1. Interferences

2. Matrix interference the matrix is the major component of the sample can affect the measured analyte response Response = constant x concentration relationship is established by the calibration standards matrix interference alters the relationship between R & C two possible ways that the relationship can be altered: Type 1 – the value of k changes – matrix causes analyte to responds differently (constant changes) Type 2 – matrix adds its own response; an extra factor (B) is added to the equation cause the response for the sample to be different to that of a standard with the same concentration

3. (a), (b) and (c) are three samples with different matrices, all giving the same response even though their concentration is different If you use a set of simple standards, each will be determined as having the same concentration = 2

4. AdvantagesDisadvantages Matrix-matched stds  deal with both Type 1 & 2 interferences  do not create extra solutions for analysis  can be hard to duplicate matrix  can be expensive to obtain Std addition  can be performed on any sample type  every sample creates 3-4 solutions for analysis  does not cope with Type 2 interferences  possibility of adding too much (outside linear range)

5. Exercise 7.1 a)nickel in steel matrix-match b)iron in Cornflakes matrix-match c)lead in soil std addition this Chapter (Ch. 7) will be available for downloading by next Monday we will be covering standard addition next week

6. Standard addition samples need to be at the lower end of the working range 2 or 3 additions can be made each giving a reasonable increase in response without going out of the top of the range eg an absorbance of 0.2-0.3 for the sample would allow three 0-15-0.2 Abs increases for the standard additions. working out the correct amount to add may be trial and error at the start

7. Std addn graph Mass of analyte added Samples with added analyte Mass of analyte in sample 0

8. Calculations extend line of best fit down to the horizontal axis this point is the mass of analyte in the analysed sample amount

9. Example 5.1 10 mL of sample is pipetted into each of four 50 mL volumetric flasks to these flasks is added, 0, 5, 10 and 20 mL of 100 mg/L analyte the flasks are made up the mark with water and the absorbance of each solution is measured: 0.226, 0.357, 0.475 and 0.683 respectively 1 mL of 1000 mg/L contains 1 mg 1 mL of 100 mg/L contains 0.1 mg 5 mL contains 0.5 mg and so on

10. mass of analyte = 0.226 ÷ 0.227 = 1.00 mg this is in the solution analysed, ie 50 mL of diluted solution equates to a concentration of 20 mg/L the dilution factor was 10 to 50 (DF = 5), so the original sample was 100 mg/L.

11. Exercise 5.2 25 mL aliquots of sample 0, 100, 200 and 300 uL of 500 mg/L standard added Sample abs: 0.118 concentration of analyte in the sample in mg/L a)Calculate the mass added in the 100 uL aliquot 1 mL of 1000 mg/L => 1 mg 1 mL of 500 mg/L => 0.5 mg 0.1 mL of 500 mg/L => 0.05 mg

12. Exercise 5.2 b)Use the value from (a) to complete the horizontal axis scale - each division is equal to this value. c)Estimate the mass of analyte in the analysed sample from the graph. 0.050.10.15 0.050.1 0.07

13. Exercise 5.2 d)The trendline slope was found to be 1.694. Calculate the exact mass of analyte Mass = sample abs ÷ slope = 0.118 ÷ 1.694 = 0.0696 mg which corresponds to the estimate e)Calculate the concentration of analyte in the sample in mg/L. Assume the added volumes do not cause a change in the volume from 25 mL. 0.0696 mg ÷ 0.025 L = 2.79 mg/L

14. Exercise 5.2 (2) 0.6922 g of sample is dissolved & made up to 100 mL 10 mL aliquots to 100 mL 5, 10 & 20 mL of 250 mg/L added Sample abs: 0.205 a)Calculate the mass added in the 5 mL aliquot of standard 1 mL of 1000 mg/L => 1 mg 1 mL of 250 mg/L => 0.25 mg 5 mL of 250 mg/L => 1.25 mg

15. Exercise 5.2 (2) b)Add scale to graph c)Estimate mass of analyte 1.25 2.5 5.0 1.25 2.5 2.0

16. Exercise 5.2 (2) d)The trendline slope was found to be 0.0964. Calculate the exact mass of analyte Mass = sample abs ÷ slope = 0.205 ÷ 0.0964 = 2.13 mg which corresponds to the estimate e)Calculate the mass of analyte in the original sample. 2.13 mg in 100 mL of analysed solution this contains 10 mL of sample 21.3 mg in 100 mL of original solution f)Calculate the %w/w = 100 x 21.3 ÷ 692.2 = 3.07%