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Mn hc: CC PHNG PHP PHN TCH HA LGVGD: T.S. NGUYN VN TRNGT:
0903543882Email: [email protected] DUNG CHNHSTTTN TI1 Phng
php ph hp th phn t UV-VIS2 Phng php ph hp th nguyn t3 Phng php ph
pht x nguyn t4 Phng php ph hng ngoi5 Phng php ph hunh quang6 Phng
php ph ln quang (b d ti ny)7 Ph khi lng8 Ph cng hng t ht nhn9 Sc k
lng cao p10 Sc k kh11 Sc k trao i ion12 Phng php o in th13 Phng php
in phn14 Phng php vn - ampe15 Phng php chun Karl Fischer 16Sc k lng
ghp khi phChng 1: i cng v phn tch ha lPhn loinh lng trong phn tch
ha l (PTHL)Mt s c trng ca phng php PTHLK thut v phng php PTHLPhn
loi cc phng php PTHLPhn tch quang hcPhn tch sc kPhn tch in haCc
phng php khcNH LNG TRONG PTHLNH LNG TRONG PTHLPhng php so snhPhng
php ng chunPhng php thm chunPhng php ni chunPhng php so snh 1
chun5/16/201413 nh lng Pb trong mu thc phm, ta tin hnh cn 5,000 g
mu, ho tan thnh dung dch, sau tin hnh to phc vi thuc th dithizon,
dng phc Pb ditizon tan trong CHCl3. Tin hnh chit bng CHCl3, dung
dch sau khi chit c nh mc thnh 25 mL. Dung dch chun c chun b tng t
nh mu, cha 10 g Pb2+ trong th tch dung dch em o l 20,00 mL. Mt
quang ca chun v mu = 545 vi l = 1 cm ln lt l Ac = 0,320 v Am =
0,225. V dPhn ng ca Pb2+ v dithizonPb2+/H2O + 2H2Dz/CHCl3 =
Pb(HDz)2 + 2H+ chit ln CHCl3 hoc CCl4So snh 2 chun
5/16/201415 nh lng Pb trong mu thc phm, ta tin hnh cn 5,000 g
mu, ho tan thnh dung dch, sau tin hnh to phc vi thuc th dithizon,
dng phc Pb ditizon tan trong CHCl3. Tin hnh chit bng CHCl3, dung
dch sau khi chit c nh mc thnh 25 mL. Dung dch chun c chun b tng t
nh mu, vi mt bnh cha 6,25 g Pb2+ trong th tch dung dch em o l 25,00
mL v mt bnh cha 12,5 g Pb2+ trong th tch dung dch em o l 25,00 mL.
Mt quang ca chun v mu = 545 vi l = 1 cm ln lt l A1 = 0,160, A2 =
0,320 v Am = 0,225.V dPhng php ng chunPha mt lot dung dch chun c Cc
tng dn mt cch u n. (thng 5 8 Cc) (Cc dung dch chun phi c cng iu kin
nh dung dch xc nh)Tin hnh o tn hiu ca dy chun.Dng th AX = f(Cx).
Vit PTHQ tuyn tnh ca ng chun.Tin hnh pha ch dung dch xc nh.Do tn
hiu ca mu.Cn c vo PTHQ tuyn tnh ca dy chun v AX m xc nh nng ca cht
X trong muQui trnh thc hinPhng php ng chun (t.t.) th A = f(Ctc) tu
theo cch o ta thu c 2 dng ng chun:+Dng 1: i qua gc ta +Dng 2: khng
i qua gc ta Khi chn vng nng xy dng ng chun phi ch :+Vng nng ca dy
chun phi bao gm c CX+Cc gi tr Ac ng vi nng chn phi sao cho khi o
trn my c lp li cao v bo m s tuyn tnh A = f(C)185/16/2014Trong phng
php o dy chun ca mt dung dch mu cho kt qu:
Nu mu phn tch c A = 0,672 th nng dung dch l:a. 0,2251b. 0,1390c.
0,1374d. 0,1928
V dPhng php thm chunTheo phng php ny th tn hiu ca dd mu cha cht
cn xc nh c so snh vi chnh dung dch c thm nhng lng xc nh ca cht cn
xc nh.Trong phng php thm chun, ta thm vo dd xc nh mt lng dd tiu
chun.C 2 cch thc hin:Dng mt dung dch chun v p dng cng thc tnhDng th
biu dinPhng php thm chunDng cng thc tnhPha mt dung dch c th tch V
cha ion cn xc nh c hm lng rt nh. Sau tin hnh pha 3 dung dch nh
sau:STTBnh 1Bnh 2Bnh 3Dung dch mu VXVx0Dung dch chun C0Vc0Thuc th
to mu, m pHthmthmThm(mu trng)215/16/2014
Phng php thm chun225/16/2014Ly 20,00mL dung dch mu c cha st cho
to phc vi thuc th 1,10 phenaltroline nh mc thnh 50,00mL. o o hp ca
dung dch = 510nm c gi tr A = 0,225 (cuvt c l = 1cm).Ly 20,00mL dung
dch mu cha st khc thm vo 4mL dung dch st chun 10 mgFe/L cho to phc
v nh mc thnh 50,00mL. o o hp ca dung dch = 510nm c gi tr A = 0,358.
Tnh nng ppm ca dung dch mu st ban u.a. 3,38b. 1,35c. 0,80 d. 27V
dPhng php thm chun(s dng th)Nng Mg sau khi nh mcXX+0.1X+0.2X+0.3100
mlDung miNo.1No.2No.4No.310 mlMu cha bit hm lng10 ml10 ml10 ml10
ml1.0 ppm X dung dch chun (ppm : mg/1000ml)20 ml30 ml10 mlng chun
ca phng php thm chun
Nng muPhng php thm chun(s dng th)16/05/2014y l phng php thng dng
trong sc k kh, phng php cho php xc nh chnh xc nng ca cht phn tch m
khng b nh hng ng k bi th tch mu gia cc ln tim. Trong phng php ny,
ta thm vo tt c cc dung dch (mu, chun) mt lng xc nh (mIS hoc CIS) ca
cht c chn lm cht ni chun. Cht ni chun c th l mt cht l v cng c th l
mt cu t no sn c trn sc k , vi iu kin l sau y Cht ni chun phi c thi
gian lu gn vi thi gian lu ca cu t cn phn tch. Pic ca cht ni chun
phi gn vi pic ca cht cn phn tchPhng php ni chun (Internal
standard)16/05/2014
Cht ni chunCht cn xc nhPhng php ni chun (Internal standard)u im
ca phng php ni chun l hn ch sai s do th tch mu vo ct sc k khng ng u
gia cc ln tim v khc phc nhng nhc im ca phng php ngoi chun.Internal
Standard MethodH s p ng - Response Factor (RF)H s RF c tnh theo cng
thc sau
Ax, Ais l din tch ca cht phn tch v cht ni chunmx, mis khi lng ca
cht phn tch v cht ni chun c bm voCh Khi phn tch mt mu khng bit hm
lng, cht ni chun bit c hm lng thm vo mu trc khi timTrong qu trnh
thc hnh, chng ta nn chun b mt dy cht chun v lp mt ng thng gia
Ax/Ais v mx/mis. Chng ta o t s Ax/Ais t sc k v tnh t s mx/mis cho
cht cn phn tch. T ta c th tnh c cht cn phn tch mx.
16/05/2014Sc k mu v kt qu tch phn
ISXAIS = 17.80AX = 27.01
16/05/2014ng chun vi cht ni chunCht chunCht ni chun c nh, cht
chun thay ing chun l ng thng di qua gc ta MuThm hm lng ca cht ni
chunTim v o c gi tr Ax/AisXc nh gi tr cx/cis cho mu ca bn t ng
chun. Vi cis bit, cx rt d dng c tnh theo cng thc
cx = (cx/cis)cis
*This expression for the response factor is not used directly in
your calculations. The following expression which accounts for the
intercept is more rigorous (in practice the intercept is very near
zero). Calculations based on the calibration data do take the
intercept into account.
16/05/2014Cch thc hin phng php ni chun Pha mt dy dung dch chun c
nng khc nhau thm cng 1 lng cht ni chun vo cc dung dch chun v mu Xy
dng th gia t s p ng v t s nng Ngoi suy t s nng mu t th bng cch s
dng h s p ng ca mu 16/05/201430
th gia t s p ng v t s nng Conc. Unknown =Conc. Ratio (from Std.
Curve) X Conc. Internal Standard16/05/201431Mt s c trng ca phng php
PTHL ng v chnh xc(Accuracy and Precision)
nhy v chn lc(sensitivity and selectivity)Gii hn pht hin v gii hn
nh lngLimit of Quantitation Limit of DetectionSignal to Noise
RationoisePeak ALODPeak BLOQBaselineLOD v LOQWorld Health
Organization16 May, 201436There are no specific criteria set for
the Limit of Quantitation (LOQ) and Limit of Detection (LOD) but
guidance is available from specifications and pharmacopeias.
The noise is measured by running the instrument at maximum gain
with no test being processed. The ripple generated is noise due to
the instruments electronics, etc. The peak is measured relative to
this noise and the ratio is calculated. This is known as the Signal
to Noise Ratio (SNR).
Generally:
LOD SNR should be greater than 2:1. Peak A is acceptable for LOD
but not for quantitation;The LOD can be calculated if the standard
deviation (SD) of the response (which is standard deviation of the
blank) and the slope is determined: LOD = 3.3 x SD slope
Similarly LOQ = 10 x SD slope
The LOQ SNR should generally be above 10:1. Peak B is suitable
for quantitation;
Precision as a percentage relative standard deviation should be
5 - 10% at the limit for LOQ.Khong nng phn tch
K thut phn tchK thutNguyn tc o lngPhn tch khi lngo khi lng cht
phn tchChun Th tch dd chun phn ng vi cht phn tchQuang ph phn t v
nguyn tBc sng v cng hp th hoc pht xKhi phT l M/zPhn tch nhitTnh cht
l ha ca cht phn tch khi thay i nhit Phn tch in haTnh cht in ca cht
phn tch trong ddPhn tch phng xBc x ht nhn ca cht phn tchPhng php
phn tchL mt bn chi tit phn tch nh sau:Thuc th, ha cht Thit b o lng,
dng c th nghimCch chun ha thuc th, thit bCch x l muCc bc phn tch
(quy trnh)Cch thu thp v x l s liuMu01234
CN00,10,150,20,25
A00,2460,3610,5120,819
Sheet1GC Calibration Curve for Cocaine with Internal
StandardStandardCocainemg/mLInt.
Std.mg/mLcx/cisAxAixAx/Ais12.505.000.5001206000.20025.005.001.0002416010.401310.005.002.0004806000.800425.005.005.00011986001.997
Sheet1
cx/cisAx/AisCocaine with Interal Standard1.0 microliter
injections
Sheet2
Sheet3
