Transcript CALIBRATION METHODS

CALIBRATION METHODS
For many analytical techniques, we need to evaluate
the response of the unknown sample against the
responses of a set of standards (have known
concentrations).
This involves a calibration!
1) Accurately make up a set of standards with a known
concentration of analyte.
2) Determine the instrumental responses for the
standards.
3) Draw a calibration curve.
4) Find the response of the unknown sample.
5) Compare the response of the unknown sample to that
from the standards to determine the concentration of the
unknown.
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 given in the table below.
Concentration
/ mg.l-1
Absorbance
Corrected
absorbance
0
0.002
0.000
1
0.078
0.076
2
0.163
0.161
4
0.297
0.295
6
0.464
0.462
8
0.600
0.598
Corrected absorbance = (sample absorbance) – (blank absorbance)
0.7
0.6
0.5
Abs
Response = dependent variable = y
Calibration curve:
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
7
Conc / mg/l
Concentration = independent variable = x
8
9
Fit best straight line:
0.7
0.6
Abs
0.5
0.4
0.3
0.2
y = 0.075x + 0.003
0.1
0
0
1
2
3
4
5
Conc / mg/l
6
7
8
9
I then measured my sample to have an absorbance of
0.418 and the blank, 0.003.
I can calculate the concentration in my sample using
my calibration curve.
y = 0.0750
x
+ 0.0029
For my unknown:
Corrected absorbance =
Check on your calibration curve!!
Absorbance = 0.415
Conc = 5.49 mg.l-1
0.7
0.6
Abs
0.5
0.4
0.3
0.2
y = 0.075x + 0.003
0.1
0
0
1
2
3
4
5
Conc / mg/l
6
7
8
9
How do we find the best straight line to pass through
the experimental points???
METHOD OF LEAST SQUARES
Minimise only the
vertical deviations
 assume that the
error in the y-values
are greater than
that in the x-values.
Abs
Assume:
 There is a linear relationship.
 Errors in the y-values (measured values) are greater
than the errors in the x-values.
 Uncertainties for all y-values are the same.
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1 0
Vertical
deviation
= yi - y
2
4
6
Conc/ppm
8
10
Recall:
Equation of a straight line:
y = mx + c
where m = slope and c = y-intercept
We thus need to calculate m and c for a set of points.
Points = (xi, yi) for i = 1 to n
m
n xi yi    xi  yi
    x 
n xi
2
2
i
(n= total number of points)

x  y   x y  x

c
n x    x 
2
i
Note: Same denominator
i
2
i
i i
2
i
i
Example 1
xi
0
1
2
4
6
8
xi = 21
yi
0.000
0.076
0.161
0.295
0.462
0.598
yi = 1.592
Slope:
m
m
xi2
0
1
4
16
36
64
(xi2) = 121
xiyi
0
0.076
0.322
1.180
2.772
4.784
xiyi = 9.134
y-intercept:
n xi yi    xi  yi
    x 
n xi
2
2
i
(6)(9.134)  (21)(1.592)
(6)(121)  (21)2
m  0.075

x  y   x y  x

c
n x    x 
2
i
i
2
i
c
i i
2
i
(121)(1.592)  (9.134)(21)
(6)(121)  (21)2
c  0.003
i
The good news:
Usually done is some kind of spreadsheet
0.7
0.6
Abs
0.5
0.4
0.3
0.2
y = 0.075x + 0.003
0.1
0
0
1
2
3
4
5
Conc / mg/l
6
7
8
9
Abs
The vertical deviation can be calculated as follows:
di = yi – (mxi + c)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1 0
Vertical
deviation
= yi - y
Some deviations
are positive (point
lies above the curve)
and some are
negative (point
2
4
6
Conc/ppm
8
10
lies below the
curve).
Our aim  to reduce the deviations
 square the values so that the sign does not play a role.
di2 = (yi – mxi - c)2
Example 1
xi
yi
di
di2
0
0.000
-0.0029
8.41 x 10-6
1
0.076
-0.0025
6.25 x 10-6
2
0.161
0.0077
5.93 x 10-5
4
0.295
-0.0079
6.24 x 10-5
6
0.462
0.0095
9.02 x 10-5
8
0.598
-0.0041
1.68 x 10-5
(di2) = 2.43 x 10-4
Abs
How reliable are the least squares parameters?
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
y = 0.075x + 0.0029
0
2
4
6
8
10
Abs
Conc/ppm
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
y = 0.0742x - 0.0081
0
2
4
6
Conc/ppm
8
10
Standard deviation for the slope (m):
2
2
sm 
sy n
 
n xi2   xi 2
Standard deviation for the intercept (c):
2
sc 
 
s y2  xi2
 
n xi2   xi 2
Estimate the standard deviation for all y values.
sy 
2

d
 i 
n2
Example 1
sy 
xi
xi2
di2
0
0
8.41 x 10-6
1
1
6.25 x 10-6
2
4
5.93 x 10-5
4
16
6.24 x 10-5
6
36
9.02 x 10-5
8
64
1.68 x 10-5
xi = 21
(xi2) = 121
(di2) = 2.43 x 10-4
 di2 
n2
2
sm 
s y2n
    x 
n xi
2
2
i
2
sc 
 
s y2  xi2
 
n xi2   xi 2
Sy = 7.79 x 10-3
Sm2 = 1.28 x 10-6
Sc2 = 2.58 x 10-5
Sy2 = 6.08 x 10-5
Sm = 0.00113
Sc = 0.00508
Sy = 7.79 x 10-3
Sm = 0.00113
Sc = 0.00508
What does this mean?
Slope =
Intercept =
The first decimal place of the standard deviation is the
last significant figure of the slope or intercept.
CORRELATION COEFFIECIENT  used as a measure
of the correlation between two variables (x and y).
The Pearson correlation coefficient is calculated as
follows:
n xi yi   xi  yi
r
2
2
2
2
n xi   xi  n yi   yi 


r = 1  An exact correlation between the 2 variables
r = 0  Complete independence of variables
In general:
0.90 < r < 0.95  fair curve
0.95 < r < 0.99  good curve
r > 0.99  excellent linearity

Example 1
xi
0
1
yi
0.000
0.076
xiyi
0
0.076
xi2
0
1
yi2
0.000
0.00578
2
4
6
8
xi =
21
0.161
0.295
0.462
0.598
yi =
1.594
0.322
1.180
2.772
4.784
xiyi =
9.134
4
16
36
64
(xi2) =
121
0.0259
0.0870
0.213
0.358
(yi2) =
0.690
Correlation coefficient:
r
r
n x
n xi yi   xi  yi
2
i

 xi 2 n yi2
(6)(9.134)  (21)(1.594)

 yi 2 
(6)(121)  (21)2 (6)(0.690)  (1.594)2 
r  0.999
Note:
 A linear calibration is preferred, although a nonlinear curve can be used.
 It is not reliable to extrapolate any calibration curve.
 With any measurement there is a degree of
uncertainty. This uncertainty is propagated as
this data is used to calculate further results.
STANDARD ADDITION
In a sample, the analyte is generally not isolated from
other components in the sample.
The MATRIX is:
Some times certain components interfere in the
analysis by either enhancing or depressing the
analytical signal  matrix effect.
BUT, the extent to which the signal is affected is
difficult to measure.
How do we circumvent the problem of matrix effects?
STANDARD ADDITION!
Add a small volume of concentrated standard solution
to a known volume of the sample.
NB: Standard must be the same as the analyte!
Assumption:
The matrix will have the same effect on the
analyte in the standard as it would on the
original analyte in the sample.
Example 2
Fe was analysed in a zinc electrolyte. The signal
obtained from an AAS for was 0.381 absorbance units.
5 ml of a 0.20 M Fe standard was added to 95 ml of the
sample. The signal obtained was 0.805.
In this case, only ONE “sample + standard”!
Conc of Fe in sample
Signal from sample

Conc of Fe in sample  std Signal from sample  std
[X]i
IX

[X]f  [S]f IX S
Fe was analysed in a zinc electrolyte. The signal obtained from an
AAS for was 0.381 absorbance units. 5ml of a 0.20 M Fe standard
was added to 95 ml of the sample. The signal obtained was 0.805.
Note that when we add the 5 ml standard solution to
the 95 ml sample solution, we are diluting both
solutions. Total volume = 100 ml.
Thus we need to take the DILUTION into account.
CiVi = CfVf
For the mixture of sample and standard:
Standard:
Sample:
Fe was analysed in a zinc electrolyte. The signal obtained from an
AAS for was 0.381 absorbance units. 5ml of a 0.20 M Fe standard
was added to 95 ml of the sample. The signal obtained was 0.805.
For the mixture of sample and standard:
Final conc of sample 0.95[x]i
Final conc of std  0.010M
Hence:
[X]i
IX

[X]f  [S]f IX S
How is this best done in practise?
More than ONE
“sample + standard”!
The solutions in all the
flasks all have the same
concentration of the matrix.
Add a quantity of standard
solution such that the
signal is increased by about
1.5 to 3 times that for the
original sample.
Analyse all solutions.
The result:
NB: take dilution
into account to
find initial conc
in sample
standard
5 mL
10 mL
15 mL
20 mL
5 mL
sample
Example 3
Gold was determined in a waste stream using
voltammetry. The peak height of the current
signal is proportional to concentration.
A standard addition analysis was done by adding specific volumes of
10 ppm Au solution to the sample as shown in the table below. All
solutions were made up to a final volume of 20 ml. The peak currents
obtained from the analyses are also tabulated below.
Calculate the concentration of Au in the original sample.
Volume of
sample / ml
10
10
10
10
Volume of std
/ ml
0
2
5
10
Peak current
/A
8
16
25
41
1 - Calculate the concentration of added Au to each sample.
Volume of std
/ ml
Conc. of added std
/ ppm
Peak current
/A
8
2
16
5
25
10
41
Peak current /  A
0
50
40
30
20
10
0
0
1
2
3
4
Conc added std / ppm
5
6
2 - Find the best straight line
m
 
n xi2   xi 2

x  y   x y  x

c
n x    x 
2
i
i
2
i i
2
i
i
Conc. Added
xi
Peak hgt
yi
xi yi
xi2
0
8
0
0
1
16
16
1
2.5
25
62.5
6.25
5
41
205
25

m
n xi yi    xi  yi
8.5

90

283.5

i
32.25
(4)(283.5) (8.5)(90)
(32.25)(90)  (283.5)(8.5)

6.50
c
 8.68
2
2
(4)(32.25)- (8.5)
(4)(32.25)- (8.5)
3 - Extrapolate to the x-axis (y=0)
y = 6.50x + 8.68
Peak current /  A
 Conc of sample
analysed =
50
40
30
20
y = 6.50x + 8.68
R2 = 1.00
10
0
-2
-10 0
2
Conc added std / ppm
4 - Take dilutions into account
CiVi = CfVf


Conc of original sample =
4
6
SPIKE RECOVERY
One way to check if there are matrix effects is to
perform a spike recovery.
Experiment:
(1) Add a small volume of concentrated standard
solution (spike) to a known volume of the
sample. (sample + spike)
(2) Add the same volume of standard solution to the
same known volume of water (+ reagent). (spike)
(3) Dilute the sample to the same extent as in (1)
(sample)
Analyse the all three samples and calculate the
concentrations using a calibration curve.
Note: dilutions for the sample and for the spike must
be the same whether combined or alone.
Concept:
conc(sample + spike) = conc(spike) + conc(sample)
IF THERE ARE NO MATRIC EFFECTS
conc(sam ple spike)
% Re cov ery 
 100
conc(sam ple)  conc(spike)
% recovery > 100% if the matrix enhances the signal
% recovery < 100% if the matrix depresses the signal
% recovery = 100% if there is no matrix effect
INTERNAL STANDARDS
An internal standard is a known concentration of a
compound, different from the analyte, that is added to
the unknown.
Why add an internal standard?
The signal from the analyte is compared to the signal
from the internal standard when determining the
concentration of analyte present.
If the instrument response varies slightly from run to
run, the internal standard can be used as an indication
of the extent of the variation.
Assumption:
If the internal standard signal increases 10% for the
same solution from one run to the other, it is most
likely that the signal from the sample also increases by
10%.
Note: If there are 2 different components in solution
with the same concentration, they need NOT have the
same signal intensity. The detector will generally give
a different response for each component.
Say analyte (X) and internal standard (S) have the
same concentration in solution.
The signal height for X may be 1.5 times greater than
that for S.
4
3
2
1
0
X
S
0
1
2
3
4
IX
 IS 
F

[X]
 [S] 
5
6
The response
factor (F) is 1.5
times greater for X
than for S.
DETECTION LIMITS
All instrumental methods have a degree of noise
associated with the measurement.
 limits the amount of analyte that can be detected.
Detection limit – the lowest concentration level that
can be determined to be statistically different from the
analyte blank.
Generally, the sample signal must be 3x the standard
deviation of the background signal