5B Calibrating the Signal (Stotal) 5C Determining the Sensitivity (kA) 5D Linear Regression and Calibration Curves5E Compensating for the Reagent… [600835]

153Chapter 5
Standardizing Analytical
Methods
Chapter Overview
5A Analytical Standards
5B Calibrating the Signal (Stotal)
5C Determining the Sensitivity (kA)
5D Linear Regression and Calibration Curves5E Compensating for the Reagent Blank (S
reag)
5F Using Excel and R for a Regression Analysis5G Key Terms5H Chapter Summary5I Problems5J Solutions to Practice Exercises
The American Chemical Society’s Committee on Environmental Improvement defines
standardization as the process of determining the relationship between the signal and the
amount of analyte in a sample.1 In Chapter 3 we defined this relationship as
Sk n S Sk C Stotal A A reag total A A ror =+ = +eeag
where Stotal is the signal, nA is the moles of analyte, CA is the analyte’s concentration, kA is the
method’s sensitivity for the analyte, and Sreag is the contribution to Stotal from sources other
than the sample. To standardize a method we must determine values for kA and Sreag. Strategies
for accomplishing this are the subject of this chapter.
1 ACS Committee on Environmental Improvement “Guidelines for Data Acquisition and Data Quality Evaluation in
Environmental Chemistry,” Anal. Chem. 1980, 52, 2242–2249.
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154 Analytical Chemistry 2.0
5A Analytical Standards
To standardize an analytical method we use standards containing known
amounts of analyte. The accuracy of a standardization, therefore, depends
on the quality of the reagents and glassware used to prepare these standards.
For example, in an acid–base titration the stoichiometry of the acid–base re-
action defines the relationship between the moles of analyte and the moles of titrant. In turn, the moles of titrant is the product of the titrant’s con-
centration and the volume of titrant needed to reach the equivalence point.
The accuracy of a titrimetric analysis, therefore, can be no better than the
accuracy to which we know the titrant’s concentration.
5A.1 Primary and Secondary Standards
We divide analytical standards into two categories: primary standards and
secondary standards. A primary standard is a reagent for which we can
dispense an accurately known amount of analyte. For example, a 0.1250-g
sample of K
2Cr2O7 contains 4.249 × 10–4 moles of K2Cr2O7. If we place
this sample in a 250-mL volumetric flask and dilute to volume, the con-
centration of the resulting solution is 1.700 × 10–3 M. A primary standard
must have a known stoichiometry, a known purity (or assay), and it must be stable during long-term storage. Because of the difficulty in establishing
the degree of hydration, even after drying, a hydrated reagent usually is not
a primary standard.
Reagents that do not meet these criteria are secondary standards.
The concentration of a secondary standard must be determined r
elative
to a primary standard. Lists of acceptable primary standards are available.2
Appendix 8 provides examples of some common primary standards.
5A.2 Other Reagen
ts
Preparing a standard often requires additional reagents that are not pri-
mary standards or secondary standards. Preparing a standard solution, for
example, requires a suitable solvent, and additional reagents may be need
to adjust the standard’s matrix. These solvents and reagents are potential
sources of additional analyte, which, if not accounted for, produce a deter-
minate error in the standardization. If available, reagent grade chemicals
conforming to standards set by the American Chemical S
ociety should be
used.3 The label on the bottle of a reagent grade chemical ( Figure 5.1) lists
either the limits for specific impurities, or provides an assay for the impuri-
ties.
We can improve the quality of a reagent grade chemical by purifying it,
or by conducting a more accurate assay. As discussed later in the chapter, we
2 (a) Smith, B. W.; Parsons, M. L. J. Chem. Educ. 1973, 50, 679–681; (b) Moody, J. R.; Green-
burg, P . R.; Pratt, K. W.; Rains, T. C. Anal. Chem. 1988, 60, 1203A–1218A.
3 Committee on Analytical Reagents, Reagent Chemicals, 8th ed., American Chemical Society:
Washington, D. C., 1993.See Chapter 9 for a thorough discussion of titrimetric methods of analysis.
The base NaOH is an example of a sec-
ondary standard. Commercially avail-able NaOH contains impurities of NaCl, Na2CO3, and Na2SO4, and readily
absorbs H2O from the atmosphere. To
determine the concentration of NaOH in a solution, it is titrated against a primary standard weak acid, such as potassium hy-drogen phthalate, KHC8H4O4.
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155 Chapter 5 Standardizing Analytical Methods
can correct for contributions to Stotal from reagents used in an analysis by
including an appropriate blank determination in the analytical procedure.
5A.3 Preparing Standard Solutions
It is often necessary to prepare a series of standards, each with a different
concentration of analyte. We can prepare these standards in two ways. If the
range of concentrations is limited to one or two orders of magnitude, then
each solution is best prepared by transferring a known mass or volume of
the pure standard to a volumetric flask and diluting to volume.
When working with larger ranges of concentration, particularly those
extending over more than three orders of magnitude, standards are best pre-
pared by a serial dilution from a single stock solution. In a serial dilution
we prepar
e the most concentrated standard and then dilute a portion of it
to prepare the next most concentrated standard. Next, we dilute a portion
of the second standard to prepare a third standard, continuing this process
until all we have prepared all of our standards. Serial dilutions must be pre-
pared with extra care because an error in preparing one standard is passed
on to all succeeding standards.
Figure 5.1 Examples of typical packaging labels for reagent grade chemicals. Label
(a) provides the manufactur
er’s assay for the reagent, NaBr. Note that potassium
is flagged with an asterisk ( *) because its assay exceeds the limits established by
the American Chemical Society (ACS). Label (b) does not provide an assay for impurities, but indicates that the reagent meets ACS specifications. An assay for the reagent, NaHCO
3 is provided.
(a)
(b)
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156 Analytical Chemistry 2.0
5B Calibrating the Signal (Stotal)
The accuracy of our determination of kA and Sreag depends on how accurately
we can measure the signal, Stotal. We measure signals using equipment, such
as glassware and balances, and instrumentation, such as spectrophotom-
eters and pH meters. To minimize determinate errors affecting the signal,
we first calibrate our equipment and instrumentation. We accomplish the
calibration by measuring Stotal for a standard with a known response of Sstd,
adjusting Stotal until
SStotal std=
Here are two examples of how we calibrate signals. Other examples are
provided in later chapters focusing on specific analytical methods.
When the signal is a measur
ement of mass, we determine Stotal using
an analytical balance. To calibrate the balance’s signal we use a reference
weight that meets standards established by a governing agency, such as the
National Institute for Standards and Technology or the American Society
for Testing and Materials. An electronic balance often includes an internal
calibration weight for routine calibrations, as well as programs for calibrat-
ing with external weights. In either case, the balance automatically adjusts
Stotal to match Sstd.
We also must calibrate our instruments. For example, we can evaluate
a spectrophotometer’s accuracy by measuring the absorbance of a carefully
prepared solution of 60.06 mg/L K2Cr2O7 in 0.0050 M H2SO4, using
0.0050 M H2SO4 as a reagent blank.4 An absorbance of 0.640 ± 0.010
absorbance units at a wavelength of 350.0 nm indicates that the spectrom-eter’s signal is properly calibrated. Be sure to read and carefully follow the
calibration instructions provided with any instrument you use.
5C Determining the Sensitivity (kA)
To standardize an analytical method we also must determine the value of
kA in equation 5.1 or equation 5.2.
Sk n Stotal A A reag=+ 5.1
Sk C Stotal A A reag=+ 5.2
In principle, it should be possible to deriv e the v
alue of kA for any analyti-
cal method by considering the chemical and physical processes generating
the signal. Unfortunately, such calculations are not feasible when we lack
a sufficiently developed theoretical model of the physical processes, or are
not useful because of nonideal chemical behavior. In such situations we
must determine the value of kA by analyzing one or more standard solutions,
each containing a known amount of analyte. In this section we consider
4 Ebel, S. Fresenius J. Anal. Chem. 1992, 342, 769.See Section 2D.1 to review how an elec-tronic balance works. Calibrating a balance is important, but it does not eliminate all sources of determinate error in measuring mass. See Appendix 9 for a discussion of correcting for the buoyancy of air.
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157 Chapter 5 Standardizing Analytical Methods
several approaches for determining the value of kA. For simplicity we will
assume that Sreag has been accounted for by a proper reagent blank, allow-
ing us to replace Stotal in equation 5.1 and equation 5.2 with the analyte’s
signal, SA.
Sk nAA A= 5.3
Sk CAA A= 5.4
5C.1 Single -P
oint versus Multiple-Point Standardizations
The simplest way to determine the value of kA in equation 5.4 is by a sin-
gle-point standardiz
ation in which we measure the signal for a standard,
Sstd, containing a known concentration of analyte, Cstd. Substituting these
values into equation 5.4
kS
CAstd
std= 5.5
gives the v alue for kA. Having determined the value for kA, we can calculate
the concentration of analyte in any sample by measuring its signal, Ssamp,
and calculating CA using equation 5.6.
CS
kAsamp
A= 5.6
A single-point standardization is the least desirable method for stan-
dardizing a method. Ther
e are at least two reasons for this. First, any error
in our determination of kA carries over into our calculation of CA. Second,
our experimental value for kA is for a single concentration of analyte. Ex-
tending this value of kA to other concentrations of analyte requires us to
assume a linear relationship between the signal and the analyte’s concentra-
tion, an assumption that often is not true.5 Figure 5.2 shows how assum-
ing a constant value of kA may lead to a determinate error in the analyte’s
concentration. Despite these limitations, single-point standardizations find
routine use when the expected range for the analyte’s concentrations is
small. Under these conditions it is often safe to assume that kA is constant
(although you should verify this assumption experimentally). This is the
case, for example, in clinical labs where many automated analyzers use only
a single standard.
The preferred approach to standardizing a method is to prepare a se-
ries of standards, each containing the analyte at a different concentration.
Standards are chosen such that they bracket the expected range for the ana-
lyte’s concentration. A multiple-point standardization should include
at least three standards, although mor
e are preferable. A plot of Sstd versus
5 Cardone, M. J.; Palmero, P . J.; Sybrandt, L. B. Anal. Chem. 1980, 52, 1187–1191.Equation 5.3 and equation 5.4 are essen-
tially identical, differing only in whether we choose to express the amount of ana-lyte in moles or as a concentration. For the remainder of this chapter we will limit our treatment to equation 5.4. You can extend this treatment to equation 5.3 by replac-ing CA with nA.
Linear regression, which also is known as the method of least squares, is one such al-gorithm. Its use is covered in Section 5D.
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158 Analytical Chemistry 2.0
Cstd is known as a calibration curve. The exact standardization, or calibra-
tion relationship is determined by an appropriate curve-fitting algorithm.
There are at least two advantages to a multiple-point standardization.
First, although a determinate error in one standard introduces a determinate
error into the analysis, its effect is minimized by the remaining standards.
Second, by measuring the signal for several concentrations of analyte we
no longer must assume that the value of kA is independent of the analyte’s
concentration. Constructing a calibration curve similar to the “actual rela-
tionship” in Figure 5.2, is possible.
5C.2 External Standards
The most common method of standardization uses one or more external
standards
, each containing a known concentration of analyte. We call
them “external” because w
e prepare and analyze the standards separate from
the samples.
SINGLE EXTERNAL STANDARD
A quantitative determination using a single external standard was described
at the beginning of this section, with kA given by equation 5.5. After de-
termining the value of kA, the concentration of analyte, CA, is calculated
using equation 5.6.
Example 5.1
A spectrophotometric method for the quantitative analysis of Pb2+ in
blood yields an Sstd of 0.474 for a single standard whose concentration of
lead is 1.75 ppb What is the concentration of Pb2+ in a sample of blood
for which Ssamp is 0.361?Figure 5.2 Example showing how a single-point
standardization leads to a determinate error in an
analyte
’s reported concentration if we incorrectly
assume that the value of kA is constant. (CA)reported CstdSstdSsamp
(CA)actualactual relationshipassumed relationship
Appending the adjective “external” to the noun “standard” might strike you as odd at this point, as it seems reasonable to as-sume that standards and samples must be analyzed separately. As you will soon learn, however, we can add standards to our samples and analyze them simultane-ously.
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159 Chapter 5 Standardizing Analytical Methods
SOLUTION
Equation 5.5 allows us to calculate the value of kA for this method using
the data for the standard.
kS
CAstd
std-1
ppbppb == =0 474
17 50 2709.
..
Having determined the value of kA, the concentration of Pb2+ in the sam-
ple of blood is calculated using equation 5.6.
CS
kAsamp
A-1ppbppb == =0 361
0 270913 3.
..
MULTIPLE EXTERNAL STANDARDS
Figure 5.3 shows a typical multiple-point external standardization. The
volumetric flask on the left is a reagent blank and the remaining volu-
metric flasks contain increasing concentrations of Cu2+. Shown below the
volumetric flasks is the resulting calibration curve. Because this is the most
common method of standardization the resulting r
elationship is called a
normal calibration curve.
When a calibration curve is a straight-line, as it is in F
igure 5.3, the
slope of the line gives the value of kA. This is the most desirable situation
since the method’s sensitivity remains constant throughout the analyte’s
concentration range. When the calibration curve is not a straight-line, the
0 0.0020 0.0040 0.0060 0.008000.050.100.150.200.25
Sstd
Cstd (M)Figure 5.3 Shown at the top is a reagent
blank (far left) and a set of five exter-nal standards for C
u2+ with concen-
trations increasing from left to right. Shown below the external standards is the resulting normal calibration curve. The absorbance of each standard, S
std,
is shown by the filled circles.
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160 Analytical Chemistry 2.0
method’s sensitivity is a function of the analyte’s concentration. In Figure
5.2, for example, the value of kA is greatest when the analyte’s concentration
is small and decreases continuously for higher concentrations of analyte.
The value of kA at any point along the calibration curve in Figure 5.2 is given
by the slope at that point. In either case, the calibration cur
ve provides a
means for relating Ssamp to the analyte’s concentration.
Example 5.2
A second spectrophotometric method for the quantitative analysis of Pb2+
in blood has a normal calibration curve for which
SCstd-1
stdppb =() ×+ 0 296 0 003..
What is the concentration of Pb2+ in a sample of blood if Ssamp is 0.397?
SOLUTION
To determine the concentration of Pb2+ in the sample of blood we replace
Sstd in the calibration equation with Ssamp and solve for CA.
CS
Asamp
-1ppb=−
=− 0 003
0 2960 397 0 003
0 296.
…
. pppbppb-1=13 3.
It is worth noting that the calibration equation in this problem includes
an extra term that does not appear in equation 5.6. Ideally we expect
the calibration curve to hav
e a signal of zero when CA is zero. This is the
purpose of using a reagent blank to correct the measured signal. The extra
term of +0.003 in our calibration equation results from the uncertainty in
measuring the signal for the reagent blank and the standards.
An external standardization allows us to analyze a series of samples
using a single calibration curve. This is an important advantage when we
have many samples to analyze. Not surprisingly, many of the most common
quantitative analytical methods use an external standardization.
There is a serious limitation, however, to an external standardization.
When we determine the value of kA using equation 5.5, the analyte is pres-Practice Exercise 5.1
Figure 5.3 shows a normal calibration curve for the quantitative analysis
of Cu2+. The equation for the calibration curve is
Sstd = 29.59 M–1 × Cstd + 0.0015
What is the concentration of Cu2+ in a sample whose absorbance, Ssamp,
is 0.114? Compare your answer to a one-point standardization where a standard of 3.16 × 10
–3 M Cu2+ gives a signal of 0.0931.
Click here to review your answer to this exercise.The one-point standardization in this ex-ercise uses data from the third volumetric flask in Figure 5.3.
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161 Chapter 5 Standardizing Analytical Methods
ent in the external standard’s matrix, which usually is a much simpler ma-
trix than that of our samples. When using an external standardization we
assume that the matrix does not affect the value of kA. If this is not true,
then we introduce a proportional determinate error into our analysis. This
is not the case in Figure 5.4, for instance, where we show calibration curves
for the analyte in the sample’s matrix and in the standard’s matrix. In this
example, a calibration curve using external standards results in a negative determinate error. If we expect that matrix effects are important, then we
try to match the standard’s matrix to that of the sample. This is known as
matrix matching. If we are unsure of the sample’s matrix, then we must
show that matrix effects are negligible, or use an alternativ
e method of stan-
dardization. Both approaches are discussed in the following section.
5C.3 Standard Additions
We can avoid the complication of matching the matrix of the standards to
the matrix of the sample by conducting the standardization in the sample.
This is known as the method of standard additions.
SINGLE STANDARD ADDITION
The simplest version of a standard addition is shown in Figure 5.5. First we
add a portion of the sample, Vo, to a volumetric flask, dilute it to volume,
Vf, and measure its signal, Ssamp. Next, we add a second identical portion
of sample to an equivalent volumetric flask along with a spike, Vstd, of an
external standard whose concentration is Cstd. After diluting the spiked
sample to the same final volume, we measure its signal, Sspike. The following
two equations relate Ssamp and Sspike to the concentration of analyte, CA, in
the original sample.The matrix for the external standards in Figure 5.3, for example, is dilute ammonia, which is added because the Cu(NH
3)42+
complex absorbs more strongly than Cu
2+. If we fail to add the same amount
of ammonia to our samples, then we will introduce a proportional determinate er-ror into our analysis.(CA)reportedSsamp
(CA)actualstandard’s
matrix
sample’s
matrix
Figure 5.4 Calibration curves for an analyte in the
standard’
s matrix and in the sample’s matrix. If the
matrix affects the value of kA, as is the case here, then
we introduce a determinate error into our analysis if we use a normal calibration curve.
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162 Analytical Chemistry 2.0
Sk CV
Vsamp A Ao
f= 5.7
Sk CVVCVV
spike A A
fo
std
fstdGQ5.8
As long as Vstd is small relative to Vo, the effect of the standard’s matrix on
the sample’s matrix is insignificant. Under these conditions the value of kA
is the same in equation 5.7 and equation 5.8. Solving both equations for
kA and equating gives
S
CV
VS
CV
VCV
Vsamp
Ao
fspike
Ao
fstdstd
f=
+5.9
which we can solv e for the concentration of analyte, CA, in the original
sample.
Example 5.3
A third spectrophotometric method for the quantitative analysis of Pb2+ in
blood yields an Ssamp of 0.193 when a 1.00 mL sample of blood is diluted
to 5.00 mL. A second 1.00 mL sample of blood is spiked with 1.00 μL of
a 1560-ppb Pb2+ external standard and diluted to 5.00 mL, yielding an add V o of C A add V std of C std
dilute to V f
CV
VAo
fr CV
VCV
VAo
fstdstd
fr rConcentration
of AnalyteFigure 5.5 Illustration showing the method of stan-
dard additions. The volumetric flask on the left con-
tains a por
tion of the sample, Vo, and the volumetric
flask on the right contains an identical portion of the sample and a spike, V
std, of a standard solution of the
analyte. Both flasks are diluted to the same final vol-ume, V
f. The concentration of analyte in each flask is
shown at the bottom of the figure where CA is the ana-
lyte’s concentration in the original sample and Cstd is
the concentration of analyte in the external standard.
The ratios Vo/Vf and Vstd/Vf account for
the dilution of the sample and the stan-dard, respectively.
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163 Chapter 5 Standardizing Analytical Methods
Sspike of 0.419. What is the concentration of Pb2+ in the original sample
of blood?
SOLUTION
We begin by making appropriate substitutions into equation 5.9 and solv-
ing for CA. Note that all volumes must be in the same units; thus, we first
covert Vstd from 1.00 μL to 1.00 × 10–3 mL.
0 193
10 0
50 00 419
10.
.
..
.C CA AmL
mL=
0 0
50 0156010 0 1 03mL
mLppbm
..+×−L L
mL 50 0.
0 193
0 2000 419
0 200 0 3120.
..
.. CCAAppb=+
0.0386CA + 0.0602 ppb = 0.0838CA
0.0452CA = 0.0602 ppb
CA = 1.33 ppb
The concentration of Pb2+ in the original sample of blood is 1.33 ppb.
It also is possible to make a standard addition directly to the sample,
measuring the signal both before and after the spike (Figure 5.6). In this
case the final volume after the standard addition is Vo + Vstd and equation
5.7, equation 5.8, and equation 5.9 become
add V std of C std
Concentration
of AnalyteVo Vo
CACV
VVCV
VVAo
os t dstdstd
os t d
Figure 5.6 Illustration showing an alternative form of the method of standard ad-
ditions. In this case we add a spike of the external standar
d directly to the sample
without any further adjust in the volume.
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164 Analytical Chemistry 2.0
Sk Csamp A A=
Sk CVVVCVVV
spike A A
o stdo
std
o stdstdGQ5.10
S
CS
CV
VVCV
VVsamp
Aspike
Ao
os t dstdstd
os t d=
+++5.11
Example 5.4
A fourth spectrophotometric method for the quantitative analysis of Pb2+
in blood yields an Ssamp of 0.712 for a 5.00 mL sample of blood. After spik-
ing the blood sample with 5.00 μL of a 1560-ppb Pb2+ external standard,
an Sspike of 1.546 is measured. What is the concentration of Pb2+ in the
original sample of blood.
SOLUTION
To determine the concentration of Pb2+ in the original sample of blood, we
make appropriate substitutions into equation 5.11 and solve for CA.
0 712 1 546
50 0
5 0051..
.
.CCA
AmL
mL=
+5 56050 0 1 0
5 0053
ppbmL
mL.
.×−
0 712 1 546
0 9990 1 558..
.. CCAAppb=+
0.7113CA + 1.109 ppb = 1.546CA
CA = 1.33 ppb
The concentration of Pb2+ in the original sample of blood is 1.33 ppb.
MULTIPLE STANDARD ADDITIONS
We can adapt the single-point standard addition into a multiple-point
standard addition by preparing a series of samples containing increasing
amounts of the external standard. Figure 5.7 shows two ways to plot a
standard addition calibration curv
e based on equation 5.8. In Figure 5.7a
we plot Sspike against the volume of the spikes, Vstd. If kA is constant, then
the calibration curve is a straight-line. It is easy to show that the x-intercept
is equivalent to –CAVo/Cstd.Vo + Vstd = 5.00 mL + 5.00×10–3 mL
= 5.005 mL
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165 Chapter 5 Standardizing Analytical Methods
Example 5.5
Beginning with equation 5.8 show that the equations in Figure 5.7a for
the slope, the y-intercept, and the x
-intercept are correct.
SOLUTION
We begin by rewriting equation 5.8 as
SkCV
VkC
VVspikeAA o
fAs t d
fstd=+ ×
which is in the form of the equation for a straight-line
Y = y-intercept +
slope × X
-4.00 -2.00 0 2.00 4.00 6.00 8.00 10.00 12.0000.100.200.300.400.500.60Sspike
CstdVstd
Vfrslope = k Ay-intercept = kACAVo
Vf
x-intercept = -CAVo
Vf00.100.200.300.400.500.60
Sspike-2.00 0 2.00 4.00 6.00
CstdVstdslope =kACstd
Vf
x-intercept = -CAVoy-intercept = kACAVo
Vf(a)
(b)(mL)
(mg/L)Figure 5.7 Shown at the top is a set of
six standard additions for the determi-nation of Mn
2+. The flask on the left
is a 25.00 mL sample diluted to 50.00 mL. The remaining flasks contain 25.00 mL of sample and, from left to right, 1.00, 2.00, 3.00, 4.00, and 5.00 mL of an external standard of 100.6 mg/L Mn
2+. Shown below are two ways to
plot the standard additions calibration curve. The absorbance for each stan-dard addition, S
spike, is shown by the
filled circles.
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166 Analytical Chemistry 2.0
where Y is Sspike and X is Vstd. The slope of the line, therefore, is kACstd/Vf
and the y-intercept is kACAVo/Vf. The x-intercept is the value of X when
Y is zero, or
0=+ ×kCV
VkC
VxAA o
fAs t d
f-intercept
xkCV
V
kC
VCV
C-interceptAA o
f
As t d
fAo
std=− =−
Practice Exercise 5.2
Beginning with equation 5.8 show that the equations in Figure 5.7b for
the slope, the y-intercept, and the x-intercept ar
e correct.
Click here to review your answer to this exercise.
Because we know the volume of the original sample, Vo, and the con-
centration of the external standard, Cstd, we can calculate the analyte’s con-
centrations from the x-intercept of a multiple-point standard additions.
Example 5.6
A fifth spectrophotometric method for the quantitative analysis of Pb2+
in blood uses a multiple-point standard addition based on equation 5.8.
The original blood sample has a volume of 1.00 mL and the standar d used
for spiking the sample has a concentration of 1560 ppb Pb2+. All samples
were diluted to 5.00 mL before measuring the signal. A calibration curve
of Sspike versus Vstd has the following equation
Sspike = 0.266 + 312 mL–1 × Vstd
What is the concentration of Pb2+ in the original sample of blood.
SOLUTION
To find the x-intercept we set Sspike equal to zero.
0 = 0.266 + 312 mL–1 × Vstd
Solving for Vstd, we obtain a value of –8.526 × 10–4 mL for the x-intercept.
Substituting the x-interecpt’s value into the equation from Figure 5.7a
− ×= − =−×−8 526 1010 0
15604..mLmL
ppCV
CCAo
stdA
b b
and solving for CA gives the concentration of Pb2+ in the blood sample
as 1.33 ppb.
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167 Chapter 5 Standardizing Analytical Methods
Since we construct a standard additions calibration curve in the sample,
we can not use the calibration equation for other samples. Each sample,
therefore, requires its own standard additions calibration curve. This is a
serious drawback if you have many samples. For example, suppose you need
to analyze 10 samples using a three-point calibration curve. For a normal
calibration curve you need to analyze only 13 solutions (three standards
and ten samples). If you use the method of standard additions, however,
you must analyze 30 solutions (each of the ten samples must be analyzed
three times, once before spiking and after each of two spikes).
USING A STANDARD ADDITION TO IDENTIFY MATRIX EFFECTS
We can use the method of standard additions to validate an external stan-
dardization when matrix matching is not feasible. First, we prepare a nor-
mal calibration curve of Sstd versus Cstd and determine the value of kA from
its slope. Next, we prepare a standard additions calibration curve using
equation 5.8, plotting the data as shown in Figure 5.7b . The slope of this
standard additions calibration curv
e provides an independent determina-
tion of kA. If there is no significant difference between the two values of
kA, then we can ignore the difference between the sample’s matrix and that
of the external standards. When the values of kA are significantly different,
then using a normal calibration curve introduces a proportional determi-
nate error.
5C.4 Internal Standards
To successfully use an external standardization or the method of standard
additions, we must be able to treat identically all samples and standards.
When this is not possible, the accuracy and precision of our standardiza-
tion may suffer. For example, if our analyte is in a volatile solvent, then its
concentration increases when we lose solvent to evaporation. Suppose we Practice Exercise 5.3
Figure 5.7 shows a standard additions calibration curve for the quantita-tive analysis of Mn
2+. Each solution contains 25.00 mL of the original
sample and either 0, 1.00, 2.00, 3.00, 4.00, or 5.00 mL of a 100.6 mg/L external standard of Mn
2+. All standard addition samples were diluted to
50.00 mL before reading the absorbance. The equation for the calibration curve in Figure 5.7a is
S
std = 0.0854 × Vstd + 0.1478
What is the concentration of Mn2+ in this sample? Compare your answer
to the data in Figure 5.7b, for which the calibration curve is
Sstd = 0.0425 × Cstd(Vstd/Vf) + 0.1478
Click here to review your answer to this exercise.
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168 Analytical Chemistry 2.0
have a sample and a standard with identical concentrations of analyte and
identical signals. If both experience the same proportional loss of solvent
then their respective concentrations of analyte and signals continue to be identical. In effect, we can ignore evaporation if the samples and standards
experience an equivalent loss of solvent. If an identical standard and sample
lose different amounts of solvent, however, then their respective concen-trations and signals will no longer be equal. In this case a simple external
standardization or standard addition is not possible.
We can still complete a standardization if we reference the analyte’s
signal to a signal from another species that we add to all samples and stan-
dards. The species, which we call an internal standard, must be different
than the analyte.
Because the analyte and the internal standard in any sample or standar
d
receive the same treatment, the ratio of their signals is unaffected by any
lack of reproducibility in the procedure. If a solution contains an analyte of
concentration CA, and an internal standard of concentration, CIS, then the
signals due to the analyte, SA, and the internal standard, SIS, are
Sk CAA A=
Sk CIS IS IS=
where kA and kIS are the sensitivities for the analyte and internal standard.
Taking the ratio of the two signals gives the fundamental equation for an
internal standardization.
S
SkC
kCKC
CA
ISAA
IS ISA
IS== × 5.12
Because K is a ratio of the analyte
’s sensitivity and the internal standard’s
sensitivity, it is not necessary to independently determine values for either
kA or kIS.
SINGLE INTERNAL STANDARD
In a single-point internal standardization, we prepare a single standard con-
taining the analyte and the internal standard, and use it to determine the
value of K in equation 5.12.
KCC
SS
AIS
std ISA
std GGQQ5.13
Having standardized the method, the analyte’s concentration is given by
CKC
SS
AIS
ISA
samp =GQ
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169 Chapter 5 Standardizing Analytical Methods
Example 5.7
A sixth spectrophotometric method for the quantitative analysis of Pb2+ in
blood uses Cu2+ as an internal standard. A standard containing 1.75 ppb
Pb2+ and 2.25 ppb Cu2+ yields a ratio of ( SA/SIS)std of 2.37. A sample of
blood is spiked with the same concentration of Cu2+, giving a signal ratio,
(SA/SIS)samp, of 1.80. Determine the concentration of Pb2+ in the sample
of blood.
SOLUTION
Equation 5.13 allows us to calculate the value of K using the data for the
standard
KCACISGQ
stdSISSAGQ
std
1 . 75 ppb Pb22 . 25 ppb Cu2.3 73.0 5
ppb Pb2ppb Cu2
The concentration of Pb2+, therefore, is
CAKCISSISSAGQ
samp
3.0 5
ppb Pb2ppb Cu22 . 25 ppb Cu2
1.8 01 . 33 ppb Cu2
MULTIPLE INTERNAL STANDARDS
A single-point internal standardization has the same limitations as a single-
point normal calibration. To construct an internal standard calibration
curve we prepare a series of standards, each containing the same concen-
tration of internal standard and a different concentrations of analyte. Under
these conditions a calibration curve of ( SA/SIS)std versus CA is linear with
a slope of K/CIS.
Example 5.8
A seventh spectrophotometric method for the quantitative analysis of Pb2+
in blood gives a linear internal standards calibration curve for which
(. ) .SSC 2 11 0 006 ppb
ISA
std1
A GQ
What is the ppb Pb2+ in a sample of blood if (SA/SIS)samp is 2.80?
SOLUTION
To determine the concentration of Pb2+ in the sample of blood we replace
(SA/SIS)std in the calibration equation with (SA/SIS)samp and solve for CA.Although the usual practice is to prepare the standards so that each contains an identical amount of the internal standard, this is not a requirement.
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170 Analytical Chemistry 2.0
CA
2.11 ppb1SISSAGQ
samp0 . 006

2 . 11 ppb12.8 00 . 0061 . 33 ppb
The concentration of Pb2+ in the sample of blood is 1.33 ppb.
In some circumstances it is not possible to prepare the standards so that
each contains the same concentration of internal standard. This is the case,
for example, when preparing samples by mass instead of volume. We can
still prepare a calibration curve, however, by plotting ( SA/SIS)std versus CA/
CIS, giving a linear calibration curve with a slope of K.
5D Linear Regression and Calibration Curves
In a single-point external standardization we determine the value of kA by
measuring the signal for a single standard containing a known concentra-
tion of analyte. Using this value of kA and the signal for our sample, we
then calculate the concentration of analyte in our sample (see Example
5.1). With only a single determination of kA, a quantitative analysis using
a single-point external standardization is straightforward.
A multiple-point standardization presents a more difficult problem.
Consider the data in Table 5.1 for a multiple-point external standardiza-
tion. What is our best estimate of the relationship between Sstd and Cstd?
It is tempting to treat this data as five separate single-point standardiza-
tions, determining kA for each standard, and reporting the mean value.
Despite it simplicity, this is not an appropriate way to treat a multiple-point
standardization.
So why is it inappropriate to calculate an average value for kA as done
in Table 5.1? In a single-point standardization we assume that our reagent
blank (the first row in Table 5.1) corrects for all constant sources of deter-
minate error. If this is not the case, then the value of kA from a single-point
standardization has a determinate error. Table 5.2 demonstrates how an
Table 5.1 Data for a Hypothetical Multiple-Point External
Standardization
Cstd (arbitrary units) Sstd (arbitrary units) kA = Sstd/ Cstd
0.000 0.00 —
0.100 12.36 123.6
0.200 24.83 124.2
0.300 35.91 119.7
0.400 48.79 122.0
0.500 60.42 122.8
mean value for kA = 122.5
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171 Chapter 5 Standardizing Analytical Methods
uncorrected constant error affects our determination of kA. The first three
columns show the concentration of analyte in the standards, Cstd, the signal
without any source of constant error, Sstd, and the actual value of kA for five
standards. As we expect, the value of kA is the same for each standard. In the
fourth column we add a constant determinate error of +0.50 to the signals,
(Sstd)e. The last column contains the corresponding apparent values of kA.
Note that we obtain a different value of kA for each standard and that all of
the apparent kA values are greater than the true value.
How do we find the best estimate for the relationship between the
signal and the concentration of analyte in a multiple-point standardiza-
tion? Figure 5.8 shows the data in Table 5.1 plotted as a normal calibration
curve. Although the data cer
tainly appear to fall along a straight line, the
actual calibration curve is not intuitively obvious. The process of math-
ematically determining the best equation for the calibration curve is called
linear regression.Table 5.2 Effect of a Constant Determinate Error on the Value of kA From a Single-
Point Standardization
CstdSstd
(without constant error)kA = Sstd/ Cstd
(actual)(Sstd)e
(with constant error)kA = (Sstd)e/ Cstd
(apparent)
1.00 1.00 1.00 1.50 1.50
2.00 2.00 1.00 2.50 1.253.00 3.00 1.00 3.50 1.174.00 4.00 1.00 4.50 1.135.00 5.00 1.00 5.50 1.10
mean k
A (true) = 1.00 mean kA (apparent) = 1.23
Figure 5.8 Normal calibration curve for the hypothetical multiple-point external
standardization in Table 5.1
.0.0 0.1 0.2 0.3 0.4 0.50102030405060
Sstd
Cstd
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172 Analytical Chemistry 2.0
5D.1 Linear Regression of Straight Line Calibration Curves
When a calibration curve is a straight-line, we represent it using the follow-
ing mathematical equation
yx=+ββ01 5.14
where y is the signal, Sstd, and x is the analyte’s concentration, Cstd. The
constants β0 and β1 are, respectively, the calibration curve’s expected y-in-
tercept and its expected slope. Because of uncertainty in our measurements,
the best we can do is to estimate values for β0 and β1, which we represent
as b0 and b1. The goal of a linear regression analysis is to determine the
best estimates for b0 and b1. How we do this depends on the uncertainty
in our measurements.
5D.2 Unweighted Linear Regression with Errors in y
The most common approach to completing a linear regression for equation
5.14 makes three assumptions:
(1) that any difference between our experimental data and the calculated
regression line is the result of indeterminate errors affecting y,
(2) that indeterminate errors affecting y are normally distributed, and
(3) that the indeterminate errors in y are independent of the value of x.
Because we assume that the indeterminate errors are the same for all stan-
dards, each standard contributes equally in estimating the slope and the
y-intercept. For this reason the result is considered an unweighted linear
regression
.
The second assumption is generally true because of the central limit the-
orem, which w
e considered in Chapter 4. The validity of the two remaining
assumptions is less obvious and you should ev
aluate them before accepting
the results of a linear regression. In particular the first assumption is always
suspect since there will certainly be some indeterminate errors affecting the
values of x. When preparing a calibration curve, however, it is not unusual
for the uncertainty in the signal, Sstd, to be significantly larger than that for
the concentration of analyte in the standards Cstd. In such circumstances
the first assumption is usually reasonable.
HOW A LINEAR REGRESSION WORKS
To understand the logic of an linear regression consider the example shown
in Figure 5.9, which shows three data points and two possible straight-lines
that might reasonably explain the data. Ho
w do we decide how well these
straight-lines fits the data, and how do we determine the best straight-
line?
Let’s focus on the solid line in Figure 5.9. The equation for this line is
ˆyb b x =+01 5.15
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173 Chapter 5 Standardizing Analytical Methods
where b0 and b1 are our estimates for the y-intercept and the slope, and ˆyis
our prediction for the experimental value of y
for any value of x. Because we
assume that all uncertainty is the result of indeterminate errors affecting y,
the difference between y andˆy for each data point is the residual error ,
r, in the our mathematical model for a par ticular value of
x.
ry yii i=−( ˆ)
Figure 5.10 shows the residual errors for the three data points. The smaller
the total residual error
, R, which we define as
Ry yii
i=−∑( ˆ)2
5.16
the better the fit between the straight-line and the data. In a linear regres-
sion analysis, we seek values of b0 and b1 that give the smallest total residual
error. If you are reading this aloud, you pro-
nounce ˆyas y-hat.
The reason for squaring the individual re-
sidual errors is to prevent positive residual error from canceling out negative residual errors. You have seen this before in the equations for the sample and popula-tion standard deviations. You also can see from this equation why a linear regression is sometimes called the method of least squares.Figure 5.9 Illustration showing three data points and two possible straight-lines that might explain the data. The goal of a linear regression is to find the mathematical model, in
this case a straight-line, that best explains the data.
F
igure 5.10 Illustration showing the evaluation of a linear regression in which we assume that all uncer-
tainty is the result of indeterminate err ors affecting y
. The points in blue, yi, are the original data and the
points in red, ˆyi, are the predicted values from the regression equation, ˆyb b x=+01.The smaller the
total residual error (equation 5.16), the better the fit of the straight-line to the data.ˆy1ˆy2ˆy3
ry y11 1
( ˆ)ry y22 2
( ˆ)ry y33 3
( ˆ)ˆyb b x01
y1y2
y3
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174 Analytical Chemistry 2.0
FINDING THE SLOPE AND Y-INTERCEPT
Although we will not formally develop the mathematical equations for a
linear regression analysis, you can find the derivations in many standard
statistical texts.6 The resulting equation for the slope, b1, is
bnx y x y
nx xii
ii
ii
i
i
ii
i1
22=−
−∑∑ ∑
∑∑5.17
and the equation for the y-intercept, b0, is
byb x
ni
ii
i
01
=−∑∑5.18
Although equation 5.17 and equation 5.18 appear formidable, it is only
necessary to evaluate the following four summations
xi
i∑
yi
i∑
xyii
i∑
xi
i2∑
Many calculators, spreadsheets, and other statistical software packages are
capable of performing a linear regression analysis based on this model. To
save time and to avoid tedious calculations, learn how to use one of these
tools. For illustrative purposes the necessary calculations are shown in detail
in the following example.
Example 5.9
Using the data from Table 5.1, determine the relationship between Sstd
and Cstd using an unweighted linear regression.
SOLUTION
We begin by setting up a table to help us organize the calculation.
xi yi xiyi xi2
0.000 0.00 0.000 0.000
0.100 12.36 1.236 0.010
0.200 24.83 4.966 0.040
0.300 35.91 10.773 0.090
0.400 48.79 19.516 0.1600.500 60.42 30.210 0.250
Adding the values in each column gives
x
i
i∑ = 1.500 yi
i∑ = 182.31 xyii
i∑ = 66.701 xi
i2∑ = 0.550
6 See, for example, Draper, N. R.; Smith, H. Applied Regression Analysis, 3rd ed.; Wiley: New
York, 1998.See Section 5F in this chapter for details
on completing a linear regression analysis using Excel and R.
Equations 5.17 and 5.18 are written in
terms of the general variables x and y. As
you work through this example, remem-ber that x corresponds to Cstd, and that y
corresponds to Sstd.
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175 Chapter 5 Standardizing Analytical Methods
Substituting these values into equation 5.17 and equation 5.18, we find
that the slope and the y-intercept are
b16 66 701 1 500 182 31
6 0 550 1 500=×() − × ()
×() −.. .
.. (()= ≈2120 706 120 71..
b0182 31 120 706 1 500
60 209 0 21 =− × ()=≈.. .
..
The relationship between the signal and the analyte, therefore, is
Sstd = 120.71 × Cstd + 0.21
For now we keep two decimal places to match the number of decimal plac-
es in the signal. The resulting calibration curve is shown in Figure 5.11.
UNCERTAINTY IN THE REGRESSION ANALYSIS
As shown in Figure 5.11, because of indeterminate error affecting our signal,
the regression line may not pass through the exact center of each data point.
The cumulative deviation of our data from the regression line—that is, the
total residual error—is proportional to the uncertainty in the regression.
We call this uncertainty the standard deviation about the regression ,
sr, which is equal to
syy
nii
i
r=−()
−∑ ˆ2
25.19
Figure 5.11 Calibration curve for the data in Table 5.1 and Example 5.9.Did you notice the similarity between the
standard deviation about the regression (equation 5.19) and the standard devia-tion for a sample (equation 4.1)?
0.0 0.1 0.2 0.3 0.4 0.5
Cstd0102030405060
Sstd
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176 Analytical Chemistry 2.0
where yi is the ith experimental value, and ˆyi is the corresponding value pre-
dicted by the regression line in equation 5.15. Note that the denominator
of equation 5.19 indicates that our regression analysis has n–2 degrees of
freedom—w
e lose two degree of freedom because we use two parameters,
the slope and the y-intercept, to calculate ˆyi.
A more useful representation of the uncertainty in our regression is
to consider the effect of indeterminate errors on the slope, b1, and the y-
intercept, b0, which we express as standard deviations.
sns
nx xs
xxb
i
ii
ii
i12
222
2=
−=
−()∑∑rr
∑ ∑5.20
ssx
nx xsx
bi
i
i
ii
ii
i
022
2222
=
−=∑
∑∑rr ∑ ∑
∑−()nx xi
i2 5.21
We use these standard deviations to establish confidence intervals for the
expected slope, β1, and the expected y-intercept, β0
β111=±bt sb 5.22
β000=±bt sb 5.23
where we select t for a significance level of α and for n–2 degrees of free-
dom. Note that equation 5.22 and equation 5.23 do not contain a factor of
()n−1because the confidence interval is based on a single regression line.
Again, many calculators, spreadsheets, and computer software packages
provide the standard deviations and confidence intervals for the slope and
y-intercept. Example 5.10 illustrates the calculations.
Example 5.10
Calculate the 95% confidence intervals for the slope and y-intercept from
Example 5.9.
SOLUTION
We begin by calculating the standard deviation about the regression. To do
this we must calculate the predicted signals, ˆyi, using the slope and y-in-
tercept from Example 5.9, and the squares of the residual error, ( ˆ) yyii−2.
Using the last standard as an example, we find that the predicted signal is
ˆ .. . . yb b x60 1 60 209 120 706 0 500 60 562 =+ = + × () =
and that the square of the residual error isYou might contrast this with equation
4.12 for the confidence interval around a sample’s mean value.
As you work through this example, re-
member that x corresponds to Cstd, and
that y corresponds to Sstd.
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177 Chapter 5 Standardizing Analytical Methods
( ˆ)( . .) . . yyii− =− = ≈2260 42 60 562 0 2016 0 202
The following table displays the results for all six solutions.
xi yiˆyi ( ˆ) yyii−2
0.000 0.00 0.209 0.0437
0.100 12.36 12.280 0.0064
0.200 24.83 24.350 0.23040.300 35.91 36.421 0.26110.400 48.79 48.491 0.08940.500 60.42 60.562 0.0202
Adding together the data in the last column gives the numerator of equa-
tion 5.19 as 0.6512. The standard deviation about the regression, therefore,
is
sr=−=0 6512
620 4035..
Next we calculate the standard deviations for the slope and the y-intercept
using equation 5.20 and
equation 5.21. The values for the summation
terms are from in
Example 5.9.
sns
nx xb
i
ii
i12
2226 0 4035
=
−=×()
∑∑r.
6 6 0 550 1 5500 9652×() −()=
…
ssx
nx xbi
i
i
ii
i022
220 4035
=
−=( ∑
∑∑r . ) )×
×() −()2
20 550
6 0 550 1 550.
..
Finally, the 95% confidence intervals ( α = 0.05, 4 degrees of fr eedom) for
the slope and
y-intercept are
β111120 706 2 78 0 965 120 7 2 7 =± = ± × () =± bt sb.. . . .
β00002 0 9 27 8 02 9 2 02 08 =± = ± × () =± bt sb.. .. .
The standard deviation about the regression, sr, suggests that the signal, Sstd,
is precise to one decimal place. For this reason we report the slope and the
y-intercept to a single decimal place.You can find values for t in Appendix 4.
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178 Analytical Chemistry 2.0
MINIMIZING UNCERTAINTY IN CALIBRATION CURVES
To minimize the uncertainty in a calibration curve’s slope and y-intercept,
you should evenly space your standards over a wide range of analyte con-
centrations. A close examination of equation 5.20 and equation 5.21 will
help you appreciate why this is tr
ue. The denominators of both equations
include the term ()xxi−∑2. The larger the value of this term—which
you accomplish by increasing the range of x around its mean value—the
smaller the standard deviations in the slope and the y-intercept. Further-
more, to minimize the uncertainty in the y-intercept, it also helps to de-
crease the value of the term xi∑ in equation 5.21, which you accomplish
by including standards for lo
wer concentrations of the analyte.
OBTAINING THE ANALYTE ’S CONCENTRATION FROM A REGRESSION EQUATION
Once we have our regression equation, it is easy to determine the concen-
tration of analyte in a sample. When using a normal calibration curve, for
example, we measure the signal for our sample, Ssamp, and calculate the
analyte’s concentration, CA, using the regression equation.
CSb
bA=−samp 0
15.24
What is less obvious is how to report a confidence interval for CA that
expresses the uncertainty in our analysis. To calculate a confidence interval
we need to know the standard deviation in the analyte’s concentration, sCA,
which is given by the following equation
ss
bmnSS
bC CCr
iiA
stdsamp std
std=+ +−()
() −()∑12
12 211
5.25
where m is the number of r
eplicate used to establish the sample’s average
signal ( Ssamp), n is the number of calibration standards, Sstd is the average
signal for the calibration standards, and C
istd and Cstd are the individual and
mean concentrations for the calibration standards.7 Knowing the value of
sCA , the confidence interval for the analyte’s concentration is
μ
AA A CC Ct s=±
where μCA is the expected value of CA in the absence of determinate errors,
and with the value of t based on the desired level of confidence and n–2
degrees of freedom.
7 (a) Miller, J. N. Analyst 1991, 116, 3–14; (b) Sharaf, M. A.; Illman, D. L.; Kowalski, B. R. Ch-
emometrics, Wiley-Interscience: New York, 1986, pp. 126-127; (c) Analytical Methods Commit-
tee “Uncertainties in concentrations estimated from calibration experiments,” AMC Technical
Brief, March 2006 (http://www.rsc.org/images/Brief22_tcm18-51117.pdf )Equation 5.25 is written in terms of a cali-
bration experiment. A more general form of the equation, written in terms of x and
y, is given here.
ss
bmnYy
bx xxr
i
i=+ +−
−()
() ( )∑ 12
12211
A close examination of equation 5.25 should convince you that the uncertainty in CA is smallest when the sample’s av-
erage signal,
S
samp, is equal to the average
signal for the standards, S
std. When prac-
tical, you should plan your calibration curve so that Ssamp falls in the middle of
the calibration curve.
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179 Chapter 5 Standardizing Analytical Methods
Example 5.11
Three replicate analyses for a sample containing an unknown concentra-
tion of analyte, yield values for Ssamp of 29.32, 29.16 and 29.51. Using
the results from Example 5.9 and Example 5.10, determine the analyte’s
concentration, CA, and its 95% confidence interval.
SOLUTION
The average signal, Ssamp, is 29.33, which, using equation 5.24 and the
slope and the y-intercept from E
xample 5.9, gives the analyte’s concentra-
tion as
CSb
bA=−=−=samp 0
129 33 0 209
120 7060 241..
..
To calculate the standard deviation for the analyte’s concentration we must
determine the values for Sstd and CCstd stdi−()∑2. The former is just the
average signal for the calibration standards, which, using the data in Table
5.1, is 30.385. Calculating CCstd stdi−()∑2looks formidable, but we can sim-
plify its calculation by recognizing that this sum of squares term is the
numerator in a standard deviation equation; thus,
CC s nstd std Cis td−() =() ×−() ∑22
1
where sCstd is the standard deviation for the concentration of analyte in
the calibration standards. Using the data in Table 5.1 we find that sCstd is
0.1871 and
CCstd stdi−() =× −= ∑220 1871 6 1 0 175(. ) ( ) .
Substituting known values into equation 5.25 gives
sCA=+ +− () 0 4035
120 7061
31
629 33 30 385
120 72
.
…
.0 06 0 1750 00242() ×=
..
Finally, the 95% confidence interval for 4 degrees of freedom is
μ
AA A CC Ct s=± = ± × () =± 0 241 2 78 0 0024 0 241 0 007.. . . .
Figure 5.12 shows the calibration curve with curves showing the 95%
confidence interv al for
CA.You can find values for t in Appendix 4.
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180 Analytical Chemistry 2.0
In a standard addition we determine the analyte’s concentration by
extrapolating the calibration curve to the x-intercept. In this case the value
of CA is
Cxb
bA -intercept==−0
1
and the standard deviation in CA isFigure 5.12 Example of a normal calibration curve with
a superimposed confidence interval for the analyte
’s con-
centration. The points in blue are the original data from
Table 5.1 . The black line is the normal calibration curve
as determined in Example 5.9. The red lines show the
95% confidence interval for CA assuming a single deter-
mination of Ssamp.
Practice Exercise 5.4
Figure 5.3 shows a normal calibration curve for the quantitative analysis of Cu
2+. The data for the calibration curve are shown here.
[Cu2+] (M) Absorbance
00
1.55×10–3 0.050
3.16×10–3 0.093
4.74×10–3 0.143
6.34×10–3 0.188
7.92×10–3 0.236
Complete a linear regression analysis for this calibration data, reporting
the calibration equation and the 95% confidence interval for the slope and the y-intercept. If three replicate samples give an S
samp of 0.114,
what is the concentration of analyte in the sample and its 95% confi-dence interval?
Click here to review your answer to this exercise.0102030405060
Sstd
0.0 0.1 0.2 0.3 0.4 0.5
Cstd
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181 Chapter 5 Standardizing Analytical Methods
ss
bnS
bC CCr
iiAstd
std std=+()
() −()∑12
12 21
where n is the number of standard additions (including the sample with no
added standard), and Sstd is the average signal for the n standards. Because
we determine the analyte’s concentration by extrapolation, rather than by
interpolation, sCA for the method of standard additions generally is larger
than for a normal calibration curve.
EVALUATING A LINEAR REGRESSION MODEL
You should never accept the result of a linear regression analysis without
evaluating the validity of the your model. Perhaps the simplest way to evalu-
ate a regression analysis is to examine the residual errors. As we saw earlier,
the residual error for a single calibration standard, ri, is
ry yii i=−( ˆ)
If your regression model is valid, then the residual errors should be ran-
domly distributed about an average residual err
or of zero, with no apparent
trend toward either smaller or larger residual errors (Figure 5.13a). T rends
such as those shown in Figure 5.13b and Figure 5.13c provide evidence that
at least one of the model’s assumptions is incorrect. For example, a trend
toward larger residual errors at higher concentrations, as shown in Figure
5.13b, suggests that the indeterminate errors affecting the signal are not
independent of the analyte’s concentration. In Figure 5.13c, the residual
Figure 5.13 Plot of the residual error in the signal, Sstd, as a function of the concentration of analyte, Cstd for an
unweighted straight-line regression model. The red line shows a residual error of zero. The distribution of the residual
error in (a) indicates that the unweighted linear regression model is appropriate. The increase in the residual errors in (b) for higher concentrations of analyte, suggest that a weighted straight-line regression is more appropriate. For (c), the curved pattern to the residuals suggests that a straight-line model is inappropriate; linear regression using a quadratic model might produce a better fit.0.0 0.1 0.2 0.3 0.4 0.5
Cstd0.0 0.1 0.2 0.3 0.4 0.5
Cstd0.0 0.1 0.2 0.3 0.4 0.5
Cstdresidual error
residual error
residual error(a) (b) (c)
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182 Analytical Chemistry 2.0
errors are not random, suggesting that the data can not be modeled with a
straight-line relationship. Regression methods for these two cases are dis-
cussed in the following sections.
5D.3 Weighted Linear Regression with Errors in y
Our treatment of linear regression to this point assumes that indeterminate
errors affecting y are independent of the value of x. If this assumption is
false, as is the case for the data in Figure 5.13b , then we must include the
variance for each value of y
into our determination of the y-intercept, bo,
and the slope, b1; thus
bwy b wx
nii
iii
i
01
=−∑∑5.26
bnw x y
w x w y
nw x w xiii
iii
iii
i
ii
iii
i1
2=−
−∑∑ ∑
∑∑2 5.27
where wi is a weighting factor that accounts for the variance in yi
wns
siy
y
ii
i=()
()−
−∑2
2 5.28
and syi is the standard deviation for yi. In a weighted linear regression,
each xy-pair’s contribution to the r
egression line is inversely proportional
to the precision of yi—that is, the more precise the value of y, the greater
its contribution to the regression.
Example 5.12
Shown here are data for an external standardization in which sstd is the
standard deviation for three replicate determination of the signal.
Cstd (arbitrary units) Sstd (arbitrary units) sstd
0.000 0.00 0.02
0.100 12.36 0.02
0.200 24.83 0.07Practice Exercise 5.5
Using your results from Practice Exercise 5.4 , construct a residual plot
and explain its significance.
Click here to review your answer to this exercise.
This is the same data used in Example 5.9
with additional information about the standard deviations in the signal.
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183 Chapter 5 Standardizing Analytical Methods
0.300 35.91 0.13
0.400 48.79 0.22
0.500 60.42 0.33
Determine the calibration curve’s equation using a weighted linear regres-
sion.
SOLUTION
We begin by setting up a table to aid in calculating the weighting factors.
xi yisyisyi()−2
wi
0.000 0.00 0.02 2500.00 2.8339
0.100 12.36 0.02 2500.00 2.83390.200 24.83 0.07 204.08 0.23130.300 35.91 0.13 59.17 0.06710.400 48.79 0.22 20.66 0.02340.500 60.42 0.33 9.18 0.0104
Adding together the values in the forth column gives
s
y
ii() =∑2
5293 09 .
which we use to calculate the individual weights in the last column. After
calculating the individual weights, w e use a second table to aid in calculat-
ing the four summation terms in
equation 5.26 and equation 5.27.
xi yi wi wi xi wi yi wi xi2 wi xi yi
0.000 0.00 2.8339 0.0000 0.0000 0.0000 0.00000.100 12.36 2.8339 0.2834 35.0270 0.0283 3.50270.200 24.83 0.2313 0.0463 5.7432 0.0093 1.14860.300 35.91 0.0671 0.0201 2.4096 0.0060 0.72290.400 48.79 0.0234 0.0094 1.1417 0.0037 0.45670.500 60.42 0.0104 0.0052 0.6284 0.0026 0.3142
Adding the values in the last four columns gives
wx wy
wx
ii
iii
i
ii
i∑∑
∑==
=0 3644 44 9499
02..
.0 0499 6 1451 wxyiii
i∑ =.
Substituting these values into the equation 5.26 and equation 5.27 gives
the estimated slope and estimated y-intercept asAs you work through this example, re-
member that x corresponds to Cstd, and
that y corresponds to Sstd.
As a check on your calculations, the sum of the individual weights must equal the number of calibration standards, n. The
sum of the entries in the last column is 6.0000, so all is well.
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184 Analytical Chemistry 2.0
b16 6 1451 0 3644 44 9499
6 0 0499 0=× − ×
× −(.) ( . .)
(.) ( .. ).
3644122 9852=
b044 9499 122 985 0 3644
60 0224 =− ×=.( . . ).
The calibration equation is
Sstd = 122.98 × Cstd + 0.02
Figure 5.14 shows the calibration curve for the weighted regression and the
calibration curve for the unweighted regression in Example 5.9. Although
the two calibration curves ar
e very similar, there are slight differences in the
slope and in the y-intercept. Most notably, the y-intercept for the weighted
linear regression is closer to the expected value of zero. Because the stan-
dard deviation for the signal, Sstd, is smaller for smaller concentrations of
analyte, Cstd, a weighted linear regression gives more emphasis to these
standards, allowing for a better estimate of the y-intercept.
Equations for calculating confidence intervals for the slope, the y-in-
tercept, and the concentration of analyte when using a weighted linear
regression are not as easy to define as for an unweighted linear regression.8
The confidence interval for the analyte’s concentration, however, is at its
8 Bonate, P . J. Anal. Chem. 1993, 65, 1367–1372.
Figure 5.14 A comparison of unweighted and w eighted normal calibration curves.
See Example 5.9 for details of the unweighted linear regression and Example 5.12
for details of the weighted linear regression.0.0 0.1 0.2 0.3 0.4 0.5
Cstd0102030405060
Sstdunweighted linear regressionweighted linear regression
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185 Chapter 5 Standardizing Analytical Methods
optimum value when the analyte’s signal is near the weighted centroid, yc
, of the calibration curve.
ynwxci i
i=∑1
5D.4 Weighted Linear Regression with Errors in Both x and y
If we r
emove our assumption that the indeterminate errors affecting a cali-
bration curve exist only in the signal (y), then we also must factor into
the regression model the indeterminate errors affecting the analyte’s con-
centration in the calibration standards ( x). The solution for the resulting
regression line is computationally more involved than that for either the
unweighted or weighted regression lines.9 Although we will not consider
the details in this textbook, you should be aware that neglecting the pres-
ence of indeterminate errors in x can bias the results of a linear regression.
5D.5 Curvilinear and Multivariate Regression
A straight-line regression model, despite its apparent complexity, is the
simplest functional relationship between two variables. What do we do if
our calibration curve is curvilinear—that is, if it is a curved-line instead of
a straight-line? One approach is to try transforming the data into a straight-
line. Logarithms, exponentials, reciprocals, square roots, and trigonometric
functions have been used in this way. A plot of log( y) versus x is a typical
example. Such transformations are not without complications. Perhaps the
most obvious complication is that data with a uniform variance in y will not
maintain that uniform variance after the transformation.
Another approach to developing a linear regression model is to fit a
polynomial equation to the data, such as y = a + bx + cx2. You can use
linear regression to calculate the parameters a, b, and c, although the equa-
tions are different than those for the linear regression of a straight line.10
If you cannot fit your data using a single polynomial equation, it may be
possible to fit separate polynomial equations to short segments of the cali-
bration curve. The result is a single continuous calibration curve known as
a spline function.
The regression models in this chapter apply only to functions containing
a single independent variable, such as a signal that depends upon the ana-
lyte’s concentration. In the presence of an interferent, however, the signal
may depend on the concentrations of both the analyte and the interferent
9 See, for example, Analytical Methods Committee, “Fitting a linear functional relationship to
data with error on both variable,” AMC Technical Brief, March, 2002 ( http://www.rsc.org/im-
ages/brief10_tcm18-25920.pdf ).
10 For details about curvilinear regression, see (a) Sharaf, M. A.; Illman, D. L.; Kowalski, B. R.
Chemometrics, Wiley-Interscience: New York, 1986; (b) Deming, S. N.; Morgan, S. L. Experi-
mental Design: A Chemometric Approach, Elsevier: Amsterdam, 1987.See Figure 5.2 for an example of a calibra-tion curve that deviates from a straight-line for higher concentrations of analyte.
It is worth noting that in mathematics, the
term “linear” does not mean a straight-line. A linear function may contain many additive terms, but each term can have one and only one adjustable parameter. The function
y = ax + bx2
is linear, but the function
y = axb
is nonlinear. This is why you can use linear regression to fit a polynomial equation to your data.
Sometimes it is possible to transform a
nonlinear function. For example, taking the log of both sides of the nonlinear func-tion shown above gives a linear function.
log(y) = log(a) + blog(x)
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186 Analytical Chemistry 2.0
Sk C k C S=+ +AA I I r e a g
where kI is the interferent’s sensitivity and CI is the interferent’s concentra-
tion. Multivariate calibration curves can be prepared using standards that
contain known amounts of both the analyte and the interferent, and mod-
eled using multivariate regression.11
5E Blank Corrections
Thus far in our discussion of strategies for standardizing analytical methods,
we have assumed the use of a suitable reagent blank to correct for signals
arising from sources other than the analyte. We did not, however ask an
important question—“What constitutes an appropriate reagent blank?”
Surprisingly, the answer is not immediately obvious.
In one study, approximately 200 analytical chemists were asked to evalu-
ate a data set consisting of a normal calibration curve, a separate analyte-free
blank, and three samples of different size but drawn from the same source.12
The first two columns in Table 5.3 shows a series of external standards and
their corresponding signals. The normal calibration curve for the data is
Sstd = 0.0750 × Wstd + 0.1250
where the y-intercept of 0.1250 is the calibration blank. A separate reagent
blank gives the signal for an analyte-free sample.
In working up this data, the analytical chemists used at least four dif-
ferent approaches for correcting signals: (a) ignoring both the calibration
blank, CB, and the reagent blank, RB, which clearly is incorrect; (b) using
the calibration blank only; (c) using the reagent blank only; and (d) using
both the calibration blank and the reagent blank. Table 5.4 shows the equa-
11 Beebe, K. R.; Kowalski, B. R. Anal. Chem. 1987, 59, 1007A–1017A.
12 Cardone, M. J. Anal. Chem. 1986, 58, 433–438.
Table 5.3 Data Used to Study the Blank in an Analytical Method
Wstd Sstd Sample Number Wsamp Ssamp
1.6667 0.2500 1 62.4746 0.8000
5.0000 0.5000 2 82.7915 1.0000
8.3333 0.7500 3 103.1085 1.2000
11.6667 0.8413
18.1600 1.4870 reagent blank 0.1000
19.9333 1.6200
Calibration equation: Sstd = 0.0750 × Wstd + 0.1250
Wstd: weight of analyte used to prepare the external standard; diluted to volume, V.
Wsamp: weight of sample used to prepare sample; diluted to volume, V.Check out the Additional Resources at the
end of the textbook for more information about linear regression with errors in both variables, curvilinear regression, and mul-tivariate regression.
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187 Chapter 5 Standardizing Analytical Methods
tions for calculating the analyte’s concentration using each approach, along
with the resulting concentration for the analyte in each sample.
That all four methods give a different result for the analyte’s concentra-
tion underscores the importance of choosing a proper blank, but does not
tell us which blank is correct. Because all four methods fail to predict the
same concentration of analyte for each sample, none of these blank correc-
tions properly accounts for an underlying constant source of determinate
error.
To correct for a constant method error, a blank must account for signals
from any reagents and solvents used in the analysis, as well as any bias re-
sulting from interactions between the analyte and the sample’s matrix. Both
the calibration blank and the reagent blank compensate for signals from
reagents and solvents. Any difference in their values is due to indeterminate
errors in preparing and analyzing the standards.
Unfortunately, neither a calibration blank nor a reagent blank can cor-
rect for a bias resulting from an interaction between the analyte and the
sample’s matrix. To be effective, the blank must include both the sample’s
matrix and the analyte and, consequently, must be determined using the
sample itself. One approach is to measure the signal for samples of differ-
ent size, and to determine the regression line for a plot of Ssamp versus the Table 5.4 Equations and Resulting Concentr ations of Analyte for Different
Approaches to Correcting for the Blank
Concentration of Analyte in…
Approach for Correcting Signal Equation Sample 1 Sample 2 Sample 3
ignore calibration and reagent blank CW
WS
kWAA
sampsamp
As a m p== 0.1707 0.1610 0.1552
use calibration blank only CW
WS
kWAA
sampsamp
As a m pCB
==−
0.1441 0.1409 0.1390
use reagent blank only CW
WS
kWAA
sampsamp
As a m pRB
==−
0.1494 0.1449 0.1422
use both calibration and reagent blank CW
WS
kWAA
sampsamp
As a m pCB RB
==−−
0.1227 0.1248 0.1261
use total Y ouden blank CW
WS
kWAA
sampsamp
As a m pTYB
==−
0.1313 0.1313 0.1313
CA = concentration of analyte; WA = weight of analyte; Wsamp = weight of sample; kA = slope of calibration curve (0.075–see
Table 5.3); CB = calibration blank (0.125–see Table 5.3); RB = reagent blank (0.100–see Table 5.3 ); TYB = total Youden blank
(0.185–see text)
Because we are considering a matrix effect of sorts, you might think that the method of standard additions is one way to over-come this problem. Although the method of standard additions can compensate for proportional determinate errors, it cannot correct for a constant determinate error; see Ellison, S. L. R.; Thompson, M. T. “Standard additions: myth and reality,” Analyst, 2008, 133, 992–997.
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188 Analytical Chemistry 2.0
amount of sample. The resulting y-intercept gives the signal in the absence
of sample, and is known as the total youden blank.13 This is the true
blank correction. The regression line for the three samples in Table 5.3 is
Ssamp = 0.009844 × Wsamp + 0.185
giving a true blank correction of 0.185. As shown by the last row of Table
5.4, using this value to correct Ssamp gives identical values for the concentra-
tion of analyte in all three samples.
The use of the total Youden blank is not common in analytical work,
with most chemists relying on a calibration blank when using a calibra-
tion curve, and a reagent blank when using a single-point standardization.
As long we can ignore any constant bias due to interactions between the
analyte and the sample’s matrix, which is often the case, the accuracy of an
analytical method will not suffer. It is a good idea, however, to check for
constant sources of error before relying on either a calibration blank or a
reagent blank.
5F Using Excel and R for a Regression Analysis
Although the calculations in this chapter are relatively straightforward—
consisting, as they do, mostly of summations—it can be quite tedious to
work through problems using nothing more than a calculator. Both Excel
and R include functions for completing a linear regression analysis and for
visually evaluating the resulting model.
5F.1 Excel
Let’s use Excel to fit the following straight-line model to the data in Ex-
ample 5.9.
yx=+ββ01
Enter the data into a spreadsheet, as shown in Figure 5.15. Depending
upon your needs, there are many ways that you can use Excel to complete
a linear regression analysis. We will consider three approaches here.
USE EXCEL ’S BUILT-IN FUNCTIONS
If all you need are values for the slope, β1, and the y-intercept, β0, you can
use the following functions:
=intercept(known_y’s, known_x’s)
=slope(known_y’s, known_x’s)
where known_y’s is the range of cells containing the signals ( y), and known_x’s
is the range of cells containing the concentrations ( x). For example, clicking
on an empty cell and entering
13 Cardone, M. J. Anal. Chem. 1986, 58, 438–445.Figure 5.15 Portion of a spread-sheet containing data from Ex-
ample 5.9
(Cstd = Cstd; Sstd =
Sstd).AB
1 Cstd Sstd
2 0.000 0.003 0.100 12.364 0.200 24.835 0.300 35.916 0.400 48.797 0.500 60.42
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189 Chapter 5 Standardizing Analytical Methods
=slope(B2:B7, A2:A7)
returns Excel’s exact calculation for the slope (120.705 714 3).
USE EXCEL ’S DATA ANALYSIS TOOLS
To obtain the slope and the y-intercept, along with additional statistical
details, you can use the data analysis tools in the Analysis ToolPak. The
ToolPak is not a standard part of Excel’s instillation. To see if you have
access to the Analysis ToolPak on your computer, select T ools from the
menu bar and look for the Data Analysis… option. If you do not see Data
Analysis…, select Add-ins… from the T ools menu. Check the box for the
Analysis T oolPak and click on OK to install them.
Select Data Analysis… from the T ools menu, which opens the Data
Analysis window. Scroll through the window, select Regression from the
available options, and press OK. Place the cursor in the box for Input Y
range and then click and drag over cells B1:B7. Place the cursor in the box
for Input X range and click and drag over cells A1:A7. Because cells A1 and
B1 contain labels, check the box for Labels. Select the radio button for
Output range and click on any empty cell; this is where Excel will place the
results. Clicking OK generates the information shown in Figure 5.16.
There are three parts to Excel’s summary of a regression analysis. At the
top of Figure 5.16 is a table of Regression Statistics. The standard error is the
standard deviation about the regression, sr. Also of interest is the value for
Multiple R, which is the model’s correlation coefficient, r, a term with which
you may already by familiar. The correlation coefficient is a measure of the
extent to which the regression model explains the variation in y. Values of r
range from –1 to +1. The closer the correlation coefficient is to ±1, the bet-
ter the model is at explaining the data. A correlation coefficient of 0 means
that there is no relationship between x and y. In developing the calculations
for linear regression, we did not consider the correlation coefficient. There Once you install the Analysis ToolPak, it will continue to load each time you launch Excel.
Figure 5.16 Output from Excel’s Regression command in the Analysis ToolPak. See the text for a discussion of how to interpret the information in these tables.SUMMARY OUTPUT
Regression Statistics
Multiple R 0.99987244
R Square 0.9997449
Adjusted R Square 0.99968113
Standard Error 0.40329713
Observations 6
ANOVA
df SS MS F Significance F
Regression 1 2549.727156 2549.72716 15676.296 2.4405E-08
Residual 4 0.650594286 0.16264857
Total 5 2550.37775
Coefficients Standard Error t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 0.20857143 0.29188503 0.71456706 0.51436267 -0.60183133 1.01897419 -0.60183133 1.01897419
Cstd 120.705714 0.964064525 125.205016 2.4405E-08 118.029042 123.382387 118.029042 123.382387Including labels is a good idea. Excel’s
summary output uses the x-axis label to identify the slope.
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190 Analytical Chemistry 2.0
is a reason for this. For most straight-line calibration curves the correla-
tion coefficient will be very close to +1, typically 0.99 or better. There is
a tendency, however, to put too much faith in the correlation coefficient’s
significance, and to assume that an r greater than 0.99 means the linear
regression model is appropriate. Figure 5.17 provides a counterexample.
Although the regression line has a correlation coefficient of 0.993, the data
clearly shows evidence of being curvilinear. The take-home lesson here is:
don’t fall in love with the correlation coefficient!
The second table in Figure 5.16 is entitled ANOVA , which stands for
analysis of variance. We will take a closer look at ANOVA in Chapter 14.
For now, it is sufficient to understand that this part of Excel’s summary
provides information on whether the linear regression model explains a
significant portion of the variation in the values of y. The value for F is the
result of an F-test of the following null and alternative hypotheses.
H0: regression model does not explain the variation in y
HA: regression model does explain the variation in y
The value in the column for Significance F is the probability for retaining
the null hypothesis. In this example, the probability is 2.5 ×10–6%, sug-
gesting that there is strong evidence for accepting the regression model. As
is the case with the correlation coefficient, a small value for the probability is a likely outcome for any calibration curve, even when the model is inap-
propriate. The probability for retaining the null hypothesis for the data in
Figure 5.17, for example, is 9.0×10
–7%.
The third table in Figure 5.16 provides a summary of the model itself.
The values for the model’
s coefficients—the slope, β1, and the y-intercept,
β0—are identified as intercept and with your label for the x-axis data, which
in this example is Cstd. The standard deviations for the coefficients, sb0 and
sb1, are in the column labeled Standard error. The column t Stat and the
column P-value are for the following t-tests.
slope H0: β1 = 0, HA: β1 ≠ 0
y-intercept H0: β0 = 0, HA: β0 ≠ 0
The results of these t-tests provide convincing evidence that the slope is
not zero, but no evidence that the y-intercept significantly differs from
zero. Also shown are the 95% confidence intervals for the slope and the
y-intercept (lower 95% and upper 95%).
PROGRAM THE FORMULAS YOURSELF
A third approach to completing a regression analysis is to program a spread-
sheet using Excel’s built-in formula for a summation
=sum(first cell:last cell)
and its ability to parse mathematical equations. The resulting spreadsheet
is shown in Figure 5.18.Figure 5.17 Example of fitting a straight-line to curvilinear data.
See Section 4F .2 and Section 4F .3 for a review of the F-test.
See Section 4F .1 for a review of the t-test.02468 1 00246810
xyr = 0.993
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191 Chapter 5 Standardizing Analytical Methods
USING EXCEL TO VISUALIZE THE REGRESSION MODEL
You can use Excel to examine your data and the regression line. Begin by
plotting the data. Organize your data in two columns, placing the x values
in the left-most column. Click and drag over the data and select Insert:
Chart… from the main menu. This launches Excel’s Chart Wizard. Select
xy-chart, choosing the option without lines connecting the points. Click
on Next and work your way through the screens, tailoring the plot to meet
your needs. To add a regression line to the chart, click on the chart and
select Chart: Add Trendline… from the main men. Pick the straight-line
model and click OK to add the line to your chart. By default, Excel displays
the regression line from your first point to your last point. Figure 5.19
shows the result for the data in Figure 5.15.
Figure 5.19 Example of an Excel scatterplot show-
ing the data and a regression line. Figure 5.18 Spreadsheet showing the formulas for calculating the slope and the y-intercept for the data in Ex-
ample 5.9. The cells with the shading contain formulas that y
ou must enter. Enter the formulas in cells C3 to
C7, and cells D3 to D7. Next, enter the formulas for cells A9 to D9. Finally, enter the formulas in cells F2 and F3. When you enter a formula, Excel replaces it with the resulting calculation. The values in these cells should agree with the results in Example 5.9. You can simplify the entering of formulas by copying and pasting. For
example, enter the formula in cell C2. Select Edit: Copy, click and drag your cursor over cells C3 to C7, and
select Edit: Paste. Excel automatically updates the cell referencing.
Excel’s default options for xy-charts do not
make for particularly attractive scientific figures. For example, Excel automatically adds grid lines parallel to the x-axis, which
is a common practice in business charts. You can deselect them using the Grid
lines tab in the Chart Wizard. Excel also
defaults to a gray background. To remove this, just double-click on the chart’s back-ground and select none in the resulting
pop-up window.ABC D E F
1 x y xy x^2 n = 6
2 0.000 0.00 =A2*B2 =A2^2 slope = =(F1*C8 – A8*B8)/(F1*D8-A8^2)
3 0.100 12.36 =A3*B3 =A3^2 y-int = =(B8-F2*A8)/F1
4 0.200 24.83 =A4*B4 =A4^2
5 0.300 35.91 =A5*B5 =A5^2
6 0.400 48.79 =A6*B6 =A6^2
7 0.500 60.42 =A7*B7 =A7^2
8
9 =sum(A2:A7) =sum(B2:B7) =sum(C2:C7) =sum(D2:D7) <–sums
010203040506070
0 0.1 0.2 0.3 0.4 0.5 0.6
x-axisy-axis
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192 Analytical Chemistry 2.0
Excel also will create a plot of the regression model’s residual errors. To
create the plot, build the regression model using the Analysis ToolPak, as
described earlier . Clicking on the option for Residual plots cr eates the plot
shown in Figure 5.20.
LIMITATIONS TO USING EXCEL FOR A REGRESSION ANALYSIS
Excel’s biggest limitation for a regression analysis is that it does not pro-
vide a function for calculating the uncertainty when predicting values of
x. In terms of this chapter, Excel can not calculate the uncertainty for the
analyte’s concentration, CA, given the signal for a sample, Ssamp. Another
limitation is that Excel does not include a built-in function for a weighted
linear regression. You can, however, program a spreadsheet to handle these
calculations.
5F.2 R
Let’s use Excel to fit the following straight-line model to the data in Ex-
ample 5.9.
yx=+ββ01
ENTERING DATA AND CREATING THE REGRESSION MODEL
To begin, create objects containing the concentration of the standards and
their corresponding signals.
> conc = c(0, 0.1, 0.2, 0.3, 0.4, 0.5)
> signal = c(0, 12.36, 24.83, 35.91, 48.79, 60.42)
The command for creating a straight-line linear regression model is
lm(y ~ x)As you might guess, lm is short for linear
model.Figure 5.20 Example of Excel’s plot of a regression model’s r
esidual errors.-0.6-0.4-0.200.20.40.6
0 0.1 0.2 0.3 0.4 0.5 0.6
CstdResiduals
Practice Exercise 5.6
Use Excel to complete the
regression analysis in Practice
Exer
cise 5.4.
Click here to review your an-
swer to this exer
cise.
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193 Chapter 5 Standardizing Analytical Methods
where y and x are the objects containing our data. To access the results of
the regression analysis, we assign them to an object using the following
command
> model = lm(signal ~ conc)
where model is the name we assign to the object.
EVALUATING THE LINEAR REGRESSION MODEL
To evaluate the results of a linear regression we need to examine the data
and the regression line, and to review a statistical summary of the model. To
examine our data and the regression line, we use the plot command, which
takes the following general form
plot(x, y, optional arguments to control style)
where x and y are objects containing our data, and the abline command
abline(object, optional arguments to control style)
where object is the object containing the results of the linear regression.
Entering the commands
> plot(conc, signal, pch = 19, col = “blue”, cex = 2)
> abline(model, col = “red”)
creates the plot shown in Figure 5.21.
To review a statistical summary of the regression model, we use the
summary command.
> summary(model)
The resulting output, shown in Figure 5.22, contains three sections.
The first section of R’s summary of the r
egression model lists the re-
sidual errors. To examine a plot of the residual errors, use the command
> plot(model, which=1)
Figure 5.21 Example of a regression plot in R showing the data and the regression line.
You can customize your plot by adjusting the
plot command’s optional arguments. The argument pch controls the symbol used for plotting points, the argument col allows you to
select a color for the points or the line, and the argument cex sets
the size for the points. You can use the command
help(plot)
to learn more about the options for plotting data in R.The name abline comes from the follow-
ing common form for writing the equa-tion of a straight-line.
y = a + bx
0.0 0.1 0.2 0.3 0.4 0.50 1 02 03 04 05 06 0
concsignalThe reason for including the argument
which =1 is not immediately obvious.
When you use R’s plot command on an
object created by the lm command, the de-
fault is to create four charts summarizing
the model’s suitability. The first of these
charts is the residual plot; thus, which=1
limits the output to this plot.You can choose any name for the object
containing the results of the regression analysis.
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194 Analytical Chemistry 2.0
which produces the result shown in Figure 5.23. Note that R plots the re-
siduals against the predicted (fitted) values of y instead of against the known
values of x. The choice of how to plot the residuals is not critical, as you can
see by comparing Figure 5.23 to Figure 5.20. The line in Figure 5.23 is a
smoothed fit of the residuals.
The second section of Figur
e 5.22 provides the model’s coefficients—
the slope, β1, and the y-intercept, β0—along with their respective standard
deviations (Std. Error). The column t value and the column Pr(>|t|) are for
the following t-tests.
slope H0: β1 = 0, HA: β1 ≠ 0
y-intercept H0: β0 = 0, HA: β0 ≠ 0
The results of these t-tests provide convincing evidence that the slope is not
zero, but no evidence that the y-intercept significantly differs from zero.See Section 4F .1 for a review of the t-test.Figure 5.22 The summary of R’s regression analysis. See the text for a discussion of how to interpret the information in the
output
’s three sections.
Figure 5.23 Example showing R’s plot of a regression model’s residual error
.> model=lm(signal~conc)> summary(model)
Call:
lm(formula = signal ~ conc)
Residuals:
1 2 3 4 5 6 -0.20857 0.08086 0.48029 -0.51029 0.29914 -0.14143
Coefficients:
Estimate Std. Error t value Pr(>|t|) (Intercept) 0.2086 0.2919 0.715 0.514 conc 120.7057 0.9641 125.205 2.44e-08 ***–Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘ . ’ 0.1 ‘ ’ 1
Residual standard error: 0.4033 on 4 degrees of freedom
Multiple R-Squared: 0.9997, Adjusted R-squared: 0.9997 F-statistic: 1.568e+04 on 1 and 4 DF, p-value: 2.441e-08
0 1 02 03 04 05 06 0-0.6 -0.4 -0.2 0.0 0.2 0.4 0.6
Fitted valuesResiduals
lm(si gnal ~ conc)Residuals vs Fitted
43
5
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195 Chapter 5 Standardizing Analytical Methods
The last section of the regression summary provides the standard devia-
tion about the regression ( residual standard error ), the square of the cor-
relation coefficient ( multiple R-squared), and the result of an F-test on the
model’s ability to explain the variation in the y values. For a discussion of
the correlation coefficient and the F-test of a regression model, as well as
their limitations, refer to the section on using Excel’s data analysis tools.
PREDICTING THE UNCERTAINTY IN CA GIVEN SSAMP
Unlike Excel, R includes a command for predicting the uncertainty in an
analyte’s concentration, CA, given the signal for a sample, Ssamp. This com-
mand is not part of R’s standard installation. To use the command you need
to install the “chemCal” package by entering the following command ( note:
you will need an internet connection to download the package).
> install.packages(“chemCal”)
After installing the package, you will need to load the functions into R
using the following command. (note: you will need to do this step each time
you begin a new R session as the package does not automatically load when you
start R).
> library(“chemCal”)
The command for predicting the uncertainty in CA is inverse.predict,
which takes the following form for an unweighted linear regression
inverse.predict(object, newdata, alpha = value)
where object is the object containing the regression model’s results, newdata
is an object containing values for Ssamp, and value is the numerical value for
the significance level. Let’s use this command to complete Example 5.11.
First, we cr
eate an object containing the values of Ssamp
> sample = c(29.32, 29.16, 29.51)
and then we complete the computation using the following command
> inverse.predict(model, sample, alpha = 0.05)
producing the result shown in Figure 5.24. The analyte’s concentration, CA,
is given by the value $Prediction, and its standard deviation, sCA, is shown
as $`Standard Error`. The value for $Confidence is the confidence interval,
±tsCA, for the analyte’s concentration, and $`Confidence Limits` provides
the lower limit and upper limit for the confidence interval for CA.
USING R FOR A WEIGHTED LINEAR REGRESSION
R’s command for an unweighted linear regression also allows for a weighted
linear regression by including an additional argument, weights, whose value is an object containing the weights.
lm(y ~ x, weights
= object)You need to install a package once, but you need to load the package each time you plan to use it. There are ways to con-figure R so that it automatically loads certain packages; see An Introduction to R for more information (click here to view a PDF version of this document). See Section 4F .2 and Section 4F .3 for a review of the F-test.
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196 Analytical Chemistry 2.0
Let’s use this command to complete Example 5.12. First, we need to create
an object containing the weights, which in R are the r
eciprocals of the stan-
dard deviations in y, (syi)–2. Using the data from Example 5.12, we enter
> syi=c(0.02, 0.02, 0.07, 0.13, 0.22, 0.33)
> w=1/syi^2
to create the object containing the weights. The commands
> modelw = lm(signal ~ conc, w
eights = w)
> summary(modelw)
generate the output shown in Figure 5.25. Any difference between the
results shown here and the results shown in Example 5.12 are the result of
round-off errors in our earlier calculations.
Figure 5.25 The summary of R’s regression analysis for
a weighted linear regr
ession. The types of information
shown here is identical to that for the unweighted linear regression in Figure 5.22. Figure 5.24 Output from R’s command for predicting the ana-
lyte’s concentration,
CA, from the sample’s signal, Ssamp.> inverse.predict(model, sample, alpha = 0.05)
$Prediction[1] 0.2412597
$`Standard Error`
[1] 0.002363588
$Confidence
[1] 0.006562373
$`Confidence Limits`
[1] 0.2346974 0.2478221
You may have noticed that this way of defining weights is different than that shown in equation 5.28. In deriving equa-tions for a weighted linear regression, you can choose to normalize the sum of the weights to equal the number of points, or you can choose not to—the algorithm in R does not normalize the weights.
> modelw=lm(signal~conc, weights = w)> summary(modelw)
Call:
lm(formula = signal ~ conc, weights = w)
Residuals:
1 2 3 4 5 6 -2.223 2.571 3.676 -7.129 -1.413 -2.864
Coefficients:
Estimate Std. Error t value Pr(>|t|) (Intercept) 0.04446 0.08542 0.52 0.63 conc 122.64111 0.93590 131.04 2.03e-08 ***–Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘ . ’ 0.1 ‘ ’ 1
Residual standard error: 4.639 on 4 degrees of freedom
Multiple R-Squared: 0.9998, Adjusted R-squared: 0.9997 F-statistic: 1.717e+04 on 1 and 4 DF, p-value: 2.034e-08 Practice Exercise 5.7
Use Excel to complete the regression analysis in Practice
Exer
cise 5.4.
Click here to review your an-
swer to this exer
cise.
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197 Chapter 5 Standardizing Analytical Methods
5G Key Terms
external standard internal standard linear regression
matrix matchingmethod of standard
additionsmultiple-point
standardization
normal calibration curve primary standard reagent grade
residual error secondary standard serial dilution
single-point
standardizationstandard deviation about
the regressiontotal Youden blank
unweighted linear
regressionweighted linear regression
5H Chapter Summary
In a quantitative analysis we measure a signal, Stotal, and calculate the
amount of analyte, nA or CA, using one of the following equations.
Sk n Stotal A A reag=+
Sk C Stotal A A reag=+
To obtain an accurate result we must eliminate determinate errors affect-
ing the signal, Stotal, the method’s sensitivity, kA, and the signal due to the
reagents, Sreag.
To ensure that we accurately measure Stotal, we calibrate our equipment
and instruments. To calibrate a balance, for example, we a standard weight of known mass. The manufacturer of an instrument usually suggests ap-
propriate calibration standards and calibration methods.
To standardize an analytical method we determine its sensitivity. There
are several standardization strategies, including external standards, the
method of standard addition and internal standards. The most common
strategy is a multiple-point external standardization, resulting in a nor-
mal calibration curve. We use the method of standard additions, in which
known amounts of analyte are added to the sample, when the sample’s
matrix complicates the analysis. When it is difficult to reproducibly handle
samples and standards, we may choose to add an internal standard.
Single-point standardizations are common, but are subject to greater
uncertainty. Whenever possible, a multiple-point standardization is pre-
ferred, with results displayed as a calibration curve. A linear regression
analysis can provide an equation for the standardization.
A reagent blank corrects for any contribution to the signal from the
reagents used in the analysis. The most common reagent blank is one in
which an analyte-free sample is taken through the analysis. When a simple
reagent blank does not compensate for all constant sources of determinate
error, other types of blanks, such as the total Youden blank, can be used.As you review this chapter, try to define a key term in your own words. Check your answer by clicking on the key term, which will take you to the page where it was first introduced. Clicking on the key term
there, will bring you back to this page so that you can continue with another key term.
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198 Analytical Chemistry 2.0
5I Problems
1. Describe how you would use a serial dilution to prepare 100 mL each
of a series of standards with concentrations of 1.00 ×10–5, 1.00×10–4,
1.00×10–3, and 1.00×10–2 M from a 0.100 M stock solution. Calcu-
late the uncertainty for each solution using a propagation of uncertainty,
and compare to the uncertainty if you were to prepare each solution by
a single dilution of the stock solution. You will find tolerances for dif-
ferent types of volumetric glassware and digital pipets in Table 4.2 and
Table 4.3. Assume that the uncertainty in the stock solution’s molarity
is ±0.002.
2. Three r
eplicate determinations of Stotal for a standard solution that is
10.0 ppm in analyte give values of 0.163, 0.157, and 0.161 (arbitrary
units). The signal for the reagent blank is 0.002. Calculate the concen-
tration of analyte in a sample with a signal of 0.118.
3. A 10.00-g sample containing an analyte is transferred to a 250-mL
volumetric flask and diluted to volume. When a 10.00 mL aliquot of
the resulting solution is diluted to 25.00 mL it gives signal of 0.235
(arbitrary units). A second 10.00-mL portion of the solution is spiked
with 10.00 mL of a 1.00-ppm standard solution of the analyte and di-
luted to 25.00 mL. The signal for the spiked sample is 0.502. Calculate the weight percent of analyte in the original sample.
4. A 50.00 mL sample containing an analyte gives a signal of 11.5 (arbi-
trary units). A second 50 mL aliquot of the sample, which is spiked with
1.00 mL of a 10.0-ppm standard solution of the analyte, gives a signal
of 23.1. What is the analyte’s concentration in the original sample?
5. An appropriate standard additions calibration curve based on equation
5.10 places S
spike×(Vo + Vstd) on the y-axis and Cstd×Vstd on the x-axis.
Clearly explain why you can not plot Sspike on the y-axis and Cstd×[Vstd/
(Vo + Vstd)] on the x-axis. In addition, derive equations for the slope
and y-intercept, and explain how you can determine the amount of
analyte in a sample from the calibration curve.
6. A standard sample contains 10.0 mg/L of analyte and 15.0 mg/L of
internal standard. Analysis of the sample gives signals for the analyte
and internal standard of 0.155 and 0.233 (arbitrary units), respectively.
Sufficient internal standard is added to a sample to make its concentra-
tion 15.0 mg/L Analysis of the sample yields signals for the analyte and internal standard of 0.274 and 0.198, respectively. Report the analyte’s
concentration in the sample.
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199 Chapter 5 Standardizing Analytical Methods
7. For each of the pair of calibration curves shown in Figure 5.26, se-
lect the calibration curve using the more appropriate set of standards.
Briefly explain the reasons for your selections. The scales for the x-axis
and y-axis are the same for each pair.
8. The following data are for a series of external standards of Cd2+ buffered
to a pH of 4.6.14
[Cd2+] (nM) 15.4 30.4 44.9 59.0 72.7 86.0
Stotal (nA) 4.8 11.4 18.2 25.6 32.3 37.7
14 Wojciechowski, M.; Balcerzak, J. Anal. Chim. Acta 1991, 249, 433–445.Figure 5.26 Calibration curves to accom-
pany Problem 7.Signal
CA
Signal
CASignal
CA
Signal
CASignal
CA
Signal
CA(a)
(b)
(c)
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200 Analytical Chemistry 2.0
(a) Use a linear regression to determine the standardization relationship
and report confidence intervals for the slope and the y-intercept.
(b) Construct a plot of the residuals and comment on their signifi-
cance.
At a pH of 3.7 the following data were recorded for the same set of
external standards.
[Cd2+] (nM) 15.4 30.4 44.9 59.0 72.7 86.0
Stotal (nA) 15.0 42.7 58.5 77.0 101 118
(c) How much more or less sensitive is this method at the lower pH?
(d) A single sample is buffered to a pH of 3.7 and analyzed for cadmium,
yielding a signal of 66.3. Report the concentration of Cd2+ in the
sample and its 95% confidence interval.
9. To determine the concentration of analyte in a sample, a standard ad-
ditions was performed. A 5.00-mL portion of sample was analyzed and
then successive 0.10-mL spikes of a 600.0-mg/L standard of the analyte
were added, analyzing after each spike. The following table shows the
results of this analysis.
Vspike (mL) 0.00 0.10 0.20 0.30
Stotal (arbitrary units) 0.119 0.231 0.339 0.442
Construct an appropriate standard additions calibration curve and use
a linear regression analysis to determine the concentration of analyte in
the original sample and its 95% confidence interval.
10. T roost and Olavsesn investigated the application of an internal stan-
dardization to the quantitative analysis of polynuclear aromatic hy-
drocarbons.15 The following results were obtained for the analysis of
phenanthrene using isotopically labeled phenanthrene as an internal
standard. Each solution was analyzed twice.
CA/CIS0.50 1.25 2.00 3.00 4.00
SA/SIS0.514
0.5220.993
1.0241.486
1.4712.044
20.802.342
2.550
(a) Determine the standardization relationship using a linear regression,
and report confidence intervals for the slope and the y-intercept.
Average the replicate signals for each standard before completing
the linear regression analysis.
(b) Based on your results explain why the authors concluded that the
internal standardization was inappropriate.
15 T roost, J. R.; Olavesen, E. Y. Anal. Chem. 1996, 68, 708–711.
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201 Chapter 5 Standardizing Analytical Methods
11. In Chapter 4 we used a paired t-test to compare two analytical methods
used to independently analyze a series of samples of variable composi-
tion. An alternative approach is to plot the results for one method ver-
sus the results for the other method. If the two methods yield identical
results, then the plot should have an expected slope, β1, of 1.00 and
an expected y-intercept, β0, of 0.0. We can use a t-test to compare the
slope and the y-intercept from a linear regression to the expected values.
The appropriate test statistic for the y-intercept is found by rearranging
equation 5.23.
tb
sb
sbbexp=−
=β00 0
00
Rearranging equation 5.22 gives the test statistic for the slope.
tb
sb
sbbexp.
=−
=− β11 1
1110 0
Reevaluate the data in problem 25 from Chapter 4 using the same
significance level as in the original pr oblem.
12.
Consider the following three data sets, each containing value of y for
the same values of x.
Data Set 1 Data Set 2 Data Set 3
xy1 y2 y3
10.00 8.04 9.14 7.46
8.00 6.95 8.14 6.77
13.00 7.58 8.74 12.74
9.00 8.81 8.77 7.11
11.00 8.33 9.26 7.81
14.00 9.96 8.10 8.84
6.00 7.24 6.13 6.08
4.00 4.26 3.10 5.39
12.00 10.84 9.13 8.15
7.00 4.82 7.26 6.42
5.00 5.68 4.74 5.73
(a) An unweighted linear regression analysis for the three data sets gives
nearly identical results. To three significant figures, each data set
has a slope of 0.500 and a y-intercept of 3.00. The standard devia-
tions in the slope and the y-intercept are 0.118 and 1.125 for each Although this is a common approach for
comparing two analytical methods, it does violate one of the requirements for an unweighted linear regression—that in-determinate errors affect y only. Because
indeterminate errors affect both analytical methods, the result of unweighted linear regression is biased. More specifically, the regression underestimates the slope, b
1,
and overestimates the y-intercept, b0. We
can minimize the effect of this bias by placing the more precise analytical meth-od on the x-axis, by using more samples
to increase the degrees of freedom, and by using samples that uniformly cover the range of concentrations.
For more information, see Miller, J. C.;
Miller, J. N. Statistics for Analytical Chem-
istry, 3rd ed. Ellis Horwood PTR Pren-tice-Hall: New York, 1993. Alternative approaches are found in Hartman, C.; Smeyers-Verbeke, J.; Penninckx, W.; Mas-sart, D. L. Anal. Chim. Acta 1997, 338,
19–40, and Zwanziger, H. W.; Sârbu, C. Anal. Chem. 1998, 70, 1277–1280.
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202 Analytical Chemistry 2.0
data set. All three standard deviations about the regression are 1.24,
and all three data regression lines have a correlation coefficients of
0.816. Based on these results for a linear regression analysis, com-
ment on the similarity of the data sets.
(b) Complete a linear regression analysis for each data set and verify
that the results from part (a) are correct. Construct a residual plot
for each data set. Do these plots change your conclusion from part
(a)? Explain.
(c) Plot each data set along with the regression line and comment on
your results.
(d) Data set 3 appears to contain an outlier. Remove this apparent
outlier and reanalyze the data using a linear regression. Comment on your result.
(e) Briefly comment on the importance of visually examining your
data.
13. Fanke and co-workers evaluated a standard additions method for a vol-
tammetric determination of Tl.
16 A summary of their results is tabu-
lated in the following table.
ppm Tl
addedInstrument Response (μA)
0.000 2.53 2.50 2.70 2.63 2.70 2.80 2.52
0.387 8.42 7.96 8.54 8.18 7.70 8.34 7.98
1.851 29.65 28.70 29.05 28.30 29.20 29.95 28.95
5.734 84.8 85.6 86.0 85.2 84.2 86.4 87.8
Use a weighted linear regression to determine the standardization rela-
tionship for this data.
5J Solutions to Practice Exercises
Practice Exercise 5.1
Substituting the sample’s absorbance into the calibration equation and
solving for CA give
Ssamp = 0.114 = 29.59 M–1 × CA + 0.015
CA = 3.35 × 10-3 M
For the one-point standardization, we first solve for kA
16 Franke, J. P .; de Zeeuw, R. A.; Hakkert, R. Anal. Chem. 1978, 50, 1374–1380.
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203 Chapter 5 Standardizing Analytical Methods
kS
CAstd
std MM ==
×=−− 0 0931
31 6 1 029 4631 .
..
and then use this value of kA to solve for CA.
CS
kAsamp
A MM == = ×−− 0 114
29 4638 7 1 013 .
..
When using multiple standards, the indeterminate errors affecting the
signal for one standard are par
tially compensated for by the indeterminate
errors affecting the other standards. The standard selected for the one-
point standardization has a signal that is smaller than that predicted by the
regression equation, which underestimates kA and overestimates CA.
Click here to return to the chapter.
Prac
tice Exercise 5.2
We begin with equation 5.8
Sk CVVCVV
spike A A
fo
std
fstdGQ
rewriting it as
VkCVkCVV0
fAA o
As t d
fstd  
which is in the form of the linear equation
Y = y-intercept + slope × X
where Y is Sspike and X is Cstd×Vstd/Vf. The slope of the line, therefore,
is kA, and the y-intercept is kACAVo/Vf. The x-intercept is the value of X
when Y is zero, or
0=+ × {}kCV
VkxAA o
fA-intercept
xkCV
V
kCV
V-interceptAA o
f
AAo
f=− =−
Click here to return to the chapter.
Prac
tice Exercise 5.3
Using the calibration equation from Figure 5.7a, we find that the x-in-
tercept is
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204 Analytical Chemistry 2.0
x-intercept
mLmL-1=− =−0 1478
0 08541 731.
..
Plugging this into the equation for the x-intercept and solving for CA gives
the concentration of Mn2+ as
x-intercept mLmL
mg/=− =−×3 4782500
100 6..
.CA
L Lmg/L =69 6.
For Figure 7b , the x-intercept is
x-intercept
mLmL-1=− =−0 1478
0 04253 478.
..
and the concentration of Mn2+ is
x-intercept mLmL
L=− =−×= 3 47825 00
50 006 ..
.CA. .96 mg/L
Click here to return to the chapter.
Prac
tice Exercise 5.4
We begin by setting up a table to help us organize the calculation.
xi yi xiyi xi2
0.000 0.00 0.000 0.000
1.55×10–3 0.050 7.750×10–52.403×10–6
3.16×10–3 0.093 2.939×10–49.986×10–6
4.74×10–3 0.143 6.778×10–42.247×10–5
6.34×10–3 0.188 1.192×10–34.020×10–5
7.92×10–3 0.236 1.869×10–36.273×10–5
Adding the values in each column gives
xi
i∑ = 2.371×10–2 yi
i∑ = 0.710
xyii
i∑ = 4.110×10–3 xi
i2∑ = 1.278×10–4
Substituting these values into equation 5.17 and equation 5.18, we find
that the slope and the y-intercept are
b1326 4 110 10 2 371 10 0 710
61=×× − ××
×−−(. ) (. ) (. )
(. 3 378 10 2 371 1029 5742 2×− ×=−−)( . ).
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205 Chapter 5 Standardizing Analytical Methods
b020 710 29 57 2 371 10
60 0015 =− ××=−.. ( . ).
The regression equation is
Sstd = 29.57 × Cstd + 0.0015
To calculate the 95% confidence intervals, we first need to determine
the standard deviation about the regression. The following table will
help us organize the calculation.
xi yiˆyi ( ˆ) yyii−2
0.000 0.00 0.0015 2.250×10–6
1.55×10–3 0.050 0.0473 7.110×10–6
3.16×10–3 0.093 0.0949 3.768×10–6
4.74×10–3 0.143 0.1417 1.791×10–6
6.34×10–3 0.188 0.1890 9.483×10–7
7.92×10–3 0.236 0.2357 9.339×10–8
Adding together the data in the last column gives the numerator of
equation 5.19 as 1.596×10–5. The standard deviation about the regres-
sion, therefore, is
sr=×
−=×−
− 1 596 10
621 997 106
3 ..
Next, we need to calculate the standard deviations for the slope and the
y-intercept using equation 5.20
and equation 5.21.
sb16 1 997 10
6 1 378 10 2 371 1032
4=××
×× − ×−
−(. )
(. ) (.− −=220 3007
).
sb01 997 10 1 378 10
6 1 378 1032 4
4=×××
××−−
−(. ) (. )
(. ))( . ).
− ×=×−−
2 371 101 441 10223
The 95% confidence intervals are
β11129 57 2 78 0 3007 29 57 0 =± = ± × = ±bt sb.( . .). . M-18 85 M-1
β003
00 0015 2 78 1 441 10 0 0015 =± = ± × × =−bt sb.{ . ( . } . ± ±0 0040 .
With an average Ssamp of 0.114, the concentration of analyte, CA, is
CSb
bAsamp
-1M=−
=−=×0
10 114 0 0015
29 5738 0 1 0..
..− −3M
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206 Analytical Chemistry 2.0
The standard deviation in CA is
sCA=×++−−1 997 10
29 571
31
60 114 0 1183
2932.
.(. . )
(.. ) ( ).
574 778 1025
××=×−
4.408 10-5
and the 95% confidence interval is
μ
AA A CC Ct s=± = × ± × ×
=×−−3 80 10 2 78 4 778 10
38 035.{ . ( . ) }
.1 10 0 13 1033−−±×MM .
Click here to return to the chapter.
Prac
tice Exercise 5.5
To create a residual plot, we need to calculate the residual error for each
standard. The following table contains the relevant information.
xi yiˆyiyyii−ˆ
0.000 0.00 0.0015 -0.0015
1.55×10–3 0.050 0.0473 0.0027
3.16×10–3 0.093 0.0949 -0.0019
4.74×10–3 0.143 0.1417 0.0013
6.34×10–3 0.188 0.1890 -0.0010
7.92×10–3 0.236 0.2357 0.0003
Figure 5.27 shows a plot of the resulting residual errors is shown here. The
residual errors appear random and do not show any significant depen-
dence on the analyte’s concentration. Taken together, these observations
suggest that our regression model is appropriate.
Click here to return to the chapter
Prac
tice Exercise 5.6
Begin by entering the data into an Excel spreadsheet, following the format
shown in Figure 5.15. Because Excel’s Data Analysis tools provide most of
the information we need, we will use it her
e. The resulting output, which
is shown in Figure 5.28, contains the slope and the y-intercept, along
with their respective 95% confidence inter
vals. Excel does not provide a
function for calculating the uncertainty in the analyte’s concentration, CA,
given the signal for a sample, Ssamp. You must complete these calculations
by hand. With an Ssamp. of 0.114, CA
CSb
bAsamp
-1M=−
=−=×0
10 114 0 0014
29 5938 0 1 0..
..− −3MFigure 5.27 Plot of the residual errors for
the data in Practice Ex
ercise 5.5.0.000 0.002 0.004 0.006 0.008-0.0100.0000.010residual error
CA
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207 Chapter 5 Standardizing Analytical Methods
The standard deviation in CA is
sCA=×++−−1 996 10
29 591
31
60 114 0 1183
2932.
.(. . )
(.. ) ( ).
594 772 1025
××=×−
4.408 10-5
and the 95% confidence interval is
μ
AA A CC Ct s=± = × ± × ×
=×−−3 80 10 2 78 4 772 10
38 035.{ . ( . ) }
.1 10 0 13 1033−−±×MM .
Click here to return to the chapter
Prac
tice Exercise 5.7
Figure 5.29 shows an R session for this problem, including loading the
chemCal package, cr
eating objects to hold the values for Cstd, Sstd, and
Ssamp. Note that for Ssamp, we do not have the actual values for the
three replicate measurements. In place of the actual measurements, we
just enter the average signal three times. This is okay because the calcula-
tion depends on the average signal and the number of replicates, and not
on the individual measurements.
Click here to return to the chapterFigure 5.28 Excel’s summary of the regression results for Practice Ex ercise 5.6.SUMMARY OUTPUT
Regression Statistics
Multiple R 0.99979366
R Square 0.99958737Ad
justed R S q0.99948421
Standard Erro r0.00199602
Observations 6
ANOVA
df SS MS F Significance F
Regression 1 0.0386054 0.0386054 9689.9103 6.3858E-08
Residual 4 1.5936E-05 3.9841E-06
Total 5 0.03862133
Coefficients Standard Erro r t Stat P-value Lower 95% Upper 95% Lower 95.0% Upper 95.0%
Intercept 0.00139272 0.00144059 0.96677158 0.38840479 -0.00260699 0.00539242 -0.00260699 0.00539242
Cstd 29.5927329 0.30062507 98.437342 6.3858E-08 28.7580639 30.4274019 28.7580639 30.4274019
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208 Analytical Chemistry 2.0
Figure 5.29 R session for completing Practice Exercise 5.7.> library("chemCal")
> conc=c(0, 1.55e-3, 3.16e-3, 4.74e-3, 6.34e-3, 7.92e-3)> signal=c(0, 0.050, 0.093, 0.143, 0.188, 0.236)> model=lm(signal~conc)> summary(model)
Call:
lm(formula = signal ~ conc)
Residuals:
1 2 3 4 5 6 -0.0013927 0.0027385 -0.0019058 0.0013377 -0.0010106 0.0002328
Coefficients:
Estimate Std. Error t value Pr(>|t|) (Intercept) 0.001393 0.001441 0.967 0.388 conc 29.592733 0.300625 98.437 6.39e-08 ***–Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘ . ’ 0.1 ‘ ’ 1
Residual standard error: 0.001996 on 4 degrees of freedom
Multiple R-Squared: 0.9996, Adjusted R-squared: 0.9995 F-statistic: 9690 on 1 and 4 DF, p-value: 6.386e-08
> samp=c(0.114, 0.114, 0.114)
> inverse.predict(model,samp,alpha=0.05)$Prediction[1] 0.003805234
$`Standard Error`
[1] 4.771723e-05
$Confidence
[1] 0.0001324843
$`Confidence Limits`
[1] 0.003672750 0.003937719
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