Interferences

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

5. Exercise 7.1 • nickel in steel • matrix-match • iron in Cornflakes • matrix-match • 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. Samples with added analyte Mass of analyte in sample 0 Mass of analyte added Std addn graph

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 • 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 • Use the value from (a) to complete the horizontal axis scale - each division is equal to this value. • Estimate the mass of analyte in the analysed sample from the graph. 0.07 0.05 0.1 0.15 0.1 0.05

13. Exercise 5.2 • 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 • 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 • 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) • Add scale to graph • Estimate mass of analyte 2.0 1.25 5.0 2.5 1.25 2.5

16. Exercise 5.2 (2) • 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 • 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 • Calculate the %w/w • = 100 x 21.3 ÷ 692.2 = 3.07%