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Understanding Matrix Interference in Analytical Chemistry

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This text explores matrix interference, a significant factor in analytical chemistry affecting analyte response. It discusses the constant relationship between response (R) and concentration (C) established by calibration standards and highlights how matrix interference can alter this relationship. Two types of matrix interference are defined: Type 1, where the value of k changes, and Type 2, which introduces an additional contributing factor. It also demonstrates how to use standard addition methods to compensate for matrix effects and provides exercises to calculate analyte concentration accurately.

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Understanding Matrix Interference in Analytical Chemistry

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  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. 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

  5. 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

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

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

  8. 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

  9. 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.

  10. 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

  11. 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

  12. 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

  13. 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

  14. Exercise 5.2 (2) • Add scale to graph • Estimate mass of analyte 2.0 1.25 5.0 2.5 1.25 2.5

  15. 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%

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