Standard addition
5
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
Concentration
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
Disadvantages
Matrix-matched
standards
• 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
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)
Adv. Spectro
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.
EXERCISE 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.
Adv. Spectro
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
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 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.
Adv. Spectro
5.4
5. Standard Addition
EXERCISE 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.
Adv. Spectro
5.5
5. Standard Addition
EXERCISE 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.
Adv. Spectro
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
Adv. Spectro
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
Answers to these questions on following page.
Answers to class exercises can be found in the Powerpoint file provided on the website.
Adv. Spectro
5.8
5. Standard Addition
Answers to Revision Questions
1. (a) std addn & int std
(b) std addn
(c) matrix match
(d) int std
(e) matrix match
2. 5 mL contains 0.5 mg
Slope = 0.238
Mass = 0.283/0.238 = 1.19 mg
1.19 mg/50 mL = 23.8 mg/L
3. Slope = 0.446
Mass = 0.284/0.446 = 0.636 mg
0.636 mg/25 mL = 25.4 mg/L
4. 100 uL contains 0.05 mg
Slope = 3.181
Mass = 0.117/3.181 = 0.0368 mg
0.0368 mg/50 mL = 0.736 mg/L
5. 50 uL contains 0.05 mg
Slope = 1.481
Mass = 0.203/1.481 = 0.137 mg
0.137 mg/10 mL diluted sample from 500 mL total
Therefore total analyte in 500 mL VF is 50 x 0.137 = 6.852 mg
This is from 5 mL of original 100 mL sample solution, so total mass is 20 x 6.852 = 137.0 mg
%w/w = 100 x 137 ÷ 2372.5 = 5.78%
6. 100 uL contains 0.1 mg
Slope = 0.706
Mass = 0.319/0.706 = 0.452 mg
0.452 mg/25 mL diluted sample from 500 mL total
Therefore total analyte in 500 mL VF is 20 x 0.452 = 9.037 mg
This is from 5 mL of original 50 mL sample solution, so total mass is 10 x 9.037 = 90.37 mg
%w/w = 100 x 90.37 ÷ 101.8 = 88.8%
7. 100 uL contains 0.1 mg
Slope = 77503
Mass = 15860/77503 = 0.205 mg
0.205 mg/50 mL analysed sample from 1 L total
Therefore total analyte in 1 L VF is 20 x 0.205 = 4.093 mg
This is from 10 mL of original sample solution, so 4.093 mg/10 mL = 409.3 mg/L = 0.4093 g/L.
Adv. Spectro
5.9