1. CALIBRATION METHODS

2. 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). 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. This involves a calibration!

3. Concentration / mg.l -1 AbsorbanceCorrected absorbance 00.0020.000 10.0780.076 20.1630.161 40.2970.295 60.4640.462 80.6000.598 Corrected absorbance = (sample absorbance) – (blank absorbance) Example 1 I prepared 6 solutions with a known concentration of Cr 6+ 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.

4. Calibration curve: Response = dependent variable = yConcentration = independent variable = x

5. Fit best straight line:

6. 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!!

7. Absorbance = 0.415 Conc = 5.49 mg.l -1

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

9. 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

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 = (x i, y i ) for i = 1 to n (n= total number of points) Note: Same denominator

11. xixi yiyi xiyixiyi xi2xi2 00.00000 10.076 1 20.1610.3224 40.2951.18016 60.4622.77236 80.5984.78464  x i = 21  y i = 1.592  x i y i = 9.134  (x i 2 ) = 121 Example 1 Slope:y-intercept:

12. The good news: Usually done is some kind of spreadsheet

13. The vertical deviation can be calculated as follows: d i = y i – (mx i + c) Our aim  to reduce the deviations  square the values so that the sign does not play a role. d i 2 = (y i – mx i - c) 2 Some deviations are positive (point lies above the curve) and some are negative (point lies below the curve).

14. xixi yiyi didi di2di2 00.000-0.00298.41 x 10 -6 10.076-0.00256.25 x 10 -6 20.1610.00775.93 x 10 -5 40.295-0.00796.24 x 10 -5 60.4620.00959.02 x 10 -5 80.598-0.00411.68 x 10 -5  (d i 2 ) = 2.43 x 10 -4 Example 1

15. How reliable are the least squares parameters?

16. Estimate the standard deviation for all y values. Standard deviation for the slope (m): Standard deviation for the intercept (c):

17. xixi xi2xi2 di2di2 008.41 x 10 -6 116.25 x 10 -6 245.93 x 10 -5 4166.24 x 10 -5 6369.02 x 10 -5 8641.68 x 10 -5  x i = 21  (x i 2 ) = 121  (d i 2 ) = 2.43 x 10 -4 S y = 7.79 x 10 -3 S y 2 = 6.08 x 10 -5 S m 2 = 1.28 x 10 -6 S m = 0.00113 S c 2 = 2.58 x 10 -5 S c = 0.00508 Example 1

18. What does this mean? Slope = Intercept = The first decimal place of the standard deviation is the last significant figure of the slope or intercept. S y = 7.79 x 10 -3 S m = 0.00113 S c = 0.00508

19. 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 < r < 0.95  fair curve 0.95 < r < 0.99  good curve r > 0.99  excellent linearity

20. xixi yiyi xiyixiyi xi2xi2 yi2yi2 00.00000 10.076 10.00578 20.1610.32240.0259 40.2951.180160.0870 60.4622.772360.213 80.5984.784640.358  x i = 21  y i = 1.594  x i y i = 9.134  (x i 2 ) = 121  (y i 2 ) = 0.690 Example 1 Correlation coefficient:

21. 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.

22. 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. The MATRIX is: STANDARD ADDITION

23. 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. Assumption: The matrix will have the same effect on the analyte in the standard as it would on the original analyte in the sample. NB: Standard must be the same as the analyte!

24. 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”!

25. 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. C i V i = C f V f Standard: For the mixture of sample and 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.

26. For the mixture of sample and standard: Hence: 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.

27. 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. More than ONE “sample + standard”!

28. The result: 5 mL sample 5 mL10 mL15 mL20 mL standard NB: take dilution into account to find initial conc in sample

29. 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 1008 216 10525 10 41

30. 1 - Calculate the concentration of added Au to each sample. Volume of std / ml Conc. of added std / ppm Peak current /  A 08 216 525 1041

31. 2 - Find the best straight line Conc. Added x i Peak hgt y i x i y i xi2 xi2 0800 116 1 2.52562.56.25 54120525  8.5  90  283.5  32.25

32. y = 6.50x + 8.68  Conc of sample analysed = Conc of original sample =  C i V i = C f V f  3 - Extrapolate to the x-axis (y=0) 4 - Take dilutions into account

33. 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.

34. Concept: conc(sample + spike) = conc(spike) + conc(sample) IF THERE ARE NO MATRIC EFFECTS % recovery > 100% if the matrix enhances the signal % recovery < 100% if the matrix depresses the signal % recovery = 100% if there is no matrix effect Note: dilutions for the sample and for the spike must be the same whether combined or alone.

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

36. 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.

37. 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.

38. 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 DETECTION LIMITS