CALIBRATION METHODS

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CALIBRATION METHODS
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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.

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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)
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Calibration curve
Response dependant variable y
Concentration independant variable x
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Fit best straight line
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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!!
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Absorbance 0.415
Conc 5.49 mg.l-1
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How do we find the best straight line to pass
through the experimental points???
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METHOD OF LEAST SQUARES

• 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.

Minimise only the vertical deviations ? assume
that the error in the y-values are greater than
that in the x-values.
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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)
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Example 1
Slope
y-intercept
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Recall
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The vertical deviation can be calculated as
follows di yi (mxi c)
Some deviations are positive (point lies above
the curve) and some are negative (point 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

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Example 1
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How reliable are the least squares parameters?
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Standard deviation for the slope (m)
Standard deviation for the intercept (c)
Estimate the standard deviation for all y values.
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Example 1
Sy 7.79 x 10-3 Sy2 6.08 x 10-5
Sm2 1.28 x 10-6 Sm 0.00113
Sc2 2.58 x 10-5 Sc 0.00508
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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.
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CORRELATION COEFFIECIENT ? used as a measure of
the correlation between two variables (x and y).
The Pearson correlation coefficient is calculated
as follows
r 1 ? An exact correlation between the 2
variables r 0 ? Complete independence of
variables
In general 0.90 lt r lt 0.95 ? fair curve 0.95
lt r lt 0.99 ? good curve r gt 0.99 ?
excellent linearity
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Example 1
Correlation coefficient
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• 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.

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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.
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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.
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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.
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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.
Therefore
Or we could use CiVi CfVf
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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

sample
of
conc
Final

std
of
conc
Final
Hence
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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.
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The result
standard
5 mL sample
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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 Volume of std / ml Peak current /?A
10 0 8
10 2 16
10 5 25
10 10 41
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1 - Calculate the concentration of added Au to
each sample.
CiVi CfVf
Volume of std / ml Conc. of added std / ppm Peak current /?A
0 8
2 16
5 25
10 41
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2 - Find the best straight line
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
? 8.5 ? 90 ? 283.5 ? 32.25
y 6.50x 8.68
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3 - Extrapolate to the x-axis (y0)
y 6.50x 8.68
4 - Take dilutions into account
Conc of original sample 2.66 ppm
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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.
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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.
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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.
The response factor (F) is 1.5 times greater for
X than for S.
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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