d3 Poltekes Pertemuan 2

Langkah Analisis Toksikologi Forensik

Interpretasi

Penulisan Laporan (Bukti Surat / Surat Keterangan / Keterangan Ahli)

Penyiapan Sampel

Analisis:Uji Penapisan (Screening test)

Uji Pemastian (Confirmation test)

Penetapan kadar (determinasi)

Data Analisis

PENETAPAN KADAR

Penetapan Kadar = KUANTITATIF

SPEKTROFOTOMETER

KROMATOGRAFI LAPIS TIPIS-SPEKTROFOTODENSITOMETER

KROMATOGRAFI CAIR KINERJA TINGGI (HPLC)

KROMATOGRAFI GAS

AAS

CALIBRATION METHODS

CALIBRATION METHODS1. Calibration Curve

Method (External Standard)

2. Standard Additions Method

3. Internal Standard Method

CALIBRATION CURVE METHOD

1. Most convenient when a large number of similar samples are to be analyzed.

2. Most common technique.3. Facilitates calculation of Figures of

Merit.

CALIBRATION CURVE PROCEDURE

1. Prepare a series of standard solutions (analyte solutions with known concentrations).

2. Plot [analyte] vs. Analytical Signal.

3. Use signal for unknown to find [analyte].

For many analytical techniques, we need to evaluate the response of the unknown sample against a set of standards (known quantities).

1) Determine the instrumental responses for the standards.

2) Find the response of the unknown sample.

3) Compare the response of the unknown sample to that from the standards to determine the concentration of the unknown.

This involves a calibration!

Concentration/ mg.l-1

Absorbance Correctedabsorbance

0 0.002 0.0001 0.078 0.0762 0.163 0.1614 0.297 0.2956 0.464 0.4628 0.600 0.598

Corrected absorbance = (sample absorbance) – (blank absorbance)

Example 1

I prepared 6 solutions with a known concentration of Cr6+ and added the necessary colouring agents. I then used a UV-vis spectrophotometer and measured the absorbance for each solution at a particular wavelength. The results are in the table below.

Fit best straight line:

y = 0.075x + 0.003

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Conc / mg/l

Abs

I then measured my sample to have an absorbance of 0.418 and the blank, 0.003. I can calculate the concentration using my calibration curve.

y = 0.0750x + 0.0029

Abs = (0.0750 x Conc) + 0.0029Conc = (Abs – 0.0029)/0.0750

For my unknown: Corrected absorbance = 0.418 – 0.003 = 0.415

Conc = (0.415 – 0.0029)/0.0750 Conc = 5.49 mg.l-1

Check on your calibration curve!!

y = 0.075x + 0.003

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Conc / mg/l

Abs

Absorbance = 0.415

Conc = 5.49 mg.l-1

0.00

0.50

1.00

1.50

2.00

2.50

-1.00 -0.50 0.00 0.50 1.00 1.50

log(Pb concentration)

log(

Sign

al)

y = 0.0865 x + 0.853

Not Linear??

Remember

S = mC + b

log(S) = log (mC + b)

b must be ZERO!!

log(S) = log(m) + log(C)

The original curve did not pass through the origin. We must subtract the blank signal from each point.

Corrected Data

[Pb] Signal(ppb) (mAbs)0.20 1.070.50 2.831.50 8.232.50 14.103.50 19.494.50 24.405.50 30.94

18.50 97.59

log[Pb] log(S)

-0.70 0.03-0.30 0.450.18 0.920.40 1.150.54 1.290.65 1.390.74 1.491.27 1.99

Linear!

y = 0.9965x + 0.7419

0.00

0.50

1.00

1.50

2.00

2.50

-1.00 -0.50 0.00 0.50 1.00 1.50

log(Pb concentration)

log(

sign

al)

Standard Addition Method

1. Most convenient when a small number of samples are to be analyzed.

2. Useful when the analyte is present in a complicated matrix and no ideal blank is available.

Standard Addition Procedure

1. Add one or more increments of a standard solution to sample aliquots of the same size. Each mixture is then diluted to the same volume.

2. Prepare a plot of Analytical Signal versus:a) volume of standard solution added, orb) concentration of analyte added.

Standard Addition Procedure

3. The x-intercept of the standard addition plot corresponds to the amount of analyte that must have been present in the sample (after accounting for dilution).

4. The standard addition method assumes:a) the curve is linear over the concentration rangeb) the y-intercept of a calibration curve would be 0

Example: Fe in Drinking Water

Sample Volume

(mL)

Standard Volume

(mL) Signal (V)

10 0 0.21510 5 0.42410 10 0.68510 15 0.82610 20 0.967

The concentration of the Fe standard solution is 11.1 ppm

All solutions are diluted to a final volume of 50 mL

Contoh

[Fe] = ?

x-intercept = -6.08 mL

Therefore, 10 mL of sample diluted to 50 mL would give a signal equivalent to 6.08 mL of standard diluted to 50 mL.

Vsam x [Fe]sam = Vstd x [Fe]std

10.0 mL x [Fe] = 6.08 mL x 11.1 ppm

[Fe] = 6.75 ppm

Internal Standard MethodA second compound, not the analyte, added at an appropriate stage in the assay to correct for systematic errors in the analysis.

WHY?1.Most convenient when variations in analytical sample size, position, or matrix limit the precision of a technique.2.May correct for certain types of noise.

Internal StandardAn internal standard must:• Be completely resolved from the known and unknown substances

in the chromatogram• Elute near to (preferably just after the last) peak(s) of interest• Have a similar detector response (peak height or area) to the

analyte(s)• Have similar chemical and physical properties to the analyte(s)• Undergo any derivatization reaction in the same way as the

analyte(s)• Be chemically and physically stable on storage in solution and

during the analysis• Be easily available with adequate purity

Internal Standard Method• Internal standard’ method is often used to reduce the

impact of systematic errors such as variations in injection volume or evaporation of extraction solvent during the analysis.

• A known amount of the internal standard that behaves similarly to the analyte during the analysis, but elutes at a different place on the chromatogram or is otherwise detected independently of the analyte is added at an appropriate stage in the analysis.

• Subsequently, the detector response of the analyte relative to the response of the internal standard is plotted against analyte concentration when constructing a calibration graph.

Internal Standard Procedure

1. Prepare a set of standard solutions for analyte (A) as with the calibration curve method, but add a constant amount of a second species (B) to each solution.

2. Prepare a plot of SA/SB versus [A].

Example: Pb by ICP EmissionEach Pb solution contains 100 ppm Cu.

[Pb] (ppm) Pb Cu Pb/Cu

20 112 1347 0.08340 243 1527 0.15960 326 1383 0.23680 355 1135 0.313100 558 1440 0.388

Signal

No Internal Standard Correction

0

100

200

300

400

500

600

0 20 40 60 80 100 120

[Pb] (ppm)

Pb E

mis

sion

Sig

nal

0.000

0.050

0.100

0.150

0.200

0.250

0.300

0.350

0.400

0.450

0 20 40 60 80 100 120

[Pb] (ppm)

Pb E

mis

sion

Sig

nal

Internal Standard Correction

Results for an unknown sample after adding 100 ppm Cu

Run Pb Cu Pb/Cu

1 346 1426 0.2432 297 1229 0.2423 328 1366 0.2404 331 1371 0.2415 324 1356 0.239

mean 325 1350 0.241σ 17.8 72.7 0.00144S/N 18.2 18.6 167

Signal

TERIMA KASIH