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 .5 6 x + 0 .9 3
m Abs
25
20
15
10
5
0
0
1
2
3
P b C o n c e n tra tio n (p p b )
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
lo g [P b ]
lo g (S )
-0 .7 0
-0 .3 0
0 .1 8
0 .4 0
0 .5 4
0 .6 5
0 .7 4
1 .2 7
0 .3 0
0 .5 8
0 .9 6
1 .1 8
1 .3 1
1 .4 0
1 .5 0
1 .9 9
Not Linear??
2 .5 0
y = 0 .0 8 6 5 x + 0 .8 5 3
lo g (S ig n a l)
2 .0 0
1 .5 0
1 .0 0
0 .5 0
0 .0 0
-1 .0 0
-0 .5 0
0 .0 0
0 .5 0
lo g (P b c o n c e n tra tio n )
1 .0 0
1 .5 0
Not Linear??
120
100
S ig n a l (m A b s )
80
60
40
20
0
0
2
4
6
8
10
12
P b C o n c e n tra tio n (p p b )
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
[P b ]
(p p b )
0 .2 0
0 .5 0
1 .5 0
2 .5 0
3 .5 0
4 .5 0
5 .5 0
1 8 .5 0
S ig n a l
(m A b s )
1 .0 7
2 .8 3
8 .2 3
1 4 .1 0
1 9 .4 9
2 4 .4 0
3 0 .9 4
9 7 .5 9
lo g [P b ]
lo g (S )
-0 .7 0
-0 .3 0
0 .1 8
0 .4 0
0 .5 4
0 .6 5
0 .7 4
1 .2 7
0 .0 3
0 .4 5
0 .9 2
1 .1 5
1 .2 9
1 .3 9
1 .4 9
1 .9 9
Linear!
2 .5 0
y = 0 .9 9 6 5 x + 0 .7 4 1 9
lo g (s ig n a l)
2 .0 0
1 .5 0
1 .0 0
0 .5 0
0 .0 0
-1 .0 0
-0 .5 0
0 .0 0
0 .5 0
lo g (P b c o n c e n tra tio n )
1 .0 0
1 .5 0
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
S a m p le
V o lu m e
(m L )
10
10
10
10
10
S ta n d a rd
V o lu m e
(m L )
S ig n a l (V )
0
5
10
15
20
0 .2 1 5
0 .4 2 4
0 .6 8 5
0 .8 2 6
0 .9 6 7
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
S ig n a l (V )
0 .8
0 .6
-6.08 mL
0 .4
0 .2
0
-1 0
-5
0
5
10
-0 .2
V o lu m e o f s ta n d a rd a d d e d (m L )
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.
S ig n a l
[P b ]
(p p m )
Pb
Cu
20
40
60
80
100
112
243
326
355
558
1347
1527
1383
1135
1440
P b /C u
0 .0 8 3
0 .1 5 9
0 .2 3 6
0 .3 1 3
0 .3 8 8
No Internal Standard Correction
600
P b E m is s io n S ig n a l
500
400
300
200
100
0
0
20
40
60
[P b ] (p p m )
80
100
120
Internal Standard Correction
0 .4 5 0
0 .4 0 0
P b E m is s io n S ig n a l
0 .3 5 0
0 .3 0 0
0 .2 5 0
0 .2 0 0
0 .1 5 0
0 .1 0 0
0 .0 5 0
0 .0 0 0
0
20
40
60
[P b ] (p p m )
80
100
120
Results for an unknown sample after
adding 100 ppm Cu
Run
Pb
S ig n a l
Cu
1
2
3
4
5
346
297
328
331
324
1426
1229
1366
1371
1356
m ean
σ
S /N
325
1 7 .8
1 8 .2
1350
7 2 .7
1 8 .6
P b /C u
0 .2 4 3
0 .2 4 2
0 .2 4 0
0 .2 4 1
0 .2 3 9
0 .2 4 1
0 .0 0 1 4 4
167