CALIBRATION METHODS

For many analytical techniques, we need to evaluate the response of the unknown sample against a set of standards (known quantities).

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.

This involves a calibration!

Concentration/ mg.l-1

Absorbance Correctedabsorbance

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)

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.

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Conc / mg/l

Ab

s

Calibration curve:R

es

po

ns

e =

dep

en

da

nt

vari

ab

le =

y

Concentration = independant variable = x

Fit best straight line:

y = 0.075x + 0.003

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Conc / mg/l

Ab

s

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!!

y = 0.075x + 0.003

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Conc / mg/l

Ab

s

Absorbance = 0.415

Conc = 5.49 mg.l-1

How do we find the best straight line to pass through the experimental points???

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.

METHOD OF LEAST SQUARES

-0.10

0.10.20.3

0.40.50.60.7

0 2 4 6 8 10

Conc/ppm

Ab

s

Vertical deviation= y - yi

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)

2i

2i

iiii

xxn

yxyxnm

2i

2i

iiii2

i

xxn

xyxyxc

xi yi xiyi xi2

0 0.000 0 01 0.076 0.076 12 0.161 0.322 44 0.295 1.180 166 0.462 2.772 368 0.598 4.784 64xi = 21 yi = 1.592 xiyi = 9.134 (xi

2) = 121

Example 1

Slope: y-intercept:

2i

2i

iiii

xxn

yxyxnm

2i

2i

iiii2

i

xxn

xyxyxc

y = 0.075x + 0.003

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0 1 2 3 4 5 6 7 8 9

Conc / mg/l

Ab

s

Recall:

The vertical deviation can be calculated as follows:

di = yi – (mxi + c)

Our aim to reduce the deviations

square the values so that the sign does not play a role.

di2 = (yi – mxi - c)2

-0.10

0.10.20.3

0.40.50.60.7

0 2 4 6 8 10

Conc/ppm

Ab

s

Vertical deviation= y - yi

Some deviations are positive (point

lies above the curve) and some are negative (point lies below the

curve).

 

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

Example 1

How reliable are the least squares parameters?

y = 0.075x + 0.0029

00.10.20.30.40.50.60.7

0 2 4 6 8 10

Conc/ppm

Ab

s

y = 0.0742x - 0.0081

-0.10

0.10.20.30.40.50.60.7

0 2 4 6 8 10

Conc/ppm

Ab

s

Estimate the standard deviation for all y values.

2n

ds

2i

y

Standard deviation for the slope (m):

2i

2i

2y2

mxxn

nss

2

i2

i

i2

y2c

xxn

xss

2

Standard deviation for the intercept (c):

xi xi2 di

2

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 (di

2) = 2.43 x 10-4

2n

ds

2i

y

Sy = 7.79 x 10-3

Sy2 = 6.08 x 10-5

2i

2i

2y2

mxxn

nss

Sm2 = 1.28 x 10-6

Sm = 0.00113

2

i2

i

i2

y2c

xxn

xss

2

Sc2 = 2.58 x 10-5

Sc = 0.00508

Example 1

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.

Sy = 7.79 x 10-3

Sm = 0.00113

Sc = 0.00508

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

2i

2i

2i

2i

iiii

yynxxn

yxyxnr

In general: 0.90 < r < 0.95 fair curve

0.95 < r < 0.99 good curve

r > 0.99 excellent linearity

xi yi xiyi xi2 yi

2

0 0.000 0 0 0.0001 0.076 0.076 1 0.00578

2 0.161 0.322 4 0.02594 0.295 1.180 16 0.08706 0.462 2.772 36 0.2138 0.598 4.784 64 0.358xi =

21

yi =

1.594

xiyi =

9.134

(xi2) =

121

(yi2) =

0.690

Example 1

Correlation coefficient:

2i

2i

2i

2i

iiii

yynxxn

yxyxnr

0.999r

(1.594)(6)(0.690)(21)(6)(121)

)(21)(1.594(6)(9.134)r

22

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.

In a sample, the analyte is generally not isolated from other components in the sample.

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.

STANDARD ADDITION

The MATRIX is:

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.

stdsamplefromSignal

samplefromSignal

stdsampleinFeofConc

sampleinFeofConc

SX

X

ff

i

II

[S][X][X]

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.

f

i

V

V

volumefinal

volumeinitialfactorDilution

concInitialV

VconcFinal

f

i

Therefore:

Or we could use: CiVi = CfVf

For the mixture of sample and standard:

Hence:

M0.086[x]i

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.

SX

X

ff

i

II

[S][X][X]

sampleofconcFinal

stdofconcFinal

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:

5 mL sample

standard

Example 3Gold 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

1 - Calculate the concentration of added Au to each sample.

Volume of std / ml

Conc. of added std / ppm

Peak current /A

0 8

2 16

5 25

10 41

CiVi = CfVf

01020304050

0 1 2 3 4 5 6

Conc added std / ppm

Pea

k cu

rren

t /

A

2 - Find the best straight line

2i

2i

iiii

xxn

yxyxnm

2i

2i

iiii2

i

xxn

xyxyxc

Conc. Added xi

Peak hgt yi

xi yi

xi

2

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

6.50(8.5)-(4)(32.25)

(8.5)(90)(4)(283.5)m

2

8.68

(8.5)-(4)(32.25)

5)(283.5)(8.)(32.25)(90c

2

y = 6.50x + 8.68

y = 6.50x + 8.68

Conc of original sample = 2.66 ppm

y = 6.50x + 8.68R2 = 1.00

-10

0

10

20

30

40

50

-2 0 2 4 6

Conc added std / ppmP

eak

curr

ent

/

A

3 - Extrapolate to the x-axis (y=0)

4 - Take dilutions into account

An internal standard is a known concentration of a compound, different from the analyte, that is added to the unknown.

INTERNAL STANDARDS

The signal from the analyte is compared to the signal from the internal standard when determining the concentration of analyte present.

Why add an internal standard?

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.

[S]

IF

[X]

I SX

0

12

3

4

0 1 2 3 4 5 6

SX

All instrumental methods have a degree of noise associated with the measurement.

limits the amount of analyte that can be detected.

DETECTION LIMITS

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