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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!

- Determine the instrumental responses for the

standards. - Find the response of the unknown sample.
- 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.

Corrected absorbance (sample absorbance)

(blank absorbance)

Calibration curve

Response dependant variable y

Concentration independant variable x

Fit best straight line

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

How do we find the best straight line to pass

through the experimental points???

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.

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)

Example 1

Slope

y-intercept

Recall

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

Example 1

How reliable are the least squares parameters?

Standard deviation for the slope (m)

Standard deviation for the intercept (c)

Estimate the standard deviation for all y values.

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

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

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

Example 1

Correlation coefficient

- 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.

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

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

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.

1 - Calculate the concentration of added Au to

each sample.

CiVi CfVf

2 - Find the best straight line

y 6.50x 8.68

3 - Extrapolate to the x-axis (y0)

y 6.50x 8.68

4 - Take dilutions into account

Conc of original sample 2.66 ppm

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.

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