Transcript Calibration Techniques 1. Calibration Curve Method 2. Standard Additions Method 3. Internal Standard Method.

Calibration Techniques
1. Calibration Curve Method
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].
Example: Pb in Blood by GFAAS
[Pb]
(ppb)
0.50
1.50
2.50
3.50
4.50
5.50
Signal
(mAbs)
3.76
9.16
15.03
20.42
25.33
31.87
Results of linear regression:
S = mC + b
m = 5.56 mAbs/ppb
b = 0.93 mAbs
35
30
y = 5.56x + 0.93
mAbs
25
20
15
10
5
0
0
1
2
3
Pb Concentration (ppb)
4
5
6
A sample containing an unknown amount
of Pb gives a signal of 27.5 mAbs.
Calculate the Pb concentration.
S = mC + b
C = (S - b) / m
C = (27.5 mAbs – 0.92 mAbs) / 5.56 mAbs / ppb
C = 4.78 ppb
(3 significant figures)
Calculate the LOD for Pb
20 blank measurements gives an average signal
0.92 mAbs
with a standard deviation of
σbl = 0.36 mAbs
LOD = 3 σbl/m = 3 x 0.36 mAbs / 5.56 mAbs/ppb
LOD = 0.2 ppb
(1 significant figure)
Find the LDR for Pb
Lower end = LOD = 0.2 ppb
(include this point on the calibration curve)
SLOD = 5.56 x 0.2 + 0.93 = 2.0 mAbs
(0.2 ppb , 2.0 mAbs)
Find the LDR for Pb
Upper end = collect points beyond the linear region
and estimate the 95% point.
Suppose a standard containing 18.5 ppb gives rise
to s signal of 98.52 mAbs
This is approximately 5% below the expected value
of 103.71 mAbs
(18.50 ppb , 98.52 mAbs)
Find the LDR for Pb
LDR = 0.2 ppb to 18.50 ppb
or
LDR = log(18.5) – log(0.2) = 1.97
2.0 orders of magnitude
or
2.0 decades
Find the Linearity
Calculate the slope of the log-log plot
log[Pb]
log(S)
-0.70
-0.30
0.18
0.40
0.54
0.65
0.74
1.27
0.30
0.58
0.96
1.18
1.31
1.40
1.50
1.99
Not Linear??
2.50
y = 0.0865 x + 0.853
log(Signal)
2.00
1.50
1.00
0.50
0.00
-1.00
-0.50
0.00
0.50
log(Pb concentration)
1.00
1.50
Not Linear??
120
100
Signal (mAbs)
80
60
40
20
0
0
2
4
6
8
10
12
Pb Concentration (ppb)
14
16
18
20
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]
(ppb)
0.20
0.50
1.50
2.50
3.50
4.50
5.50
18.50
Signal
(mAbs)
1.07
2.83
8.23
14.10
19.49
24.40
30.94
97.59
log[Pb]
log(S)
-0.70
-0.30
0.18
0.40
0.54
0.65
0.74
1.27
0.03
0.45
0.92
1.15
1.29
1.39
1.49
1.99
Linear!
2.50
y = 0.9965x + 0.7419
log(signal)
2.00
1.50
1.00
0.50
0.00
-1.00
-0.50
0.00
0.50
log(Pb concentration)
1.00
1.50
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, or
b) 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 range
b) the y-intercept of a calibration curve would be 0
Example: Fe in Drinking Water
Sample
Volume
(mL)
10
10
10
10
10
Standard
Volume
(mL)
Signal (V)
0
5
10
15
20
0.215
0.424
0.685
0.826
0.967
The concentration
of the Fe standard
solution is 11.1
ppm
All solutions are
diluted to a final
volume of 50 mL
1.2
1
Signal (V)
0.8
0.6
-6.08 mL
0.4
0.2
0
-10
-5
0
5
10
-0.2
Volume of standard added (mL)
15
20
25
[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 Method
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 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].
Notes
1. The resulting measurement will be
independent of sample size and
position.
2. Species A & B must not produce signals
that interfere with each other. Usually
they are separated by wavelength or
time.
Example: Pb by ICP Emission
Each Pb solution contains
100 ppm Cu.
Signal
[Pb]
(ppm)
Pb
Cu
20
40
60
80
100
112
243
326
355
558
1347
1527
1383
1135
1440
Pb/Cu
0.083
0.159
0.236
0.313
0.388
No Internal Standard Correction
600
Pb Emission Signal
500
400
300
200
100
0
0
20
40
60
[Pb] (ppm)
80
100
120
Internal Standard Correction
0.450
0.400
Pb Emission Signal
0.350
0.300
0.250
0.200
0.150
0.100
0.050
0.000
0
20
40
60
[Pb] (ppm)
80
100
120
Results for an unknown sample after
adding 100 ppm Cu
Run
Pb
Signal
Cu
1
2
3
4
5
346
297
328
331
324
1426
1229
1366
1371
1356
mean
σ
S/N
325
17.8
18.2
1350
72.7
18.6
Pb/Cu
0.243
0.242
0.240
0.241
0.239
0.241
0.00144
167