Result Validation
This resources applies to a majority of subjects in your course, but particularly Workplace Systems and
Practical Project.
Introduction
Before defining what “result validation” means, let’s first consider an important question.
EXERCISE 1
You’ve done an analysis to the best of your ability using the correct procedure. Is your answer correct?
Thus, result validation is the process by which you provide “evidence” regarding the accuracy of your
sample analysis. The basic methods for validating the accuracy of a method are:

calibration – not just making up standards, but keeping a check on the performance on the
instrument to make sure that it is maintaining a consistent output

blanks – “samples” known to contain no analyte which are taken through various parts of the
procedure to determine sources of contamination

controls – standards with similar matrices to the samples

spikes – addition of known amounts of analyte to “samples” of known concentration; the spiked
sample is measured to determine how much of the addition is picked up by the measurement

retesting – duplicate samples or analysis by another method or laboratory
Calibration
Calibration standards provide a day-to-day measure of an instrument response, and allow you to
determine the concentration of the analyte in samples. However, they do not necessarily provide a
measurement of the ongoing performance of the instrument. Therefore, calibration also means the
running of special standards which do not relate to sample analysis, but to instrument performance.
This is also known as Performance Qualification.
Blanks
Blanks – “samples” with no analyte – are used to test the process at any or all stages of the process.
There are:

field blanks – to check on contamination from the collection procedure, the blank is carried
through the same process (except actually collecting real sample)

transport blanks – kept in the transport vehicle at the site, and checks for problems caused by
storage and transport

container blanks – prepared at the laboratory from a container which is identical to those used
for field collection; it has the same preservation and storage time as the samples; particularly
useful for extended storage of samples

matrix blanks – materials of similar matrix to the samples, but with no analyte; are used to check
for interfering species causing a response, but are very difficult to obtain

reagent blanks – some analytical methods require a number of sample treatment steps involving
addition of different chemicals; while AR grade chemicals should be 99%+ pure, trace and ultratrace analysis can detect very low levels of analyte from the reagents (see example)
Result Validation
EXAMPLE 1
10 mL of concentrated HCl is used to acidify 50 mL of a river water sample (known to be around 25 ug/L
Pb), and the solution made up to 100 mL with ultra-pure water. The acid contains 0.1 mg/L (0.00001
%w/v) of lead. The sample is analysed by electrothermal AAS, which is capable of analysis down to 10
ug/L Pb. Is there a problem?
Using C1V1 = C2V2, the additional lead from the acid is 10 mL x 0.1 mg/L ÷ 100 mL = 10 ug/L. If the
sample was around 25 ug/L, then the reagent has added about 40% error to the answer!
CLASS EXERCISE 2
How can blanks be used?
Controls
Even if no measurable contamination was to occur, there are other sources of error:

matrix interference

method error

instrument malfunction

operator error
The only way to measure these errors is to know the right answer! However, if you already knew the
right answer for your samples, you wouldn’t be testing them. However, for certain sample types, it is
possible to have standard samples, known as controls or certified reference materials (CRM), which
are samples of the same matrix, which have known levels of the analytes. These are either purchased or
prepared in the laboratory.
The control is taken through the same procedure as the normal sample, and the result compared with
the “true value”. This can be used to judge the accuracy of the real samples.
EXAMPLE 2
A standard sample of flour known to contain 450 mg/kg of sodium is analysed together with real
samples. The standard sample is found to contain 425 mg/kg by the analysis. What does this
mean for the real samples?
The standard sample is reading 5.6% low [(425-450)/450]. It is reasonable to assume that their
results are similarly low. However, you would not simply subtract 5.6% from their results!!
Unfortunately, in environmental testing, this is not easy, because every body of water, piece of ground
or volume of air has a different matrix, and also changes over time.
Standard samples can also be used to check the accuracy of a method or a technician’s work. In this
case, there is no concern about matrix similarity. These “samples” can be inserted into a normal batch
of samples, with or without the technician’s knowledge.
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Result Validation
Spikes
If standard samples with matched matrices cannot be obtained (which will be the case for most
environmental testing), then an alternative approach is to “spike” a real sample or calibration standard.
This means adding a known amount of analyte to the “sample” at some point in the analysis, then
proceeding to analyse this spiked sample as normal. This is often called a recovery check.
For example, if your analysed portion of sample contained 5 mg of analyte, and another 5 mg was
added as a spike, you would expect when the spiked sample was analysed, it would contain 10 mg. If
less (or more) is recovered than was put in, then something is going wrong in the procedure.
This requires the addition of a known amount of analyte (known as a spike) to the sample, which is then
analysed as normal. The increase in concentration as analysed is then compared to how much was due
to the sample and how much was added. If a procedure works perfectly, then 100% of the added
analyte will be recovered. In general, a recovery of between 90-110% is considered perfectly
reasonable.
The key conclusion to make here is that the percent recovered is a measure of how accurate the answer
for your sample is. If only 50% of the added analyte is recovered, it suggest that 50% of the analyte in
the sample is being missed as well. However, you don’t then adjust the sample answer accordingly, just
report both results.
The recovery check process is similar in performance to standard addition, but is done for totally
different reasons. Standard addition is done to determine the sample concentration (it is an analysis
procedure), while the recovery check requires that you know the concentration in the sample (it is a
check on the analytical procedure).
Spiked samples check for matrix problems as well as method and operator errors, while spiked
standards check only for the latter errors.
The calculations associated with recovery checks are covered in a separate section in this resource. The
recovery check is a valuable way of determining whether a new method was accurate. You might
wonder why it is necessary to run recovery checks afterwards, when it is known that the method is
accurate. The reason is to keep proving that.
Retesting
Equally valuable in result validation is the retesting of samples in various ways:

analysing a duplicate sample placed in the batch (routine)

re-running a calibration standard after a certain number of samples (known as a re-slope) – this
checks for instrumental drift; if minor, then the calibration graph could be adjusted; if major,
then re-calibration and re-running of some of the samples may be required

analysing the sample by an alternative method – this is useful in checking new methods

another person (or laboratory) analysing the sample – this checks the performance of the
technician; inter-laboratory proficiency checks are commonly used as part of quality registration
Overall
A large batch of samples (e.g. 20 or more) should contain the following extra samples:

blanks of each type where appropriate

control

spiked sample

duplicate sample

re-slope standard

spiked standard/blank
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Result Validation
Recovery check calculations
The %recovery is calculated as below.
(
)
where
S is the amount of analyte found in the sample
SP the amount in the spike
SPS the amount found in the spiked sample
Amount can refer to mass, volume or concentration. It’s worthwhile noting that found means in
analysis, and added means as standard.
The final calculation itself could not be much easier. However, getting the three numbers to plug into
the equation is less easy, because it is easy to put the wrong number in when there are so many
numbers to choose from in a typical multi-step analysis.
Remember the two rules below and it will help you avoid some of the common errors.
1.
All three values must be in exactly the same unit, e.g. mg, mg/L
2.
All three values must be from the same stage of the analysis, e.g. the original sample, the first
solution, the analysed solution
When looking at these calculations, there are a lot of numbers and it is easy to get them mixed up. It is
a good idea, especially when you are learning, to identify which numbers belong to which solution:
sample, spike, spiked sample, or the procedure itself (details such as dilution volumes). You will find
that no number should belong to more than one of these categories.
The following relationship is useful as well:
1 mL of 1000 mg/L solution contains 1 mg of analyte
EXAMPLE 4
A 20 mL sample of river water is diluted to 100 mL, and found to contain an average of 2 mg/L of Na.
Another 20 mL aliquot of the water is spiked with 2 mL of 50 mg/L Na, made up to 100 mL and then
analysed. It was found to contain 2.9 mg/L.
S
SP
SPS
Procedure
2 mg/L
1 mL of 50 mg/L
2.9 mg/L
20 mL to 100 mL
Fairly obviously, the best unit to choose is mg/L because it is the only one in common.
This procedure has only two stages: the original river water and the diluted solution which is analysed.
Which stage do these mg/L numbers belong to?
S : 2 – diluted
SP: 50 – neither (it is a standard in another bottle)
SPS: 2.9 – diluted
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Result Validation
It would seem simplest to work out what the mg/L of SP in the diluted solution is, because that
completes the set of three. This is simply the dilution equation you know very well: C1V1 = C2V2.
50 x 2 = ? x 100, so SP = 1 mg/L
 2.9  2 
%RC  100 x 
 = 90%
 1 
There is a much less obvious approach to this calculation: by the mass of Na in the original solution.
S: 10 mL of 10 mg/L = 0.2 mg
SPS: 100 mL of 2.9 mg/L = 0.29 – while this is the final solution, all the Na in this came from the sample
and the spike
SP: 2 mL of 50 mg/L = 0.1
(
)
Why even think about this way, when each of the three values has to be worked out? Simply, the
“mass in the sample portion” method works regardless of the type of sample (solid or liquid) and
whether there are any dilutions.
EXAMPLE 5
Potato chips are analysed and found to contain 5.0%w/w of Na. A 1 g sample is spiked with 5 mL of
5000 mg/L Na, dissolved in 500 mL and analysed. It is found to contain 140 mg/L of Na.
S
SP
SPS
Procedure
A little tougher now as there are different units (%w/w for solid, mg/L for solution). However, there is
only one stage.
Approach 1 - concentration based
Use mg/L as the common unit, so SPS is already done.
For S, how many mg are in 1 g of 5%w/w? What is the concentration when dissolved in 500 mL?
For SP, what is the final concentration?

%RC  100 x 



 =

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Result Validation
Approach 2 – mass based
Use mg as the common unit, all 3 have to be calculated (but we already did S in Approach 1: 50 mg)
For SPS, how many mg are in 500 mL of140 mg/L?
For SP, how many mg are in 5 mL of 5000 mg/L?

%RC  100 x 



 =

Same answer. Is one approach better? No.
EXERCISE 3
(a) 10 mL of sample (known concentration 100 mg/L Ca) is spiked with 500 uL of 1000 mg/L Ca and
diluted to 250 mL. This solution, when analysed, has a concentration of 6.1 mg/L Ca.
S
SP
SPS
Procedure
(b)
5 g of soil (known concentration 50 mg/kg lead) is spiked with 5 mL of 50 mg/L Pb, and made to
100 mL. The concentration of this solution is 4.4 mg/L.
S
SP
SPS
Procedure
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Result Validation
EXERCISE 3
(c) A beer sample is diluted 1:1 and analysed for its sodium content by flame photometry, using
simple Na standards. The concentration of the beer is found to be 52 mg/L. A 50 mL aliquot of
the beer is spiked with 2 mL of 1000 mg/L Na, made up to 100 mL and this solution found to
contain 47 mg/L Na.
S
SP
SPS
Procedure
EXERCISE 4
Now for some real values, rather than nice simple whole numbers.
(a)
The concentration of Vitamin C in an orange juice sample was determined by HPLC to be 41.2
mg/100 mL. Another 25 mL aliquot of the sample was spiked with 10 mL of 1000 mg/L Vitamin C
standard and the solution made up to 50 mL. The spiked sample was analysed under the same
conditions and found to contain 398 mg/L.
(b)
The ethanol content of a wine sample was determined to be 12.3 %v/v. A 5 mL aliquot of the
wine was spiked with 200 uL of pure ethanol and diluted to 10 mL. When analysed it was found
to contain 8.2%v/v.
(c)
The sodium content of a butter sample was determined to be 475 mg/100 g. A recovery check
was prepared by weighing out 2.4376 g of sample, dispersing it in hot water, adding a 5 mL
aliquot of 1000 mg/L sodium and making up to 100 mL. A 10 mL aliquot was then diluted to 100
mL for analysis. The sodium content of this solution was 15.9 mg/L.
(d)
A 5 mL aliquot of bitter lemon (quinine concentration 27.4 mg/L) is spiked with 200 uL of 200
mg/L quinine standard and diluted to 25 mL. A 20 mL aliquot of this solution is determined to
contain 145.9 ug of quinine.
(e)
A sample toothpaste is analysed for its fluoride content and found to contain 0.115 %w/w. A
5.0762 g sample is spiked with 5 mL of 500 mg/L fluoride and dissolved in 100 mL of solution. A
25 mL aliquot is diluted to 100 mL. and analysed: it contains 21.6 mg/L.
(f)
A soluble fertiliser is analysed for its nitrate content, and found to have 10.1% w/w. A 0.5682 g
sample is dissolved in water, 5 mL of 5 g/L nitrate added and the mixture made to 500 mL. A 10
mL aliquot is diluted to 50 mL, and this solution is found to contain 32.4 mg/L.
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Result Validation
(g)
A 1.3825 g sample of peanuts (2.69%w/w K) is spiked with 20 mL of 1000 mg/L K, ashed and
dissolved up to 100 mL. A 10 mL aliquot is diluted to 100 mL, analysed and found to contain 52.3
mg/L.
(h)
25 mL of lemonade (385 mg/L benzoic acid) is spiked with 5 mL of 1000 mg/L benzoic acid and
made up to 100 mL. When analysed, this solution has a concentration of 138 mg/L.
(i)
2.5346 g of powdered milk (395 ug/kg Pb) is spiked with 5 mL of 100 ug/L Pb, ashed and dissolved
in 25 mL. This solution is analysed and found to have a lead concentration of 62.3 ug/L.
Worked answers to these are available from the website.
Other aspects of recovery checks
These include how much to add and when in the procedure, will covered in the Practical Project
subject.
Assessment
Your ability to perform recovery check calculations will be assessed in the theory exam associated with
the Practical Project subject later in the year.
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