Transcript CALIBRATION METHODS - Wits Structural Chemistry

CALIBRATION METHODS
For many analytical techniques, we need to evaluate the
response of the unknown sample against a set of
standards (known quantities).
This involves a calibration!
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.
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.
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 = dependant 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 = independant 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 using my calibration curve.
y = 0.0750x + 0.0029
Abs = (0.0750 x Conc) + 0.0029
Conc = (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!!
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
= y - yi
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
(n= total number of points)
m
n xi yi    xi  yi
 
n xi2   xi 2

x  y   x y  x

c
n x    x 
2
i
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
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

x  y   x y  x

c
n x    x 
2
i
i
2
i
i i
2
i
i
Recall:
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
= y - yi
Some deviations
are positive (point
lies above the curve)
and some are
negative (point
2
4
6
8
Conc/ppm
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 = 0.075  0.001
Intercept = 0.003  0.005
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 non-linear
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 unknown.
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
0.2 M Fe standard was added to 95 ml of the sample. The
signal obtained was 0.805.
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
Note that when we add the 5 ml standard solution to the
95 ml sample solution, we are diluting the solutions.
Total volume = 100 ml.
Thus we need to take the DILUTION FACTOR into
account.
initial volume Vi
Dilution factor 

final volume Vf
Therefore:
Vi
Final conc 
 Initial conc
Vf
Or we could use: CiVi = CfVf
Fe was analysed in a zinc electrolyte. The signal obtained from an
AAS for was 0.381 absorbance units. 5ml of a 0.2 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 
Final conc of std 
Hence:
[X]i
IX

[X]f  [S]f IX S
[x]i  0.086 M
How is this best done in practise?
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:
standard
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.
CiVi = CfVf
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)
y = 6.50x + 8.68
3 - Extrapolate to the x-axis (y=0)
y = 6.50x + 8.68
Peak current /  A
50
40
30
20
y = 6.50x + 8.68
R2 = 1.00
10
0
-2
-10 0
2
4
Conc added std / ppm
4 - Take dilutions into account
Conc of original sample = 2.66 ppm
6
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
IX
 IS 
F

[X]
 [S] 
2
3
4
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