Result validation

1. Result validation

2. Exercise 1 • You’ve done an analysis to the best of your ability using the correct procedure. Is your answer correct? • possibly, hopefully • you can’t ever be sure that the answer is correct because the sample is an unknown • all you can do is provide evidence that the answer “should” be correct • calibration • blanks • controls • spikes • retesting

3. Calibration • standards provide a day-to-day measure of an instrument response • allow you to determine concentration • do not necessarily provide a measurement of the ongoing performance of the instrument • also means the running of special standards which do not relate to sample analysis, but to instrument performance • also known as Performance Qualification

4. Blanks • “samples” with no analyte • field blanks – to check on contamination from the collection procedure • transport blanks – kept in the transport vehicle at the site • container blanks – prepared at the laboratory from a container which is identical to those used for field collection • matrix blanks – materials of similar matrix to the samples • reagent blanks – some analytical methods require a number of sample treatment steps involving addition of different chemicals Example 1 • 10 mL of conc. HCl added to river water sample (25 ug/L Pb) • the solution made up to 100 mL with ultra-pure water • acid contains 0.1 mg/L (0.00001 %w/v) of lead • the additional lead from the acid is 10 mL x 0.1 mg/L ÷ 100 mL = 10 ug/L • the reagent has added about 40% error to the answer

5. Exercise 2 How can blanks be used? • a zero (egspectrostd) (only if a low value) • an early sample to check level • if high – find a different source of the reagent

6. Controls • sources of error other than contamination: • matrix interference • method error • instrument malfunction • operator error • for certain sample types, it is possible to have standard samples, known as controls or certified reference materials (CRM) • samples of the same matrix, which have known levels of the analytes • purchased or prepared in the laboratory. • taken through the same procedure as the normal sample • the result compared with the “true value” • used to judge the accuracy of the real samples

7. Example 2 • standard sample of flour known to contain 450 mg/kg of sodium • analysed together with “real” samples of flour • found to contain 425 mg/kg • standard sample is reading 5.6% low [(425-450)/450], • reasonable to assume that their results are similarly low • you would not simply subtract 5.6% from their results!!

8. Spikes (recovery checks) • the addition of a known amount of analyte (known as a spike) to the sample, which is then analysed as normal • increase in concentration compared to how much in sample and how much was added • if a procedure works perfectly, then 100% of the added analyte will be recovered • e.g. sample contains 5 mg of analyte, added spike contains 5 mg, so spiked sample should contain 10 • if not, then the value for the sample is wrong • 90-110% recovery generally considered OK

9. the percent recovered is a measure of how accurate the answer for your sample is • if 50% of the added analyte is recovered, it suggest that 50% of the analyte in the sample is being missed as well • you don’t then adjust the sample answer accordingly, just report both results. • recovery check process is similar in performance to standard addition • done for totally different reasons • standard addition is done to determine the sample concentration • an analytical procedure • recovery check requires that you know the concentration in the sample • a check on the analytical procedure

10. Retesting • analysing a duplicate sample placed in the batch • re-running a calibration standard after a certain number of samples (known as a re-slope) • checks for instrumental drift • analysing the sample by an alternative method • another person (or laboratory) analysing the sample

11. The equation itself • SPS is amount found in spiked sample • S is amount found in sample • SP is amount added as spike • found means in analysis, added means as standard • amount can refer to mass, volume or concentration. • the final calculation is easy • getting the three numbers to plug into the equation is not easy, • it is easy to put the wrong number in when there are so many numbers to choose from

12. Two basic rules & something to remember • All three values must be in exactly the same unit, eg mg, g/L, %w/w • All three values must be from the same stage of the analysis, eg the original sample, the first solution, the analysed solution • there is no one way to do these, just one correct answer • 1 mL of 1000 mg/L contains 1 mg of analyte

13. Example 4 • 20 mL sample of river water is diluted to 100 mL • found to contain an average of 2 mg/L of Na • another 20 mL aliquot spiked with 2 mL of 50 mg/L Na, made up to 100 mL • found to contain 2.9 mg/L

14. Example 4 • the best unit to choose is mg/L • the only one in common • this procedure has only two stages: • the original river water • the diluted solution which is analysed • Which stage do these mg/L numbers belong to? • S : 2 – diluted • SP: 50 – neither (it is a standard in another bottle) • SPS: 2.9 – diluted

15. Example 4 • simplest to work out what the mg/L of SP in the diluted solution is • use the dilution equation C1V1 = C2V2. • 50 x 2 = ? x 100 • SP = 1 mg/L

16. Example 4 • less obvious approach – mass • S: 20 mL of 10 mg/L = 0.2 mg • SPS: 100 mL of 2.9 mg/L = 0.29 • SP: 2 mL of 50 mg/L = 0.05 • why even consider this way? • each of the three values has to be worked out • the “mass in the sample portion” method works regardless of the type of sample (solid or liquid) and whether there are any dilutions

17. Example 5 • potato chips contain 5.0%w/w of Na • 1 g sample is spiked with 5 mL of 5000 mg/L Na • dissolved in 500 mL • contains 140 mg/L of Na

18. Example 5 Approach 1 – concentration based • Use mg/L as the common unit - SPS is already done • For S, how many mg are in 1 g of 5%w/w? The conc. in 500 mL? • 5g/100g means 0.05 g/1 g or 50 mg • dissolved in 500 mL => 100 mg/L • For SP, what is the final concentration? • 5000 mg/L x 5 mL = ? x 500 mL • ? = 50 mg/L

19. Example 5 Approach 2 – mass based • Use mg – all 3 have to be worked out • SPS: 500 mL of 140 mg/L = 70 mg • S: done in Approach 1: 50 mg • SP: 5 mL of 5000 mg/L = 25 mg

20. Exercise 3(a) • 10 mL of sample (known concentration 100 mg/L) is spiked with 0.5 mL of 1000 mg/L and diluted to 250 mL. This solution, when analysed, has a concentration of 6.1 mg/L.

21. Exercise 3(a) Approach 1 – diluted soln, mg/L • SPS already known: 6.1 • S: 10 mL of 100 mg/L to 250 mL of 4 • SP: 0.5 mL of 1000 mg/L to 250 mL of 2

22. Exercise 3(a) Approach 2 – mass, mg • S: 10 mL of 100 mg/L is 1 mg • SP: 0.5 mL of 1000 mg/L is 0.5 mg • SPS: 250 mL of 6.1 mg/L is 1.525 mg

23. Exercise 3(b) • 5 g of sample (known concentration 50 mg/kg) is spiked with 5 mL of 50 mg/L, and made to 100 mL. The concentration of this solution is 4.4 mg/L.

24. Exercise 8(b) Approach 1 – soln, mg/L • SPS already known: 4.4 • S: 5 g of 50 mg/kg is 0.25 mg, in 100 mL is 2.5 • SP: 5 mL of 50 mg/L to 100 mL of 2.5

25. Exercise 8(b) Approach 2 – mass, mg • S only known from calcs in #1: 0.25 • SPS: 100 mL of 4.4 mg/L is 0.44 • SP: 5 mL of 40 mg/L is 0.25

26. Exercise 3(c) • The concentration in the beer is found to be 52 mg/L. A 50 mL aliquot of the beer is spiked with 2 mL of 1000 mg/L Na, made up to 100 mL and this solution found to contain 47 mg/L Na.

27. Exercise 8(c) Approach 1 – analysed soln, mg/L • SPS already known: 47 • S: 50 mL of 52 diluted to 100 mL of 26 • SP: 2 mL of 1000 mg/L to 100 mL of 20

28. Exercise 8(c) Approach 2 – mass, mg • S: 50 mL of 52 mg/L: 2.6 • SPS: 100 mL of 47 mg/L is 4.7 • SP: 2 mL of 1000 mg/L is 2