5 Standard addition PDF

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5.1 Introduction An interference is anything that causes an analysis to be incorrect, which means in practice, that the measured concentration for the presented sample is wrong. There are four basic sources of error:  operator  equipment  method  sample There are various ways of reducing these errors and some of these approaches are dealt with other subjects; here we will concentrate on modification of the calibration standards to deal with two basic interferences:  those caused by the matrix (i.e. sample)  variations in the equipment 5.2 Matrix interference In most samples, the matrix is the major component of the sample, the analyte having a relatively small concentration. Therefore, it is not surprising that the chemical substances that make up the matrix may have an effect on the analysis, and most importantly, the measured response which we use to determine the concentration of the analyte. Equation 1 is a very simple statement of what all analytical techniques rely on to determine analytical concentration. You may never have seen it expressed quite like this before, but no matter the method, this is its basis. R = kC

Eqn 5.1

where R is the measured response (mass, volume, absorbance, peak height etc), k is a constant for the analyte under the analysis conditions and C is the concentration of analyte. This relationship is established by the calibration standards. When matrix interference occurs, it alters the relationship between R and C. There are two possible ways that the relationship can be altered:  Type 1 – the matrix causes the analyte to respond differently, therefore changing the value of k (Equation 2)  Type 2 – the matrix causes a response of its own; therefore, an extra factor (B) is added to the equation (Equation 3) R = k’C

Eqn 5.2

R = kC + B

Eqn 5.3

Whichever occurs (or possibly both in combination), the answer obtained from the analysis of the sample will be incorrect. This is shown in Figure 1.

5. Standard Addition

Standards Added B Changed K

Response

Sample

(b)

(c) 0

0.5

1

1.5

(a)

2

2.5

3

3.5

4

4.5

Conce ntration FIGURE 5.1 Effect on analyte answer due to matrix interference

What does Figure 5.1 tell you? Sample (a) is responding in the same way as the standards, sample (b) has something in the matrix that changes the analyte’s response and sample (c) has something in the matrix that is causing a response. If you are using simple standards, you will determine the sample concentration as 2; but if matrix interference is occurring, the real answer could be very different. Another way of looking at this – three samples each with an analyte concentration of 1, but different matrices. They will give three very different responses (look at the left hand arrow), and if you are using a simple sets of standards, only one will be correct. The only way to deal with these problems is to modify the standards to make them more like the matrix:  matrix-matched standards – add matrix components to the standards to make them as close as possible in composition (or obtain matrix-matched standards)  standard addition – add known amount of analyte to aliquots of sample TABLE 5.1 Comparison of methods for countering matrix interference

Advantages Matrix-matched standards

 deal with both Type 1 & 2 interferences

Disadvantages  can be hard to duplicate matrix  can be expensive to obtain

 do not create extra solutions for analysis Standard 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)

5.2

5. Standard Addition

The matter about extra solutions requires some comment. With matrix-matched standards, the situation is as per normal – 3 or 4 standards plus however many samples you happen to have. With standard addition, each sample has its own set of standards, meaning that 100 samples would in principle create 300-400 solutions to analyse. The only out from this is if you can be absolutely sure that the matrix is the same across the samples. In this case, you could apply the graph from one to the others – but this is making a big assumption. E XERCISE 5.1 Which would be the better option - matrix matched standards or standard addition - in the following analyses? nickel in steel iron in Cornflakes lead in soil

Standard addition Samples requiring standard addition need to be at the lower end of the working range, so that two or three additions can be made, each giving a reasonable increase in response without going out of the top of the range. For example, if measuring an analyte on the AAS, 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 may be trial and error at the start, since the response can’t be judged from the calibration standards. The graph associated with standard addition is very different to that for normal calibration standards, as it doesn’t go through 0. This is because the point at 0 on the x-axis is the sample itself, and clearly it has some response. Figure 5.2 shows a typical standard addition graph, showing the key features.

Samples with added analyte

Mass of analyte in sample 0 Mass of analyte added FIGURE 5.2 Standard addition graph

The line of best fit created from the sample and standard additions is extended down to the horizontal axis (the response is 0). At this point is the mass of analyte in the analysed sample amount, which is the answer you want. It isn’t difficult to prove why this should be the case, but not something we will worry about here.

5.3

5. Standard Addition

Using Excel to determine this point on the graph is very simple. Add a line of best fit to the standard addition points, and get the slope. The answer from the graph is calculated as per Equation 5.4. Mass of analyte in sample 

sample response slope

Eqn 5.4

E XAMPLE 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 and the absorbance measured. Calculating the mass added Remember the good old 1 mL of 1000 mg/L contains 1 mg rule. Let's adapt that for the 100 mg/L solution used here. 1 mL of 100 mg/L contains 0.1 mg. Therefore 5 mL contains 0.5 mg etc. Volume added (mL) 0 5 10 20

Mass added (mg) 0.0 0.5 1.0 2.0

Absorbance 0.226 0.357 0.475 0.683

0.8 0.7

Slope = 0.227 0.6 0.5 0.4 0.3 0.2 1.00 mg 0.1 0 -1.5

-1

-0.5

0

0.5

1

1.5

2

2.5

Using the equation, the mass of analyte = 0.226 ÷ 0.227 = 1.00 mg. This is in the solution analysed, i.e. 50 mL of diluted solution. This 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. IMPORTANT - the extension of the graph to the horizontal axis shown above is for illustration purposes. The only time you would use this method for an "accurate" answer is if you were handdrawing the graph (and that should be an absolutely last resort). You can, however, use it to provide an estimate of the correct answer.

5.4

5. Standard Addition

E XERCISE 5.2 1. 25 mL aliquots of sample are pipetted into four beakers, 0, 100, 200 and 300 uL of 500 mg/L standard added. The absorbances are measured, and a graph prepared (see below). The sample absorbance was 0.118. (a) Calculate the mass added in the 100 uL aliquot.

(b) Use the value from (a) to complete the horizontal axis scale - each division is equal to this value.

0.4

Absorbance

0.3

0.2

0.1

0 Mass added (mg) (c) Estimate the mass of analyte in the analysed sample from the above graph.

(d) The trendline slope was found to be 1.694. Calculate the exact mass of analyte.

(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.

5.5

5. Standard Addition

E XERCISE 5.2 ( CONT'D) 2. 0.6922 g of sample is dissolved and made up to 100 mL. 10 mL aliquots of this solution are pipetted into four 100 mL vol. flasks, 0, 5, 10 & 20 mL of 250 mg/L standard added. The solution are made up to the mark and analysed. The sample absorbance was 0.205. (a) Calculate the mass added in the 5 mL aliquot of standard.

(b) Use the value from (a) to complete the horizontal axis scale - each division is equal to this value.

0.8 0.7

Absorbance

0.6 0.5 0.4 0.3 0.2 0.1 0 Mass added (mg) (c) Estimate the mass of analyte in the analysed sample from the above graph.

(d) The trendline slope was found to be 0.0964. Calculate the exact mass of analyte.

(e) Calculate the mass of analyte in the original sample.

(b) Calculate the concentration of analyte in the sample in %w/w.

5.6

5. Standard Addition

Summary At this point, it is probably worthwhile making a clear distinction between the techniques of standard addition and internal standards (covered in Chapter 8, but you may have already encountered it). Table 2 sets out the important features of each. TABLE 2 Features of standard addition and internal standards

Feature

Standard Addition

Internal Standards

Errors compensated

Matrix interference

Instrumental or technique variations

Species added

Analyte

Chemically/physically similar to analyte, but not present in sample

Method of addition

Increasing concentrations to fixed amount of sample

Fixed amount to all standards and samples

Graphical method

Response vs concentration added, extrapolate to horizontal axis

Ratio of analyte to internal standard responses vs concentration

Reason for use

Perfectly duplicates matrix

Response of internal standard changes in same proportion as analyte

What You Need To Be Able To Do     

define important terminology describe causes of analysis interferences compare methods for reducing interferences explain aspects relating to use of standard addition perform standard addition calculations

Revision Questions 1.

Which would be the appropriate choice of interference correction in the following analyses – matrix-matched standards, standard addition, internal standards: (a) sodium in soil by flame photometry (b) iron in soil by flame AAS (c) copper in brass by flame AAS (d) benzene in petrol by GC (e) ethanol in petrol by IR

2.

10 mL aliquots of sample are pipetted into 50 mL flasks, 5 & 10 mL of 100 mg/L standard added, the solutions made up to the mark, and the absorbances measured. Determine the concentration in mg/L in the original sample. mL added Sample 5 10

Absorbance 0.283 0.402 0.521

5.7

5. Standard Addition

3.

25 mL aliquots of effluent were analysed for copper by standard addition. Determine the concentration in mg/L. mg Cu added 0 0.25 0.5 0.75

4.

100, 200 and 400 uL of a 500 mg/L standard chromium solution were added to 50 mL aliquots of a sample solution. Determine the concentration of Cr in the sample solution, given the following absorbance data. uL Cr std added 0 100 200 400

5.

Absorbance 0.203 0.279 0.351 0.574

0.1018 g of a gold sample was dissolved in oxidising acid and diluted to 50.0 mL. A 5.0 mL aliquot was diluted to 500 mL, and volumes of 1000 mg/L stock solution added to 25 mL aliquots of the dilute gold solution. Calculate the percentage of gold in the sample. uL Au std added 0 100 200 300

7.

Absorbance 0.117 0.279 0.427 0.755

2.3725 g of an iron sample were dissolved in acid, and made up to 100 mL. A 5.0 mL aliquot of this solution was diluted to 500 mL. 50, 100 and 250 uL of a 1000 mg/L Fe standard solution were added to 10.0 mL aliquots of the diluted sample solution. Determine the % of Fe in the original sample, given the following absorbance data. uL Fe std added 0 50 100 250

6.

Absorbance 0.284 0.392 0.503 0.619

Absorbance 0.319 0.387 0.463 0.529

10 mL of a zinc plating bath solution was diluted to 1.00 L, and volumes of 1000 mg/L stock solution added to 50 mL of diluted sample. Calculate the g/L of zinc in the sample. uL Zn std added 0 100 200 400

Intensity 15860 23450 31070 46840

5.8...