sai s ca phng php chit

6.4.1 Trong m hnh phn tch ho hc cc phng php chit l thng c s dng phn chia cc cht. Qui tc ca cc phng php l s phn b mt cht trong hai dung mi. Th tc thc hin bao gm vic lc dung dch, thng mt dung dch m vic lc khng ho trn c m cht c chit d ho tan hn theo s phn chia c hc ca hai gia on. 6.4.2 Trong phn tch ho hc hu c dung dch - s chit dung dch c s dng cho thi gian di; trong phn tch ho hc v c, hp cht phc tp (thnh thong mu) ca cc kim loi khc nhau c chit trong dung mi phn cc thp. S phc tp ca mu c th o c ngay lp tc bi my o mu. Hoc dung mi c th c bay hi sau khi chia, v thnh phn chit c a tr li dung dch nc v xc nh c im chia. 6.4.3 Kh nng trng hp khc trong vch v d thc hin chit li vo dung dch nc ca mt thnh phn khc (thng pH khc nhau) t dung dch/ho tan khi s chit bc u kt thc. Trong cc phn tch vt s chit l thng bao gm quang ph hc hoc trc quang s ht v pht ra nguyn t. 6.4.4 Nng ca cht trong dung dch gim i bi s chit t nng nguyn C0 ti nng C1 l: VA C1 = C0 ---------------DVB + VA Nu D l h s phn b v VA, VB l dung tch ca giai on A v B s dng trong chit. Trong hot ng chit tip theo nng ca cht c chit gim so vi nng ban u lAGL 18 Ln ban hnh : 1.04 Trang : 22


C2 = C0

VA ------------------DVB + VA

2

v sau n ln thc hin VA Cn = C0 ------------------DVB + VA n

6.4.5 T nhng mi lin quan ny c tnh theo: a. Th tch dung mi s dng c th ln; b. Nhng chit hon ton s t c bng trung bnh ca chit lp li vi th tch dung mi nh hn vi chit mt ln vi th tch ln; c. Tuy nhin i lng chit hon ton c th khng bao gi t c bi vy chng ta phi tho mn vi s chit hon ton thc nghim. d. Vai tr quan trng c vn hnh bng gi tr ca i lng phn b, D nu cha ln, vic chit hon ton c th t c vi s chit n; tuy nhin h s ny thp vic chit phi c lp li mt vi ln. 6.4.6 V d 4 Khi chit 0.20g iot t 100mL ca dung dch nc vi 50mL phn CCl4 (D=85) phn cn li sau ln chit u tin l C1 = 0.2 100 ------------------85 x 50 + 100

= 0.0046 g

v khng m bo chun gy ra bi s chit khng hon ton ln u tin l: 0.0046 x 100 = 1 ------------------- = 2.3 % (mt sai s ng k) 0.2 Sau ln chit th hai cn li l:=

C2 0.2

100 ------------------85 x 50 + 100

2 = 0.0001057 g

v gy ra khng m bo l: 2=

0.0001057 x 100 ------------------0.2

= 0.05 % (khng c tnh n)

Do khi thc hin chit CCl2 l chit hon ton iot t dung dch nc.

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7.

CHUN B CHUN HIU CHUN

Gii thiu: V d ny tho lun v vic chun b mt dung dch chun cho thit b quang ph hp ph nguyn t (AAS), t kim loi c tinh khit cao (trong v d ny l 1000mg/L Cd pha trong HNO3). D v d ny khng i din cho mt php phn tch nhng vic s dng chun hiu chun l mt phn ca hu ht cc php phn tch v php phn tch thng thng l cc php o m c lin quan ti cc chun cung cp vic lin kt chun ti h n v SI. Bc 1: Yu cu k thut Mc tiu ca bc 1 l vit ra tuyn b r rng l i lng no c o. Yu cu k thut v d ny l m t vic chun b chun hiu chun v php ton th hin mi lin quan ca i lng o v cc thng s m i lng o ph thuc. Qui trnh/th tc Chun b chun hiu chun bao gm cc bc sau: Chun b chun cadmi (Cd)Lm sch b mt kim loi

Cn kim loi

Ho tan v pha long

Kt qu Cc bc tin hnh l: B mt ca kim loi c tinh khit cao c x l vi hn hp axit loi b cc oxit kim loi. Phng php lm sch c nh cung cp kim loi a ra v cn c th thu c kim loi tinh khit nh cng b trong chng ch. Bnh nh mc (100mL) c cn theo 2 cch l khng c kim loi tinh khit v c kim loi tinh khit bn trong. Cn s dng c phn gii 0.01mg. 1mL axit nitric (HNO3) (65%m/m) v 3mL nc kh ion c thm vo bnh nh mc ho tan Cd (tng ng 100mg, c cn chnh xc). Sau bnh nh mc c y nc kh ion n vch nh mc v lc bnh trn u t nht 30 ln. Tnh ton i lng o trong v d ny l nng ca dung dch hiu chun v ph thuc vo khi lng tinh khit ca kim loi c tinh khit cao (Cd), Dung tch ca dung dch c pha long. Nng ca dung dch c th hin theo cng thc:

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C Cd =Trong : CCd: 1000: m : P V : :

1000.m.P mg / L V

nng ca chun hiu chun [mg/L] thng s chuyn i t [mL] sang [L] khi lng kim loi tinh khit [mg] tinh khit ca ca kim loi dung tch ca dung dch chun hiu chun [mL]

Bc 2: Xc nh cc ngun KB trong phn tch Mc tiu ca bc th 2 l lit k tt c cc ngun gy ra KB ca tng thng s m nh hng ti gi tr ca i lng o. tinh khit: tinh khit ca kim loi (Cd) c nh cung cp a ra trong chng ch l 99.99 0.01 %. do P l 0.9999 0.0001. Cc gi tr ph thuc vo hiu qu ca vic lm sch b mt kim loi tinh khit. Theo th tc ca nh cung cp th khng c KB trong vic nhim b mt kim loi do cc oxit kim loi v do khng c KB no c ghi trong trong chng ch. Tuy nhin c th kim tra lp li v thy rng phn b KB trong phn ny c th b qua. Khi lng (m) Giai on th hai ca vic chun b bao gm vic cn kim loi tinh khit. Khi lng ca Cd c xc nh bng vic cn c b thu c m = 0.10028g Ti liu ca nh cung cp xc nh l c 3 ngun KB trong vic cn c b: lp li; kh nng c ( phn gii s/ digital resolution) ca cn v phn b KB trong chc nng hiu chun ca cn. V chc nng hiu chun c 2 ngun KB c xc nh l nhy v tnh tuyn tnh. nhy c th c b qua v cc khi lng khc nhau c cn trn cng mt cn trong khong hp. Th tch (V) Th tch ca dung dch cha trong bnh nh mc c 3 ngun gy ra KB: - KB ca chng ch ca bnh nh mc - Phng sai ca cc ln cn v nh mc ti vch - Bnh v dung dch trong bnh trong mt iu kin v nhit khc vi nhit hiu chun bnh v dung tch. Cc nh hng khc c th hin trong s xng c di y:

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Trang : 25


V Nhit Hiu chun lp li

tinh khit (P)

c(Cd) phn gii m(b) Tuyn tnh nhy Hiu chun m Bc 3: nh lng cc thnh phn KB Trong bc 3 l nh lng cc ngun gy ra KB cho i lng o, s dng cc kt qu thc nghim hoc t cc phn tch ton hc c lng. tinh khit tinh khit ca Cd trong chng ch ca Cd l 0.9999g 0.0001g. V khng c thm mt thng tin no khc v KB nn gi thit l phn b hnh ch nht. c c KB u(P) l chia 0.0001 cho 3 u ( P) = 0.0001 3 = 0.000058 lp li lp li Hiu chun phn gii m(tng) Tuyn tnh nhy

Khi lng m KB lin quan vi khi lng ca Cd c c lng s dng d liu t chng ch hiu chun cn v khuyn co ca nh sn xut v vic c lng KB l 0.05mg. Vic c lng ny l t 3 ngun KB c nu trong bc 2. Th tch V Chun b th tch c 3 ngun nh hng chnh: hiu chun, lp li v nhit . - Hiu chun: nh sn xut cung cp s liu cho bnh nh mc l 100mL 0.1mL ti iu kin 20 0C. Gi tr KB c cung cp khng c mc tin cy hoc thng tin v phn b, do d on phn b theo hnh tam gic v KB c tnh theo cng thc: 0.1mL 6 = 0.04mL

- lp li: KB qua c lng thc nghim t vic cn v y 10 ln ti 100mL cho lch chun l 0.02mL v c s dng trc tip thnh KB chun. - Nhit : theo nh cung cp bnh nh mc th bnh c hiu chun 20 0C do nhit PTN dao ng trong khong 4 0C. KB do nh hng ny c th c tnh t khongAGL 18 Ln ban hnh : 1.04 Trang : 26


dao ng nhit v h s n ca dung tch. H s n dung tch ca dung dch ln hn ca bnh nn ch cn nhc n h s n ca dung dch. T h s n ca nc l 2.1x10-40C-1, th tch thay i l: (100 x 4 x 2.1 x 10-4) = 0.084 mL Gi thit phn b hnh ch nht nn KB chun l: 0.084mL 3 = 0.05mL

Ba phn b KB trn c tng hp thnh KB chun u(V): u (V ) = 0.04 2 + 0.02 2 + 0.05 2 = 0.07mL Bc 4: Tnh ton KB tng hp Cng thc tnh CCd

C Cd =

1000.m.P mg / L V

S dng cc gi tr thu c th hin trong bng trn tnh c nng ca chun hiu chun CCd nh sau: C Cd = 1000 100.28 0.9999 = 1002.7mg / L 100.00

Ngun KB, gi tr v gi tr KB lin quan v KB tng i c th hin qua bng sau: M t tinh khit ca kim loi P Khi lng ca kim loi m Dung tch ca bnh nh mc V (mL) Gi tr x 0.9999 100.28 100.00 u(x) 0.000058 0.05 mg 0.07 mL u(x)/x 0.000058 0.0005 0.0007

Da vo cng thc tnh KB tng hp vi trng hp cng thc tnh i lng o l php nhn v chia ta thu c KB tng hp: u c (Cd ) u (V ) u( P) u (m) 2 2 2 = + + = 0.000058 + 0.0005 + 0.0007 = 0.0009 C Cd P m V u(cCd) = CCd x 0.0009 = 1002.7mg/L x 0.0009 = 0.9 mg/L KB m rng U(cCd) c tnh bng cch nhn KB tng hp vi h s ph k = 2: U(cCd) = 2 x 0.9mg/L = 1.8 mg/L Qua phn tnh KB tng hp chng ta nhn thy KB chun do nh hng ca vic chun b dung tch bnh nh mc l ln nht sau n nh hng do cn kim loi. nh hng do tinh khit ca kim loi gy ra KB rt nh, khng ng k.2 2 2

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Trang : 27


* Vic tnh ton KB tng hp c th c tnh theo cch s dng bng tnh di y: A 1 2 3 4 5 6 7 8 9 10 11 12 13 u(c(Cd)) 0.9 c(Cd) u(y,xi) u(y) ,u(y,xi)2 2

B P Gi tr KB

C m 0.9999 0.000058 0.999958 100.28 100.00 1002.75788 0.05816 0.00338

D V 100.28 0.05 0.9999 100.33 100.00 1003.19966 0.49995 0.24995

E 100.00 0.07 0.9999 100.28 100.07 1001.99832 -0.7014 0.49196

P m V

0.9999 100.28 100.00 1002.69972 0.74529

Ch thch: Cc gi tr ca cc tham s c in t C2 n E2. KB tng ng c in vo dng di t C3 - E3. Copy cc gi tr t C2 E2 vo ct th 2 t B5 B7. Kt qu tnh nng C(Cd) tnh c t cc gi tr trn v a ra trong B9. C5 ch gi tr ca P t C2 cng thm KB C3. Kt qu tnh C(Cd) s dng cc gi tr t C5 - C7 th hin trong C9. Ct D v E tnh C(Cd) cng theo cch trn. Cc gi tr ch ra dng 10 (C10 E10) tnh bng cch ly cc gi tr t dng (C9 E9) tr i gi tr B9. Dng 11 (C11 E11) l cc gi tr ca dng 10 (C10 E10) c bnh phng ln v tng cc gi tr ny c th hin trong B11. C13 a ra gi tr KB chun tng hp tnh bng cch ly cn ca B11.

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III. C LNG KB CHO MT S CH TIU TH NGHIM C TH 1. 1.1

CHUN AXIT/ BAZ: XC NH NNG CA DUNG DCH AXIT CLOHIRIC (HCL) Phng php. Xc nh dung dch NaOH 0,1 M bng chun trn c s dung dch KHP 0,1 M (potassium hydrogen phthalate); Sau xc nh nng ca 0,1M HCl bng chun trn c s dung dch 0,1M NaOH. Th tc. Bc 1: Cn 5 g KHP Bc 2: Ho tan KHP trong nc v nh mc ti 250 ml th tch Bc 3: Tnh ton nng phn t gam ca KHP Bc 4: Ho tan 2 g NaOH ht trong nc v nh mc ti 500 ml th tch Bc 5: a 25 ml dung dch NaOH vo 1 bnh tam gic Bc 6: Chun dung dch NaOH bng KHP t mt Burette 50 ml Bc 7: Tnh ton nng ca dung dch NaOH Bc 8: a 25 ml dung dch NaOH vo bnh tam gic Bc 9: Chun dung dch NaOH c chun ho bng dung dch HCl t mt Burette 50 ml Bc 10: Tnh ton nng ca dung dch HCl S nguyn nhn v kt qu xc nh nng dung dch NaOH

1.2

1.3

tinh khit (KHP)

Khi lng (KHP) Nng (NaOH)

Th tch (KHP)

Trng lng phn t (MW - KHP)

1.4

nh gi cc thnh phn khng m bo

Bc 1: Cn KHP Cng vic Hp ng + KHP Hp rng Khi lng KHP 33,5895 g 28,5130 g 5,0765 g

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Ln ban hnh : 1.04

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Cc ngun khng m boKhi lng (KHP) Hiu chun Hiu chun nhy tinh khit (KHP) lp li Khi lng (th - c b) lp li nhy tuyn tnh

tuyn tnh Khi lng (b)

c (NaOH) Th tch (KHP) Trng lng phn t (KHP)

a) Kt hp vi vic hiu chun cn c s dng. Giy chng nhn hiu chun ch ra rng ti tin cy 95%, mt ln cn t c bng s khc nhau trong cng mt phm vi l 0,1 mg ca gi tr c hin th. Thnh phn khng m bo c th c biu din nh lch chun bng cch chia 0,1/ 2 ta c 0,05 mg. b) lch chun ca nhng ln cn lp li (1- 50 g) theo bo co QA ca PTN ch ra rng mt lch chun 0,09 mg Nhng khng m bo phi hp khi cn u (Wk) u (Wk) = (0,05 2 + 0,09 2 ) = 0,103 mg

Bc 2: Chun b dung dch KPH chun Cng vic: Ho tan 5 g KPH v nh mc ti 250 ml trong bnh nh mc. Cc ngun khng m boKhi lng (KHP) nhy Hiu chun Hiu chun lp li nhy Tuyn tnh

tinh khit (KHP lp li Tuyn tnh

Cn (tng) Cn (b) lp li Hiu chun Nhit chch lp li c (NaOH)

Dung tch (KHP)

im kt thc

Khi lng phn t (KHP)

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a.

khng m bo ca th tch i vi bnh nh mc c s dng: Catalog ca nh sn xut ni rng bnh nh mc 250 ml c khng m bo 0,15 ml khng cp ti tin cy. Do , mt phn b hnh ch nht ca sai s c chp nhn vi 3. lch chun ca dung tch khi l 0,15/ 3 = 0,087 mL.

b.

khng m bo o trong vic nh mc n th tch thit k: lch chun i vi nhng th tch khc nhau ca bnh nh mc c tnh ton sau mt lot cc ln nh mc lp li (10 ln) v cn nc n vch nh mc c a ra l 0,014 mL. Gi tr s c s dng trong ln tnh ton cui cng v khng m bo o trong php o th tch.

c.

S khc nhau gia nhit dung dch v nhit hiu chun ca bnh nh mc. Coi nh ch h s gin n th tch ca dung dch bi v n l con s ng k ln hn dung tch gin n ca bnh nh mc thu tinh, cho mc ch thc hin. Ly h s gin n i vi nc l 2.1 * 10-4 trn mt C v nu s khc nhau v nhit gia dung dch v nhit hiu chun l 5 .. i vi dung tch 250 ml c s dng s a ra mt khong tin cy 95% ca: 250 x 5 x 2.1 x 10-4 mL = 0.263 mL Do s khc nhau lch chun ca nhit l: 0.263/ 2 = 0.132 ml khng m bo o phi hp ca th tch chun KHP u (Vk) l: u (Vk) = (0,087 + 0,014 + 0,132 ) = 0,16 mL2 2 2

Bc 3: Tnh ton nng phn t gam KHP Cng vic: Nng ca dung dich KHP, Mk c tnh ton theo cng thc: Mk = (Wk x P x 1000)/ (Vk x MW) Trong : Wk : Khi lng ca KHP c s dng (5,0765 g) P: tinh khit ca KHP (99,8 0,2%) Vk : Th tch ca dung dch c nh mc (250 mL) MW: Khi lng phn t ca KHP theo cng thc C8H5O4K B sung cho khng m bo ca Wk v Vk c kim tra sm hn, c hai khng m bo c xc nh,: a. khng m bo i vi tinh khit ca KHP tinh khit ca KHP do nh cung cp a ra l 99.8% 2%, c ngha P l 0.998 0.002. Khi tin cy khng c cng b cho khng m bo, Chng ta phi a ra mt phn b hnh ch nht v sai s ng vi u (P) = 0,002/ b. 3 , v ta c.

3 = 0,0012

khng m bo ca khi lng phn t KHPLn ban hnh : 1.04 Trang : 31

AGL 18


Cng thc phn t ca KHP l C8 H5O4K . Xem bng trng lng nguyn t ca cc nguyn t, C, H, O, K bao gm cc c lng khng m bo c ban hnh bi IUPAC tp ch Ho hc tinh khit v ng dng, s 66. Nguyn t C H O K Khi lng nguyn t 12.011 1.00794 15.9994 39.0983 KB 0.001 0.00007 0.0003 0.0001 KB chun 0.00058 0.00004 0.00017 0.000058 3

Ch : Cc khng m bo chun c tnh bng cch chia khng m bo cho

S tp hp khng m bo ca mi nguyn t i vi khi lng phn t ca KHP c tnh bng cch tng cng tng khng m bo chun bng s lng cc nguyn t ca mi nguyn t trong cng thc phn t, v kt qu c lp bng nh di y: S nguyn t trong cng thc C8 H5 O4 K Trng lng c tnh ton 8 x 12.011 5 x 1.00794 4 x 15.9994 1 x 39.0983 Kt qu c tnh 96.088 5.0397 63.9976 39.0983 204.2236 Do khi lng phn t ca KHP l 204,2236 v khng m bo tng hp ca u(MW) l cn bc 2 ca tng bnh phng ca cc khng m bo ring bit. V d u(MW) = (0,0046 + 0,0002 + 0,00068 + 0,000058 ) = 0,00472 2 2 2

KB c tp hp 0.0046 0.00020 0.00068 0.000058

Tt c nhng tp hp khng m bo o, chng ta c th tm tt chng nh di y: Nhn t KB Wk P Vk MW Gi tr c s dng, V 5.0765 g 0.998 250 ml 204.2236 KB 0.000104 0.0012 0.16 0.0047 lch chun lin quan 2.05 * 10 -5 1.202 * 10-3 6,4 * 10-4 2.3 * 10-5

Do khng m bo chun ca nng (Mk) trong dung dch chun KHP c biu din nh sau: u (Mk)/ Mk = (0,0000205 + 0,001202 + 0,00064 + 0,000023 ) = 0,0014 mol L-12 2 2 2

Nng dung dch KHP, Mk, c tnh theo cng thc (1) nh sau: Mk = (5,0765 * 0,998 * 1000) / (250 * 204,2236) = 0,0992 mol L-1AGL 18 Ln ban hnh : 1.04 Trang : 32


Do ta c khng m bo chun u (Mk) trong dung dch KHP l: u (MK) = 0,0014 * 0,0992 = 0,00014 T ta c nng dung dch KHP l 0.0992 mol L-1 vi khng m bo chun 0,00014 mol L-1. Bc 4: Chun b dung dch NaOH Dung dch NaOH c chun b c chun ho bng phn tch ho trc tip bi dung dch chun KHP, khng m bo lin quan ti vic chun b dung dch NaOH khng c lu mc d tinh khit ca dung dch NaOH v th tch dung dch c chun b c khng m bo. Bc 5: Cho 25 ml dung dch NaOH (bng Pipet) vo bnh nh mc Nh trong bc 2, cc phn di y c lu khi chuyn 25 mL NaOH chun : a. khng m bo trong th tch nh mc pipet c s dng Nh sn xut pipet tuyn b rng pipet c s dng c khng m bo 0,03 ml. Gn ng vi phn b hnh ch nht bi v tin cy khng c bit, lch chun ca th tch pipet c tnh l 0,03/ 3 hoc 0,017 mL. b. khng m bo trong nh mc pipet n 25 mL Cc php o cn lp li ca th tch 25 mL vi pipet a ra mt lch chun 0,001 mL, N c s dng trc tip trong vic tnh ton cui cng ca khng m bo chun. c. khng m bo trong thay i th tch do tc ng ca nhit . (Nhit ca php phn tch Vs nhit hiu chun ca pipet) ly s thay i nhit c th ca 5 0C v h s gin n th tch ca thu tinh l 2.1 x 10-4, do vi tin cy ca php phn tch th tch l 95% th nhn t nhit l: 25 x 5 x 2,1 x 10-4 = 0,0263 mL T , lch chun i vi s thay i nhit l 0,0263/ 2 = 0,013mL. Phi hp tt c 3 ngun khng m bo trn, chng ta c khng m bo u (Vs) trong vic chuyn i dung dch NaOH nh cn bc 2 tng bnh phng 3 lch chun trn, c kt qu sau: u (Vs) = (0,017 + 0,001 + 0,013 ) = 0,0212 2 2

Bc 6: Chun dung dch NaOH trn c s dung dch KHP chun (Va) 25 dung dch NaOH c chun trn c s dung dch chun KHP t mt burette 50 mL. Mt khc, chng ta cn lu cc ngun khng m bo t quan im ca cc 3 yu t tng ng c tho lun trn: a. khng m bo trong th tch c cng bo ca Buretter l 50 mL

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Nh sn xut cng b rng buret c s dng c khng m bo 0.05 mL. Gn ng vi phn b hnh ch nht bi v tin cy khng c bit, lch chun th tch ca pipet c tnh l 0,05/ 3 = 0,029 mL b. khng m bo trong th tch dung dch chun KHP c s dng chun .

Khi mong mun s dng khong 25 mL dung dch chun trong vic chun , vic cn v c lp li ca th tch 25 mL t buret c kim tra v a ra mt lch chun 0,012 mL. Chng ta phi s dng con s ny nh khng m bo chun ca th tch c s dng. c. khng m bo ca nh hng nhit gia nhit chun ti nhit phng v nhit hiu chun ca buret . Ly s thay i nhit c th 5 0C nh trc v h s gin n th tch ca thu tinh l 2,1 x 10-4 0C, Vi tin cy 95% ca php o th tch th yu t nhit l: 25 x 5 x 2,1 x 10-4 ml = 0,0263 mL Do , lch chun i vi s thay i nhit l 0,0263/2 = 0.013 ml. Trong vic chun ny, 25,20 ml dung dch KHP c dng s dng trong vic t c im cui vi dung dch NaOH. T y, s dng nhng s c c trn, khng m bo chun phi hp u (Va) c tnh ton nh sau: u (Va) = (0.0292 + 0.0122 + 0.0132 ) hoc 0.034 ml

Bc 7: Tnh ton nng dung dch NaOH. Cng thc c s dng trong vic tnh ton ca dung dch NaOH l: Ms = (Mk x Va)/Vs Ms: Nng dung dch NaOH Mk: Nng dung dch KHP chun Va: Th tch dung dch KHP chun c s dng Vs: Th tch dung dch NaOH c chun Lu ton b nhng tp hp khng m bo trong b 3 n bc 6, chng ta c th tm tt chng nh di y: Nhn t khng m bo Mk Vk Va Gi tr c s dng, V 0.0992 mol l-1 25.2 ml 25.0 ml khng m bo 0.00014 0.034 0.021 lch chun lin quan 1.41 x 10 -3 1.35 x 10 -3 8,40 x 10 -4

khng m bo chun trong nng dung dch NaOH (Ms) c tnh nh sau: u (Ms)/ Ms =AGL 18

(0.00141 + 0.00135 + 0.00084 ) = 0.0021 mol L-12 2 2

Ln ban hnh : 1.04

Trang : 34


Nng ca dung dch NaOH, Ms, c tnh theo cng thc (2) nh sau: Ms = (25.20 x 0.992)/ 25.0 = 0.100 mol L-1 T , khng m bo chun u (Ms) trong nng dung dch NaOH l: u (Ms) = 0.0021 x 0.1000 = 0.00021 mol L-1 Bc 8: Ly 25 ml dung dch NaOH chun HCL (Vb) 25 ml th tch dung dch NaOH chun c s dng cho chun , nhng lu tng t c th c p dng nh trong bc 5, 25 ml dung dch vi khng m bo chun l u (Vb) 0.021 ml. Bc 9: Chun NaOH trn c dung dch HCL (Vc) Nh trong bc 6, dung dch HCL chun 25 ml NaOH t buret 50 ml. Do khng m bo chun tng hp, u(Vc) l 0.034ml. Bc 10: Tnh ton nng dung dch HCL (Mh) Cng thc c s dng trong vic tnh ton dung dch NaOH l: Mh = (Ms x Vb)/Vc Mh: Nng dung dch HCL Ms: Nng dung dch NaOH chun Vb: Th tch dung dch NaOH chun c s dng Vc: Th tch dung dch HCL c chun Lu ton b nhng tp hp khng m bo trong bc 3 n bc 9, chng ta c th tm tt chng nh di y: Nhn t khng m bo Ms Vb Va Gi tr c s dng, V 0.100 mol l-1 25.00 ml 25.10 ml khng m bo 0.00021 0.021 0.034 lch chun lin quan 2.13 x 10 -3 8.4 x 10-4 1,35 x 10 -3

khng m bo o chun lin quan trong nng dung dch HCL (Mh) c tnh nh sau: u( M h ) Mh = (0,002132 + 0,001352 + 0,000842 ) = 0,0027 mol L-1

Nng ca dung dch HCL, Mh, c tnh theo cng thc (3) nh sau: Ms = (25,00 * 0,100)/ 25,30 = 0,0988 mol L-1 T , khng m bo tng hp chun u (Mh) v nng dung dch HCL l: u (Mh) = 0,0027 x 0,0988 = 0,00027 mol L-1

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Bc 11: Tnh ton khng m bo m rng khng m bo m rng U (Mh) c tnh bng cch nhn khng m bo tng hp chun vi h s ph k, k = 2. U(Mh) = 0,00027 x 2 = 0,00054 mol L-1 T ,nng dung dch HCL c phn tch l: 0,0984 mol l-1 0,00054 mol L-1 1.5 Nhng lu :

Trong chun axit/bazo ny, cc vn di y l nhng ngun c th gy ra KB v chng c a ra xem xt trong v d ny. Nu cc ra KB b sung c nghi th cng nn lu . Cn khi lng Hiu chun cn v lp li;

Cn trng lng - nh hng bn ngoi ca khng kh trong PTN, c th khi mt cn vi lng c s dng; nh hng nhit s khc nhau gia nhit phng v nhit hiu chun; tinh khit ca ho cht c s dng trong chun ho; khng m bo ca khi lng phn t ca ho cht c s dng; khng tinh khit c th , v d cht kim khc trong cc vin xt NaOH; Sai s h thng trong bnh thu tinh dung tch; Thay i trong vic tm kim im kt thc, v d vic nh gi c nhn; Cc phn ng cnh tranh, nh s dnh bm ca carbon dioxide t khng kh.

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2. 2.1 a. b. c. d. e. f.

XC NH AXIT LINOLEIC TRONG SA BO. PHN CHIT T SA BT BI GC-FID Chi tit k thut Cn chnh xc 10g mu sa bt Mu c chit di s c mt ca NH4OH vi ru cn, ete etilic v ete du. Cc este metyl axit bo ca vic chit cht bo c chun b bng s dng NaOCH3 v este ho BF3CH3OH Hn hp dung dch thu c sau lm bay hi ti gn kh di s bc hi ca nit. Phn cn li sau c ho tan vo heptan v dung dch thu c chuyn vo bnh nh mc 25ml v thm heptan (C4H10) cho ti vch trn cng ca bnh. Methyl linoleic c chia v xc nh s lng bi GS ch s ion ho vch hng. ng cong hiu chun tuyn tnh s dng xc nh s lng c xy dng da vo 4 mc tp trung ca methyl linoleic vi s y n im 0 (s dng im trng/ mu trng).

Nh phng php th bao gm cc qu trnh chit v dn xut bi vy xc nh chnh xc ca mt phn tch l rt c lp trong hiu qu ca cc qu trnh ny. Theo tng bc nh gi s hiu qu, nghin cu thu hi c tin hnh song song vi cc php phn tch bnh thng. Vi t l thu hi R nng ca axit linoleic trong mu sa bt (Cspl) c th c tnh nh sau: Cspl = (Cml x V x Fla) / (R x Fml x Wspl) Trong : Cspl: nng axit linoleic trong mu sa bt (mg/g) Cml: nng thu c t ng cong hiu chun methyl linoleic (mg/mL) V: dung tch cn li ca dung dch trc khi phun 25mL/ injection Fla: khi lng axit linoleic (g/mol) Fml: khi lng methyl linoleic (g/mol) Wspl: 2.2 khi lng mu sa bt (g)

Xc nh ngun khng m bo Nguyn nhn v kt qu c th c xy dng nh sau;

Fml

Fla

V

Cml Cspl

Thu hi(R) chnh xc(r)

Wspl

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2.3

Xc nh khng m bo thnh phn

Cml kt qu thu c t ng cong hiu chun ca mu l 7.15mg/mL cc phn tch ny 4 mc ca chun hiu chun vi nng l 1mg/mL, 2mg/mL, 4mg/mL v 10mg/mL tng ng c chun b t 1000 2mg/mL methyl linoleic trong chun chnh heptan. 4 chun hiu chun v mu trng ca sau: Nng Phm vi phn (mg/mL), xi ng 0 2 1 135 2 280 4 560 10 1194 xiyi b= x2 Nng xobs ca phn tch mt mu vi cc kt qu quan st thu c phn ng yobs sau a ra bi xobs/b. Khng m bo u (xobs,y) trong gi tr d on xobs vi s thay i ca y c th c lng t gi tr khc nhau ca phn cn li S nh v d 4 trn x 0 1 2 4 10 Tng Do xiyi b= x x 0 1 2 4 10 0 133 278 558 11922

chng c o v thu c kt qu Phm vi phn ng thc 0 133 278 558 1192

Mi quan h ca im 0 bnh phng thch hp l y = bx a n ng cong hiu chun l:

y 0 133 278 558 1192 2161

xy 0 133 556 2232 11920 14841

x2 0 1 4 16 100 121

17

= 14841/121 = 122.6528926 y = 122.6528926x. V th:

y 0

Tnh yc 0 122.652893 245.305785 490.61157 1226.52893

(y - yc)2 107.06263 1068.9117 4541.2005 1192.2467

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Ln ban hnh : 1.04

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Do , S2 = (yi - yc)2/(n-2) = (107.06263 + 1068.9117 + 4541.2005 +1192.2467)/ (5-2) = 2303.14 var(x)= S2/b2 = 2303.14/122.65289262 = 0.1531 u(xobs,y) = var(x) = 0.1531 = 0.391 Kt qu l u(Cml) = 0.391mg/mL V: Bnh dung tch s dng y dung dch cui cng n 25mL c chng ch gi tr 25.040 0.015 mL ti 200C thu c t ti liu k thut ca nh cung cp. Trong nghin cu ny phn b dng hnh ch nht c la chn. Bi vy khng m bo hiu chun l 0.0153 = 0.00866mL Bnh thu tinh c s dng trong mi trng m nhit thay i 4.00C (mc tin cy 95%, nu khng nu mc tin cy th phn b hnh ch nht c tha nhn) H s n ca heptan theo s thay i nhit l khng c bit n, c tha nhn n khong gp hai nh h s n ca nc 2.1x10-4 mL/0C. Trong trng hp ny h s n ca heptan c tha nhn l 4.2x10 -4 mL/0C. T kinh nghim ca chng ta chng ta bit rng s tha nhn l trong c ln hn phm vi nhit ca PTN ni m tin hnh phn tch. khng m bo theo s thay i nhit cho mt dung tch 25.04mL l t ti im ln nht ca 25.04x(4.0/2)x4.2x10-4 = 0.0210mL Bi vy, u (V) = (0.008662 + 0.02102 ) = 0.0227

Fla: nh v d trc trng lng phn t (NW) ca axit linoleic (C18H32O2) v khng m bo ca n c tnh nh sau:

C Gi tr KB C H O Fla 12.0107 1.00794 15.9994 280.44548 7.091E-05 u (Fla) 0.008421 12.0107 0.00046 12.01116 1.00794 15.9994 280.454 0.00832 6.9E-05

H 1.00794 0.00004 12.0107 1.00798 15.9994 280.447 0.00128 1.6E-06

O 15.9994 0.00017 12.0107 1.00794 15.99957 280.446 0.00035 1.2E-07

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Fml: Tng t, NW ca mehtyl linoleic (C19H34O2) v khng m bo l: C Gi tr KB C H O Fml 12.0107 1.00794 15.9994 294.47206 12.0107 0.00046 12.01116 1.00794 15.9994 294.481 0.00878 7.902E-05 u (Fml) 0.0088895 7.7E-05 H 1.00794 0.00004 12.0107 1.00798 15.9994 294.473 0.00136 1.8E-06 O 15.9994 0.00017 12.0107 1.00794 15.99957 294.472 0.00035 1.2E-07

Wspl: khi lng ca mu l 10.0232g vi nhiu ln cn khc nhau. Bo co hiu chun ch ra lch ln nht l 0.4mg t gi tr nu ca cn chun a ra l 0.4/ 3 = 0.231mg ca khng m bo chun cho mi ln cn. Nh cn ca mu bao gm c trng lng ca mu v bao b khng m bo tnh t lch c nhn hai: u (Wspl) = (0.231 + 0.231 ) = 0.327 mg 0.000327g2 2

Thu hi (R): theo cc bc nghin cu trc trn mt mu tng t th thu hi lp c th hin trn mt mu n v tm thy c trung bnh thu hi l 91.3% vi lch chun l 5.4%. V th lch chun lin quan l 5.4%/91.3% = 0.0591. i vi mu hin ti s thu hi l 0.950 (95%). V th khng m bo chun theo thu hi l 95%x0.0591 = 5.61%. chnh xc(r): trc, mt mu c th lp li. Qua m nghin cu tnh chnh xc nng axit linoleic c tm thy trung bnh l 5.65mg/g vi lch chun l 0.96mg/g. V th khng m bo chun lin quan l 0.96/5.65 = 0.170 2.4 Tnh ton khng m bo tng Cc khng m bo chun theo Cml, V, Fla, R, Fml v Wspl c tng hp u tin bi phng php chia bng sau:

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Cml Gi tr KB Cml V Fla R Fml Wspl Cspl 7.15 25.04 280.44548 95.0% 294.47206 10.0232 17.906665 1.9561078 u(Cspl) 1.3986092 7.15 0.391 7.541 25.04 280.445 95.0% 294.472 10.0232 18.8859 0.97923 0.95889

V 25.04 0.0227 7.15 25.0627 280.445 95.0% 294.472 10.0232 17.9229 0.01623 0.00026

Fla 280.44548 0.008421 7.15 25.04 280.4539 95.0% 294.47206 10.0232 17.907203 0.0005376 2.89E-07

R 95.0% 5.61% 7.15 25.04 280.445 100.61% 294.472 10.0232 16.9082 -0.9985 0.99695

Fml 294.47206 0.00889 7.15 25.04 280.44548 95.0% 294.48095 10.0232 17.906125 -0.000541 2.922E-07

Wspl 10.0232 0.00033 7.15 25.04 280.445 95.0% 294.472 10.0235 17.9061 -0.00058 3.4E-07

khng m bo chun trn thu c l tng hp m t s chnh xc c a ra cho tng khng m bo thnh phn. S chnh xc thu c trn l lch chun lin quan, v th u(Cspl)/Cspl =

[(1.3986/17.9067) + 0.170 ] = 0.1872 2

V th, tng khng m bo thnh phn l u(Cspl) = 0.187xCspl = 0.187 x 17.90667 = 3.35mg/g khng m bo m rng U(Cspl) vi mc tin cy 95% thu c t khng m bo tng hp vi h s ph k=2 l: U(Cspl) = 2x3.35 = 6.70mg/g Nng ca axit linoleic trong th nghim sa bt tnh c l: 17.9 6.7 mg/g* Bo co khng m bo l khng m bo m rng c tnh bng cch s dng h s ph k=2 vi mc tin cy tng ng l 95% 2.5 Nhn xt nh gi cc thnh phn quan trng Theo qui lut nhn ln ca sai s s thm vo v bt i lin quan, v d x = x1 + x2 khng m bo tng hp l cn ca bnh phng ca tng cc khng m bo thnh phn. v d

AGL 18

Ln ban hnh : 1.04

Trang : 41


u(x) = u x1 + u x2 . V th s tp trung ca tng thnh phn ti khng m bo tng hp c th so snh trc tip gia u(x1) v u(x2). Tuy nhin trong trng hp khi mi lin quan l nhn v/hoc chia. v d x= x1/x2 khng m bo chun tng hp c tnh l:22

[

( )

( )]

u(x)/x =

{[u(x )/ x ] +[u(x )/ x ] }2 2 1 1 2 2

Bi vy trnh t so snh phn b ca mi thnh phn ti khng m bo tng mt l so snh mi lin quan ca khng m bo chun ca cc thnh phn v d: u(x1)/x1 v u(x2)/x2 i vi v d tho lun trn mi lin quan l nhn v chia. V th s so snh phn b mi thnh phn , mi thnh phn khng m bo phi c th hin mi lin h vi khng m chun. Gi tr Cml V Fla Fml Wspl R r Cspl 17.90667mg/g 3.35 mg/g 7.15mg/g 25.04mL 280.44548g/mol 294.47206g/mol 10.0232 95.0 % KB chun 0.391 mg/g 0.0227 mL 0.00842 g/mol 0.00889 g/mol 0.000327 g 5.61 % KB lin quan 0.0547 0.0009 0.00003 0.00003 0.00003 0.0591 0.170 0.192

T bng trn r rng s gp phn chnh l t Cml, R v r. Thc t mt c th trnh by khng m bo tng hp ch t ba ngun bng cch l i khc ngun nh khc. Trong trng hp ny, u(Cml)/Cml = (0.05472 + 0.05912 + 0.1702 ) = 0.188. V th u(Cml) = 0.188x17.90667 = 3.37mg/g. V vy r rng l khng m bo ch c tnh t cc ngun chnh l khng c ngha khc nhau t s thu c sau khi cn nhc tt c cc ngun.

AGL 18

Ln ban hnh : 1.04

Trang : 42


3.

XC NH HM LNG AXITS TRONG DU

3.1 Phng php c s ca mu du c ho tan trong s trung ho IPA v chun da vo cht chun l dung dch KOH 0.1mol/L vi cht ch th phenolphthalein. 3.2 S nguyn nhn v kt qu CKOH MW (KOH) W (du)

Gi tr ca axit Chun (V)

3.3 Xc nh cc khng m bo 3.3.1 Dung dch chun KOH 0.1M vi HCl 0.5M A Chun b HCl 0.5M V phng din thng mi chun b dung dch HCl cha 18.230 g HCL (mHCl) c s dng chun b Cst HCl 0.5M dung tch1000mL Bnh nh mc s dng chun b dung dch c dung tch l 1000mL 0.4mL ti 200C. lch chun thch hp ca s hiu chun dung tch s dng phn b hnh ch nht l 0.4/ 3 hoc 0.23mL T nhit thc v nhit hiu chun bnh nh mc l -3 0C vi mc tin cy 95% h s n dung tch ca nc l 2.1 x 10-4/0C, s bin i th tch c th l 1000 x 3 x 2.1 x 10-4 hoc 0.63mL lch chun tng ng l 0.63/1.96 hoc 0.32mL Tng hp hai phn b khng m bo u(V), chng ta c: u(V)/V =

(0 .23

2

+ 0 . 32

2

)/ 1000

= 0.00039

Nng ca HCL l mHCL/ MHCL. V ni MHCL l khi lng phn t ca HCL. Nh sn xut dung dch HCL ch dn mt lch c th ca chun 0.02% cho mi0

C. Ly mt s khc nhau c th trong PTN ca nh sn xut - 20 C (vi tin cy 95%), khng m bo chun ca m HCL l: u(mHCL) = 18.230 x 0.02 x 2 / (100 x 1.96) = 0.004 g u(mHCL)/ m HCL = 0.00022

AGL 18

Ln ban hnh : 1.04

Trang : 43


khng m bo chun ca trng lng phn t HCL, theo khi lng nguyn t IUPAC v phn b hnh ch nht, l: u(mHCL) = 0.000043 Cn lu rng u(MHCL)/ MHCL l khng ng k khi so snh vi u(V)/V v u(mHCL)/ mHCL , khng m bo chun lin quan l: u(Cst)/ Cst = B. Xc nh CKOH Nng chnh xc ca dung dch KOH c thit lp trc khi s dng bng chun trn c s dung dch HCL chun. Do , CKOH = Cst * Vst/ VKOH Ni, Vst l th tch (ml) dung dch HCL chun c s dng cho chun th tch VKOH (ml) ca dung dch KOH. Nh c trnh by trn, u(Cst)/ Cst = 0.00045 i vi vic chuyn dung dch KOH n bnh tam gic, mt pipet thu tinh dung tch 5 0.01 ml c s dng. Ly s bin i nhit c th l 30C vi tin cy 95%, v lp li ca pipet ( lch chun) 0.0033 mL, n c th tnh u(VKOH)/VKOH = 0.0015 Vic chun c hon thnh s dng mt buret vi lng 5 ml c vch chia 0.01 ml ( chnh xc hiu chun ca nh sn xut l 0.01 ml) S sai khc nhit c th l tng t nh nu trn, lch chun ca vic ong l 0.0033 mL v lch chun ca im ch cui cng tng ng vi c ca buret (0.017mL) l 0.0098mL V th gi tr ln nht ca u(Vst)/Vst = 0.013 nu CKOH = 0.1 mol/L v tng ng Vst = 1 mL khng m bo u(Cst)/Cst v u(VKOH)/VKOH l khng ng k trong so snh ti u(Vst)/Vst V vy u(CKOH)/CKOH = u(Vst)/Vst = 0.013 3.3.2 Xc nh gi tr ca axit Gi tr ca axit l: AV = MKOHVKOHCKOH / m Lu phng php th s dng v khi lng phn t KOH l 56.1 thay th hon ton gi tr MKOH = 56.10564; bi th trong trng hp ny u(MKOH)/MKOH = 0.00564/(56.1 x 3 ) = 0.00006 chit axit bo t do bng dung dch KOH chng ta s dng buret 5mL ni trc u(Vst)/Vst = u(Cst)/Cst = u(Vst)/Vst = 0.013AGL 18 Ln ban hnh : 1.04 Trang : 442 2 (0.00039 + 0.00022 ) = 0.00045.


khng m bo ca mu du trng lng 2.5g l u(m)/m = 0.0023 Tht r rng khng m bo ca trng lng phn t KOH v ca mu du l khng ng k, v th

u( Av) / Av =

[u(VKOH ) /VKOH ] + [u(CKOH ) / CKOH ]2 = 0.018U(AV)/AV = 2 x 0.018 hoc 0.04

khng m bo m rng vi h s ph k 2 l

Ghi ch: s ch ca im kt thc ca chun l ngun nh hng khng m bo ln. Nu buret v d c c l 0.043mL khng m bo m rng c th tng n 0.07 Hn na mu ca du v s thay i c th trong cht ch th chy ti gn im kt thc trong dung dch du ho tan l khng c tnh ti. Tng t mi lin quan ti nh hng ca CO2 khng kh trong CKOH

AGL 18

Ln ban hnh : 1.04

Trang : 45


4.

C LNG KB T D LIU PH DUYT PHNG PHP. XC NH THUC TR SU PHOSPHO HU C TRONG BNH M Tm tt

4.1

4.1.1 Mc ch Xc nh d lng thuc tr su phospho hu c trong bnh m dng phng php chit v sc k kh (GC) 4.1.2 Qu trnh o Qu trnh xc nh d lng thuc tr su phospho hu c trong bnh m c th hin hnh 4.a Hnh 4.a. Phn tch thuc tr su phospho hu c ng nht

Tch chit

Lm sch Chun b chun hiu chun

C c

Xc nh GC

Hiu chun GC

Kt qu

4.1.3 i lng o D lng thuc tr su phospho hu c trong bnh m c tnh theo cng thc sau: Pop = I op .cref .Vop I ref . Re c.msample .Fhom .10 6 mg.kg-1

Trong : Pop: hm lng thuc tr su c trong mu (mg.kg-1) Iop : Cng ti a ca phn chit ca mu Cref : Nng khi lng ca cht chun (g.ml-1) Vop : Th tch cui cng ca dch chit 106 : chuyn t g.g-1 sang mg.kg-1 Iref : Cng ti a ca cht chun

AGL 18

Ln ban hnh : 1.04

Trang : 46


Rec : phn thu hi/ thu hi msample : khi lng mu Fhom : h s hiu chnh cho mu khng ng nht 4.1.4 Xc nh cc ngun gy ra khng m bo o Cc ngun gc ca khng m bo o c cc nguyn nhn c th hin hnh 4.b Hnh 4.b Phn tch ngun gc ca khng m bo o ti lpI(op) I(ref) m(ref) V(ref) pha long V(op)m (sample)

I (op)

hiu chun m(ref) nhit V(ref)

tuyn tnh

c(ref) tinh khit (ref)

V(op)

hiu chun

nhit

hiu chunPha long

hiu chun

hiu chun

P(op)m(gross)

m(tare)

hiu chun

tuyn tnhhiu chun hiu chun

tuyn tnh

F(hom)

thu hi

I(ref)

m mu

4.1.5 nh lng cc thnh phn ca khng m bo o Da trn d liu ph duyt phng php th ni b, c 3 yu t to nn KB o. Cc yu t ny c lit k trong bng 4.c v th hin trn s 4.d (cc gi tr c suy ra t bng 4.e) Bng 4.c : khng m bo o trong phn tch thuc tr suM t cc ngun Gi tr KB chun u(x) j KB chun tng i u(x)/x Ghi ch

lp li chch Cc ngun gc khc ( ng nht...) u(Pop)/Pop

1.0 0.9 1.0 --

0.27 0.043 0.2 --

0.27 0.048 0.2 0.34

Da trn 2 ln th ca 2 mu khc nhau mu thm nh gi trn c s phng php s dng KB chun tng i

AGL 18

Ln ban hnh : 1.04

Trang : 47


Hnh 4.d khng m bo o trong phn tch thuc tr su u(y,xi) (mg/kg) ng nht chch lp li P(op)0 0.1 0.2 0.3 0.4

Cc gi tr ca u(y, xi) = (y/xi).u(xi) ly t bng 4.k 4.2 Hng dn chi tit

4.2.1 Gii thiu V d ny minh ha mt mt phng php tnh KB bng cch s dng cc d liu ph duyt phng php th ni b. Mc ch ca php o l xc nh tng d lng thuc tr su phospho hu c trong bnh m. Qui trnh ph duyt v cc lin kt thit lp nghin cu thc nghim bng cch tin hnh trn cc mu thm. Gi nh KB do s khc nhau trong kt qu o ti s thm v phn tch trong mu l nh so vi KB o tng. 4.2.2 Bc 1. c im k thut c trng ca php o i vi cc phng php phn tch c m rng s tt hn nu c m t cc bc khc nhau ca qu trnh phn tch v m t cng thc ton hc tnh kt qu o/th. Qui trnh: Qu trnh o c m t trong s hnh 4.a Qu trnh ny c chia thnh cc giai on sau: i) ng nht: Mu c chia thnh nhng phn nh (khong 2cm), ly ngu nhin 15 phn trong s ny v ng ho chng. Nu c phn no nghi ng khng c ng ho tt phi c ly li trc khi trn u. ii) iii) iv) v) vi) Cn trng lng ca khi mu ny c khi lng l msamble Chit : lng dch chit ca mu phn tch vi dung mi hu c, em gn v sy qua ct Na2SO4 v c c dch chit bng cch s dng thit b Kuderna-Danish. Ho lng Axeton nitril/hexane lng, ra axeton nitril bng hexane, sy hexane qua ct natri sulphate C c dch ra bng ga vi ngn la nh cho n khi dch chit gn nh kh hon tonLn ban hnh : 1.04 Trang : 48

AGL 18


vii) Pha long n th tch tiu chun Vop (khong 2ml) trong ng nghim chia vch 10ml viii) o : Tim 5l dch chit v o GC c c cng mi (peak) Iop ix) x) Chun b dung dch chun 5l .ml-1 (nng khi lng Cref) Hiu chun GC s dng 5l dung dch chun, tim v o GC c cng mi chun Iref Hnh 4.a. Phn tch thuc tr su phospho hu c ng nht

Tch chit

Lm sch Chun b chun hiu chun

Bulk up

Xc nh GC

Hiu chun GC

Kt qu

Tnh ton kt qu Nng khi lng Cop ca mu cui cng c tnh nh sau C op = c ref . I op I ref g.ml-1

v nh gi Pop ca hm lng thuc tr su trong mu c (mg.kg-1) c tnh bi: Pop = Hoc, thay Cop vo cng thc c Pop = I op .c ref .Vop I ref . Re c.m sample .10 6 mg.kg-1 cop .Vop Re c.msample . 106 mg.kg-1

Trong : Pop: Mc thuc tr su trong mu (mg.kg-1) Iop: Cng mi ca mu chit Cref: Nng khi lng ca mu chun (g.ml-1)

AGL 18

Ln ban hnh : 1.04

Trang : 49


Vop: Th tch cui cng ca dch chit (ml) 106: i t (g.g-1) ra (mg.kg-1) Iref : Cng pick ca mu chun Rec: thu hi msample: Khi lng ca chia [g] Phm vi: Phng php phn tch ny c kh nng p dng cho cc thuc tr su c bn cht ho hc tng t vi nng t 0.01 n 2 mg.kg-1 vi cc loi mu bnh m khc nhau. 4.2.3 Bc 2: Xc nh cc ngun KB o Xc nh v phn tch cc nguyn nhn gy ra khng m bo o l mt qu trnh phn tch tng hp, cch tt nht l phn tch theo s nguyn nhn v kt qu. Cc thng s trong biu thc ton hc ca php th c th hin bi cc nhnh chnh trn s . Cc thng s mi c thm vo s theo tng bc trong qu trnh phn tch (hnh 4.b) cho n khi cc thng s c lit k y . S khng ng nht ca mu khng phi l mt thng s trong cng thc ton hc tnh kt qu ca php th nhng n li nh hng ng k n qu trnh phn tch. V th mt nhnh chnh mi cho thng s F(hom), i din cho s khng ng nht ca mu c thm vo trong s nguyn nhn v kt qu (hnh 4.e) Hnh 4.e: S phn tch cc ngun gc KB o trong trng hp mu khng ng nht

hiu chun

I (op)m(ref) nhit V(ref)

tuyn tnh


V(op)

chm hiu

hiu chun chm

nhit hiu chun

hiu chun Pha long

chm

chm P(op)m(khng b)

chmhiu chun

m(b)

tuyn tnh nhy

chmhiu chun m mu

tuyn tnh

nhy

F(hom)

thu hi

I(ref)

hiu chun

AGL 18

Ln ban hnh : 1.04

Trang : 50


Tm li s th hin cc nguyn nhn dn n khng m bo o th s khng ng nht ca mu bao gm trong s tnh ton ca php th. ch ra cc kt qu ca khng m bo o mt cch r rng, tt nht nn vit cng thc: Pop = I op .cref .Vop I ref . Re c.msample .Fhom .10 6 mg.kg-1

Trong F(hom) l mt s hiu chnh gi thit cho tnh thng nht trong php tnh gc. iu ch r rng khng m bo o trong h s hiu chnh phi c bao gm trong s nh gi khng m bo o tng. Cng thc trn cng ch ra khng m bo o s c p dng.Ch thch: Cc h s hiu chnh: cch tip cn ny kh l khi qut v c th rt c gi tr. V mt nguyn l, mi php th u phi c cc h s hiu chnh thng nht. V d: khng m bo o Cop c th c din t bng khng m bo o chun i vi Cop hoc khng m bo o chun i din cho khng m bo o trong cho h s hiu chnh. Trong trng hp sau, gi tr xc nh khng m bo o Cop c din t bi lch tiu chun tng i.

4.2.4 Bc 3 : nh lng cc thnh phn ca khng m bo o nh lng cc thnh phn khc nhau ca KB o t d liu nghin cu phng php th ni b v nghin cu ph duyt phng php: - S c lng tt nht ca phng sai tng ca ton b qu trnh phn tch qua nhiu qu trnh th nghim (th nghim lp li, ti lp) S c lng tt nht ca tng chch (Rec)v khng m o ca n S nh lng ca bt c khng m bo o no lin quan n kt qu tnh ton khng y i vi ton b qu trnh.

Mt vi s sp xp li cc nguyn nhn v biu hiu qu rt c gi tr to thnh mi lin h v a ra nhng thng tin r rng hn. (hnh 4.f)Hnh 4.f: S nguyn nhan v kt qu sau khi phn b li c cc d liu nghin cu ti lp(1)I(op) I(ref) m(ref) V(ref) pha long V(op)m (sample)

hiu chun(2)

I (op)m(ref)

tuyn tnh


V(op)

nhit (2) hiu chun(2) V(ref) hiu chun(2)

nhit (2)

hiu

hiu chun(2)

pha long

P(op)m(c mu v b)

tuyn tnhhiu chun(3) hiu chun(2)

m(b)

tuyn tnhhiu chun(2)

F(hom) (3)

thu hi (2)

I(ref)

m mu

AGL 18

Ln ban hnh : 1.04

Trang : 51


Ch thch: Thng thng mu c chy tng l mu nh, mi l mu chy u phi hiu chun thit b, mu kim tra thu hi kim sot chch v mu p ngu nhin kim tra chm. S phi thc hin hnh ng khc phc nu trong trng hp cc ln th ny c du hiu cho cc kt qu lch nhau ng k. Kim sot cht lng c bn ny p ng c cc yu cu chnh ca vic nh gi KB t d liu ph duyt phng php.

Vic chn thm mt yu t nh hng lp li trong s cho ta cng thc tnh Pop nh sau: Pop = I op .c ref .Vop I ref . Re c.m sample .Fhom .10 6 .Fref mg.kg-1 cng thc *

iu ch ra lp li ging nh ng nht. Cch tnh ton theo dng ny thun tin hn v c trnh by di y. Cc nh gi ca cc nh hng khc nhau s c cn nhc Nghin cu chm Ton b phng sai ca qu trnh phn tch qua nhiu ln phn tch c thc hin i vi cc mu th p (cng mt mu ng nht, tch chit v qu trnh xc nh) i vi d lng thuc tr su phospho hu c c tm thy trong cc mu bnh m khc nhau. Kt qu c trnh by trong bng 4.g Bng 4.g: Kt qu nghin cu th nghim lp thuc tr su D lng Malathion Malathion Malathion Malathion Malathion Pirimiphos Methyl Chlorpyrifos Methyl Pirimiphos Methyl Chlorpyrifos Methyl Pirimiphos Methyl Chlorpyrifos Methyl Chlorpyrifos Methyl Pirimiphos Methyl Chlorpyrifos Methyl Pirimiphos Methyl D1 [mg.kg-1] 1.30 1.30 0.57 0.16 0.65 0.04 0.08 0.02 0.01 0.02 0.03 0.04 0.07 0.01 0.06 D2 [mg.kg-1] 1.30 0.09 0.53 0.26 0.58 0.04 0.09 0.02 0.02 0.01 0.02 0.06 0.08 0.01 0.03 Trung bnh [mg.kg-1] 1.30 1.10 0.55 0.21 0.62 0.04 0.085 0.02 0.015 0.015 0.025 0.05 0.75 0.10 0.045 D1 D2 0.00 0.40 0.04 -0.10 0.07 0.00 -0.01 0.00 -0.01 0.01 0.01 -0.02 -0.10 0.00 0.03 D1-D2/ trung bnh 0.000 0.364 0.073 -0.476 0.114 0.000 -0.118 0.000 -0.667 0.667 0.400 -0.400 -0.133 0.000 0.667

AGL 18

Ln ban hnh : 1.04

Trang : 52


Cc d liu thu c khc nhau (khc so vi trung bnh) cung cp mt kt qu cho phng sai tng ca qu trnh th nghim lp li. a ra s c lng KB chun tng i cho tng th nghim n l, lch chun phi chia cho 2 thnh lch chun cho KB chun cho cc gi tri n. T suy ra gi tr ca KB chun t phng sai ca qu trnh th nghim lp li tun theo ton b cc bc trong qu trnh phn tch nhng loi tr nh hng ng nht 0.382/ 2 = 0.27.Ch thch: Ban u c th cc php th p ny khng a ra bc t do. Nhng n khng c mc ch l a ra s lng chnh xc i v chm ca qu trnh phn tch i vi mt loi thuc tr su c bit trong mt loi bnh m c bit no . iu quan trng hn trong qu trnh phn tch php th l nghin cu th nghim c trn nhiu loi mu th v mc nng d lng trong mu a ra s la chn i din ca cc loi thuc tr su phospho hu c. Cch thc c hiu qu nht l th nghim lp li trn cc loi mu bnh khc nhau nghin cu bc t do cho mi loi mu nghin cu.

Nghin cu v chch chch ca phng php phn tch c nghin cu ni b trong phng th nghim bng cch nghin cu cc mu thm (ng nht mu v chia mu thnh cc mu nh v thm mt hm lng vo mi mu nh) bng 4.h a ra cc kt qu ca mt qu trnh nghin cu cc mu thm trong mt thi gian di trn cc dng khc nhau. Bng 4.h. Kt qu nghin cu thu hi thuc tr su C cht Du thi B Thc n gia sc tng hp 1 Du thc vt v ng vt Brassicas 1987 (ci bp) Bnh m Bnh biscot thc phm ch bin t xng v tht Gluten ng ci du La m tng La mch (1): s nghin cu thc nghim (2): Trung bnh v lch chun mu s tnh % thu hiAGL 18 Ln ban hnh : 1.04 Trang : 53

Loi d lng PCB OC OC OC OC OP OP OC OC OC OC OC OC

Nng [mg.kg-1] 10.0 0.65 0.325 0.33 0.32 0.13 0.325 0.325 0.325 0.325 0.325 0.325 0.325

N1) 8 33 100 34 32 42 30 8 9 11 25 13 9

Trung bnh 2) (%) 84 109 90 102 104 90 84 95 92 89 88 85 84

s2) (%) 9 12 9 24 18 28 27 12 9 13 9 19 22


ng tuyn tnh c suy ra khi lm th nghim trn cc mu bnh m (c nh du bng mu ghi) ch ra rng H s thu hi trung bnh t 42 mu l 90% vi lch chun (s) l 28%. KB chun c tnh t lch chun ca trung bnh u (Re c) = 0.28 / 42 = 0.0432 Mt ngha ca php th c s dng xc nh cho h s thu hi trung bnh t so snh s khc nhau ng k so vi 1.0. Theo th nghim thng k t c tnh ton theo cng thc sau: t= 1 Re c u (Re c) = (1 0.9) = 2.315 0.0432

Gi tr ny c so snh vi gi tr ti hn l 2, i vi bc t do n-1 v tin cy l 95% (trong n l s lng kt qu dng c lng Re c ). Nu t >=t k so vi 1. t = 2.31 tcrit,41 = 2.021.crit

th Rec l khc ng

Trong v d ny, h s hiu chnh (1/ Re c ) c s dng v do vy r rng trong cc kt qu tnh ton Re c c th hin. Cc ngun gc khc ca khng m bo o. Trong s 4.i ch ra cc ngun gc khc ca khng m bo o bao gm cc d liu v chm (1), d liu v h s thu hi (2) hoc cc ngun gc khc (3) c kim tra v tnh ton trong khng m bo o. Tt c cc cn v dng c o th thch quan trng phi c kim sot nghim ngt. chm v h s thu hi phi c tnh n trong s nh hng ca vic hiu chun cc dng c dng o th tch bi v trong sut qu trnh th nghim u phi dng cc dng c o th tch v pippet. Nghin cu cc phng sai m rng trong gn na nm cho thy c nh hng ca nhi mi trng ln kt qu. Mt s cc ngun ng gp khc vo KB l tinh khit ca mu chun, s khng tuyn tnh ca thit b GC khi tr li kt qu (i din bi vic hiu chun cho cc thng s Iref v Iop) v vic mu ng nht cng l mt thnh phn thm vo trong yu cu ca vic nghin cu. Hnh 4.i nh gi cc ngun gc khc ca khng m bo ohiu chun(2)

lp li (1)I(op) I(ref) m(ref) V(ref) pha long V(op)m (sample)

I (op)m(ref)

tuyn tnh


V(op)

nhit (2) hiu chun(2) V(ref) hiu chun(2)

nhit (2)

hiu chun(2)

hiu chun(3)

pha long

P(op)m(gross) tuyn tnh hiu chun(3) hiu chun(2) I(ref) m mu

m(tare)

tuyn tnhhiu chun(2)

F(hom) (3)

thu hi (2)

AGL 18

Ln ban hnh : 1.04

Trang : 54


(1) ti lp c tnh n trong qu trnh nghin cu phng sai ca th tc phn tch (2) tnh n trong qu trnh nghin cu chch ca th tc nghin cu (3) c tnh n trong s nh gi cc ngun gc khc ca khng m bo o tinh khit ca chun m nh sn xut cung cp l 99.53% 0.06%. tinh khit cng l mt ngun gc ca KB o vi KB o chun l 0.0006/ 3 = 0.00035 (phn b hnh ch nht). Nhng gi tr ny qu nh so vi nghin cu KB t nghin cu chm nn c th b qua. tuyn tnh ca kt qu thuc tr su photpho hu c trong phm vi gii hn nng c thit lp trong sut qu trnh nghin cu. Hn na vi cc ch s trong bng 4.g v 4.h khng tuyn tnh cng ng gp vo phn b chm. ng nht ca mu bnh m l nguyn nhn cui cng gy ra khng m bo o. Khng c d liu v phn b d lng thnh phn hu c trong cc sn phm bnh m. ( l mt iu ngc nhin v phn tch i vi a s thc phm ng nht him hn i vi cc mu khng ng nht). ng thi thc t khng thc hin cc php o trc tip ng nht. Bi vy, s nh gi khng m bo o da trn c s ca phng php ly mu. Mt s phn b d lng thuc tr su c cng b v mt vi php thng k n gin c s dng tnh ton KB o chun cho ton th mu phn tch v vic tnh ton lin quan n khng m bo o tiu chun trong ca thuc tr su c trong mu cui cng l : D lng phn b pha trn cng b mt ch c: 0.58 D lng phn b ph u trn b mt ch c: 0.20 D lng phn b u trong mu nhng b gim nng do bay hi hoc bi phn hu t bn trong ln b mt : 0.05 0.10 (ph thuc vo dy b mt). Trng hp (a) l c bit cung cp vi t l ly mu hoc ng nht hon ton : n s xy ra trong trng hp thm cht ph gia ln b mt. Trng hp (b) bi vy c coi nh l trng hp xu nht. Trng hp (c) c xem nh c kh nng nht nhng khng th phn bit d dng vi (b), trn c s gi tr 0.20 c chn. 4.2.5. Bc 4 : Tnh ton khng m bo o tng hp Trong qu trnh nghin cu phng php ni b ca thao tc phn tch, ti lp, chch v tt c cc ngun gc khc ca khng m bo o c nghin cu k lng. Cc gi tr ny c ghi trong bng 4.c Cc gi tr tng i c kt hp bi cng thc: u c(Pop) / Pop = (0.27 2 + 0.048 2 + 0.2 2 = 0.34 uc (Pop) = 0.34 x Pop Bng tnh i vi trng hp ny c ch ra trong bng 4.k Ch rng KB o tuyt i (0.373) i vi kt qu hiu chnh danh ngha l 1.1111, t c gi tr 0.373 / 1.11 = 0.34

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Bng 4.c : khng m bo o trong phn tch thuc tr suM t cc ngun Gi tr KB chun u(x) j KB chun tng i u(x)/x Ghi ch

lp li chch Cc ngun gc khc ( ng nht...) u(Pop)/Pop

1.0 0.9 1.0 --

0.27 0.043 0.2 --

0.27 0.048 0.2 0.34

Da trn 2 ln th ca 2 mu khc nhau Mu thm nh gi trn c s phng php s dng KB chun tng i

Kch thc tng i ca 3 thnh phn ny c th c so snh bng cch s dng s Hnh 4.d ch ra cc gi tr [u(y, xi)] c suy ra t bng 4.k. Hnh 4.d khng m bo o trong phn tch thuc tr su

u(y,xi) (mg/kg) ng nht chch lp li P(op)0 0.1 0.2 0.3 0.4

Cc gi tr ca u(y, xi) = (y/xi).u(xi) ly t bng 4.k lp li gp phn ln nht trong phn b KB o. Thnh phn ny c ngun gc t tng phng sai ca phng php nn cc nghin cu thm cn phi tin hnh ngy cng ci tin cho phng php v gim KB o. V d KB o c th gim ng k nu ng nht mu c tnh n v thc hin trong ton b cng on trc khi thc hin ly mt mu phn tch. khng m bo o m rng U(Pop) c tnh ton bng cch nhn KB tng hp vi h s ph k = 2. Ta c: U(Pop) = 0.34 x Pop x 2 = 0.68 x Pop

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Bng 4.k. khng m bo o trong php phn tch thuc tr su A 1 2 3 4 5 6 7 8 9 10 11 12 13 B Gi tr KB o ti lp chch ng nht Pop u(y, xi) u(y)2, u(y, xi)2 u(Pop) 1.0 0.9 1.0 1.1111 0.1420 0.377 C lp li 1.0 0.27 1.27 0.9 1.0 1.4111 0.30 0.09 D chch 0.9 0.043 1.0 0.943 1.0 1.0604 - 0.0507 0.00257 E ng nht 1.0 0.2 1.0 0.9 1.2 1.333 0.222 0.04938

(0,377/1.111 = 0.34 tng ng vi khng m bo o tng i)

Cc gi tr ca cc thng s c a vo dng th 2 t C2 n E2. KB chun ca chng c th hin dng di t C3 n E3. a cc gi tr t C2 n E2 vo ct B5 n B7. Kt qu ca cc gi tr c th hin B9 (=B5 x B7/B6, da trn tnh ton cng thc *). C5 a ra gi tr lp li t C2 tr i KB ca n C3. Kt qu tnh ton s dng cc gi tr C5:C7 c C9. Ct D v E theo cc bc tng t. Gi tr a ra dng 10 (C10 n E10) l gi tr dng t C9 n E9 tr i gi tr B9. Ti dng 11 (C11 n E11). l nhng gi tr ca dng 10 (C10:E 10) l bnh phng v tng ca gi tr th hin B11. B13 l gi tr KB tng hp bng cch khai cn bc hai ca gi tr ca B11.

4.2.6. Kha cnh c bit: S dng mu khng ng nht tnh khng m bo o ca php th xc nh hm lng thuc tr su photpho hu c. Cho rng tt c cc vt liu c trong mu, u c th tch chit c tin hnh phn tch khng k n trng thi ca chng, trng hp xu nht l trng hp mu khng ng nht, mi phn ca mu li cha mt c cht khc nhau. Ti thiu l trng hp c 2 loi vt liu trong cc phn khc nhau ca mu (k hiu l L1 v L2). H qu ca vic khng ng nht trong trng hp mu ph c th c c lng bng thng k nh thc. Gi tr mong mun l trung bnh v lch tiu chun ca lng vt liu trong n phn bng nhau c chn ngu nhin sau khi chia nh mu. Cc gi tr ny c tnh bi: = n.(p1.l1 + p2.l2) = n.p 1.(l1 l2) + n.l2 [1] 2 = n.p1. (1 p1) . (l1 l2)2 [2]

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Trong l1 v l2 l lng c cht trong cc phn bng nhau thu c t mu L1 & L2 , tng trng lng l X. p1 , p2 l kh nng xy ra ca cc phn chia c chn t mu (n phi nh khi i chiu vi tng s cc phn chia c chn) Coi rng kch thc ca mu l 12 x 12 x 24 cm, chia thnh 432 phn bng nhau c kch thc 2 x 2 x 2cm v ly 15 phn bng nhau ngu nhin v ng nht trong s . Cng thc trn c tnh nh sau: Trng hp (a) Vt liu c xc nh l n v c b mt rng trn nh ca mu. V th L2 = 0 ging nh l2 v L1 = 1. Mi phn chia bao gm mt phn ca b mt s nhn c mt lng l1 ca vt liu. Vi kch thc ni trn bng 1/6 ca kch thc mu v th, p1 = 1/6 hoc 0.167 v l1 l X/72 (v c 72 phn chia) Ta c: = 15 x 0.167 x l1 = 2.5 l1 2 = 15 x 0.167 x (1-0.17) x l12 = 2.08 l12 = 1.44l1 RSD = / = 0.58Ch thch : tnh ton lng X trong ton b mu, gi tr trung bnh ca 432/15 v th X s l:

X = 432/15 x 2.5 x l1 = 72 x X/72 = X.Kt qu ny ca mt dng ly mu ngu nhin; Cc gi tr trung bnh ny l chnh xc v c th i din cho gi tr trung bnh ca ton b tp hp. i vi ly mu ngu nhin, th phn b ca khng m bo o ti khng m bo o tng khc vi phng sai ca vic th nghim lp li v c din t bi hoc RSD

Trng hp (b): Vt liu c phn phi trn ton b b mt. Lp lun tng t nh trn, v coi trn ton b b mt cc phn chia cha cng mt lng l1 ca vt liu, l2 = 0 v p1 c kch thc c tnh bi: p 1 = ((12 x 12 x 24) (8 x 8 x 20)) / (12 x 12 x 24) = 0.63 p1 l phn chia ca mu ngoi 2 cm. l1 = X / 272Ch thch: c s thay i gi tr so vi trng hp (a)

Ta c: =15 x 0.63 x l1 = 9.5 l1 = 15 x 0.63 x (l1-0.63) x l12 = 3.5 l12 = 1.87 l1 RSD = / = 0.2 Trng hp (c) Lng vt liu gn b mt gim xung gn n 0 do c s bay hi hoc cc tht thot khc. Trng hp ny c th tnh ton mt cch n gin nht l coi n nh mt trng hp ngc li vi trng hp (b), vi p 1 = 0.37 v l1 = X/160AGL 18 Ln ban hnh : 1.04 Trang : 58


Ta c: =15 x 0.37 x l1 = 5.6 l1 2 = 15 x 0.37 x (1-0.37) x l12 = 3.5 l12 = 1.87 l1 RSD = / = 0.33 Tuy nhin, s tht thot ny t n mt su nh hn kch thc ca phn chia, mi phn s cha mt t lng vt liu l1 v l2 v th, c hai s khng bng 0. Ta ly trng hp m phn ngoi l 50% v phn trung tm mu l 50%. Ta c: l1 = 2 x l2 l1 = X / 296 = 15 x 0.37 x (l1 - l2) + 15 x l2 = 20.6l2 2 =15 x 0.37 x (1 - 0.37) x (l1 - l2) 2 = 3.5 l22 RSD = 1.87/20.6 = 0.09 Trong dng ny, tng ng vi su 1cm m vt liu b mt. Kim tra cc mu bnh m c v dy 1cm hoc nh hn v s dng. Trong cc dng hin nay th tng ng vi su 1cm ci m vt liu b mt. Kim tra cc dng mu bnh m c v dy 1 cm hoc nh hn v to ra su cho vt liu v quan tm n s mt (v cng bn thn n l b ngn cn s mt di chiu su ny). do phng sai thc trong trng hp (c) s a ra gi tr ca , khng nh gi tr 0.09 trnCh thch: Trong trng hp trn s gim cc ny sinh KB trong vic mu khng ng nht l thng s nh hn l ng nht. Ni chung s u mnh vic gim phn b KB. Cn thm vo m hnh nghin cu cc trng hp s lng ln hn l s lng nh ( ging nh s kt hp cht ch ca cc ht trong cc khi ln) cha ng gi tr khng t l vi vt liu quan tm. Cung cp rng kh nng tng hp tr nn khng kt hp vo cc phn mu ng nht l ln, phn b KB s khng vt qu bt c tnh ton no trong cc trng hp trn.

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5. 5.1

XC NH VITAMIN A V E TRONG SA BT Gii thiu phng php

Phng php c dng xc nh nhng ng phn ca tocopherol, retinol v caroten trong mt s loi thc phm. V d ny xem xt vic ph duyt v tnh ton khng m bo o c bit cho vic xc nh cc dn xut ca all-trans retinol (vitamin A) v tocopherol (vitamin E) trong sa bt. Mu ng nht c thu phn gii phng cc ng phn ca retinol v tocopherol. Cc sn phm ny sau c chit bng mt hn hp ete v sau khi chit c c c. Dng nhng h sc k lng hiu nng cao (HPLC) khc nhau vi nhng hn hp cht chit c t l khc nhau tch v nh lng nhng ng phn theo yu cu. Mt h pha thng dng xc nh -tocopherol, trong khi mt h pha ngc c s dng xc nh all-trans retinol. Trong mi trng hp mu u c chun bng mt dung dch chun pha long t dung dch gc. Nng ca cht phn tch trong mt mu, Cvit, c tnh theo cng thc: Cvit = As V f C STD ASTD Ws

Trong : AS : l din tch mi (nh) ca dung dch mu ASTD : l din tch mi ca dung dch chun VF : l lng mu bm (ml) WF : l trng lng mu dng phn tch (g) CSTD : l nng ca dung dch chun (gml-1) Mt s dng chy minh ha cc giai on trong phng php c trnh by hnh 5.1 5.2 Nghin cu chm Mt nghin cu v nhn hiu thc phm ca cc sa bt sn c trn th trng ch ra rng tt c u c cc thnh phn c th so snh v cht bo, protein, cacbonhyrat v cha nhng nng vitamin A v vitamin E tng t. Ba mu c chn kim tra bao ph ht khong nng c th ca vitamin trong mu. Mi mu c phn tch 4 ln, trong 4 ln chit v chy sc k lng hiu nng cao (HPLC) khc nhau. Vi mi ln chy HPLC u dng ng hiu chun mi v dung mi pha ng. Thm vo phn tch mt cht chun c chng nhn (SRM 1846) c lng Rm . Nhng kt qu ny c tng kt trong nghin cu chm. Kt qu c trnh by bng 5.1

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Hnh 5.1 S minh ha cc bc trong phng php xc nh all-trans retinol v -tocopherol Butylat hydroxytoluen * Lng thch hp ca mi dung dch chy HPLC nn la chn da theo nng ca mi vitamin gn vi chun hiu chun tng ng. Cn mu Thm 1g pyrolgalol vo 150ml dung dch etanolic KOH X phng ho trong vng 30 pht bng h hi lu

Chuyn ton b dung dch vo mt phu tch Thm 500ml hn hp ete Chit vitamin bng hn hp ete

Ra dch chit ete vi nc sch n khi dung dch sch trung tnh vi phenolphatalein

Chuyn dch ete sch sang bnh ct quay Thm 2ml dung dch BHT

Ct n kh ti 40 C nh lng alltrans retinol bng HPLC pha ngc

o

Ho tan vi mt lng thch hp metanol*

Ly mt lng dung dch thch hp bng pipet

Ct n kh bng my ct quay Thm 2 ml dung dch BHT nh lng -tocopherol bng HPLC pha thng Ho tan bng mt lng thch hp dung dch pha ng

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Bng 5.1 Tng kt cc kt qu nghin cu chm phng php xc nh all - trans retinol v -tocopherolall-trans retinol (vitamin A) Mu Trung bnh (mg kg-1) 9.67 10.04 6.67 5.32 lch chun (mg kg-1) 0.146 0.247 0.217 0.285 lch chun tng i 0.0151 0.0238 0.0325 0.0536 n 4 4 4 6 Trung bnh (mg kg-1) 64.08 69.05 171.93 278.51 -tocopherol (vitamin E) lch chun (mg kg-1) 2.411 2.156 7.959 10.663 lch chun tng i 0.0376 0.0312 0.0463 0.0383 n 4 4 4 6

Sa bt A Sa bt B Sa bt C SRM 1846

lch chun quan st c vi vitamin A l theo th t ln dn. iu ny ch ra rng trong di nng nghin cu, lch chun c lp vi nng phn tch. Trong trng hp ny, khng m bo lin quan vi chm ca phng php, u(P), cho nhng mu c nng all-trans retinol thay i xp x khong 5mgkg-1 10mgkg-1, c c lng theo lch chun chung (pool) theo cng thc sau: (3 0.146 2 ) + (3 0.247 2 ) + (3 0.217 2 ) + (5 0.285 2 ) = 0.238 S pool = 3+ 3+ 3+ 5 KB u(P) cho vitaminA l 0.238mgkg-1Ch thch: Kt qu ca v d ny c tnh ton khi s dng bng tnh. Nhng gi tr th hin trong cc bng v phng trnh c lm trn. Tnh ton li cc gi tr c th c kt qu sai khc.

i vi -tocopherol, lch chun tng i quan st c i vi cc mu t l thun vi nng -tocopherol ca mu, tc l hm lng cng ln th lch chun cng tng theo. iu ny ch ra rng lch chun l mt t l xp x vi nng phn tch trong di nng nghin cu. Trong nhng trng hp nh vy c lng u(P) s dng cng thc tnh lch chun pool theo lch chun tng i, c kt qu nh sau: (3 0.0376 2 ) + (3 0.0312 2 ) + (3 0.04632 )(5 0.03832 ) = 0.0387 RSD pool = 3+ 3+ 3+ 5 KB tng i u(P)/P cho -tocopherol l 0.0387 tng ng lch chun tng i 5.3 Nghin cu ng

Tnh Rm v u ( Rm) Trong trng hp s dng mu chun c chng nhn SRM 1846 do vin Quc Gia v tiu chun v cng ngh ca Nht (NIST) sn xut. CRM l mu sa bt chun c chng nhn v s lng vitamin, bao gm vitamin A v vitamin E. Su mu CRM c phn tch. Kt qu, cng vi nhng gi tr c chng nhn ca mu chun c tng kt bng 5.2

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Bng 5.2: Kt qu t nghin cu mu chun sa bt SRM 1846 Gi tr c cng nhn khng m Nng khngm bo chun CCRM bo CRM u(CCRM) (mgkg-1) (mgkg-1)2 (mgkg-1)1 5.84 271 0.68 25 0.35 12.8 Gi tr quan st c Trung bnh C obs (mgkg-1) 5.32 278.51 lch chun sobs (mgkg-1) 0.285 10.663

Ch tiu phn tch

VitaminA vitaminE1 2

khng m bo l khng m bo m rng vi mc tin cy l 95% khng m bo chun l khng m bo m rng chia cho 1.96

Tnh Rm v u ( Rm) i vi vitamin A (all-trans retinol): Rm = 5.32/5.84 = 0.911 0.2852 0.35 u ( R m ) = 0.911 6 5.32 2 + 5.84 = 0.0581 2

xc nh phn b ca Rm ti KB tng hp, gi tr c tnh ton trn c so snh vi 1, t = 1 Rm u(R m ) = 1 0.911 = 1.53 0.0581

Trong trng hp ny t oc so snh vi h s ph, k=2. Khi t nh hn 2 khng c bng chng no cho thy Rm khc 1 ng k. V th Rm c th

do khong dam bao do 1

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