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 (eg spectro std) (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 Sample2 mg/L Spike2 mL of 50 mg/L Spiked sample2.9 mg/L Procedure20 mL to 100 mL 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

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 S5.0 %w/w SP5 mL of 5000 mg/L SPS140 mg/L Procedure1 g, 500 mL

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. Sample100 mg/L Spike500 uL of 1000 mg/L Spiked sample6.1 mg/L Procedure10 mL, 250 mL

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. Sample50 mg/kg Spike5 mL of 50 mg/L Spiked sample4.4 mg/L Procedure5 g, 100 mL

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. Sample50 mg/L Spike2 mL of 1000 mg/L Spiked sample40 mg/L Procedure50 mL to 100 mL

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