Solution Manual to Analytical Chemistry 2.1

250 Pages·2016·8.29 MB·English

(a) (b)
Phase 2
Phase 1
2.95 3.00 3.05 3.10 3.15 3.20 3.25
Mass of Pennies (g)
Analytical Chemistry 2.1
Solutions Manual
1.0 pH
stoichiometric mixture
0.8
absorbance 00..46 metal in excess ligand in excess In– is cinodloicra otof Irn–
indicator’s
0.2
pH = pK color transition
a,HIn
0.0 range
0.0 0.2 0.4 0.6 0.8 1.0
XL
indicator
HIn is color of HIn
Production History
Print Version
Solutions Manual to Modern Analytical Chemistry by David Harvey
ISBN 0–697–39760–3
Copyright © 2000 by McGraw-Hill Companies
Copyright transferred to David Harvey, February 15, 2008
Electronic Version
Solutions Manual to Analytical Chemistry 2.1 by David Harvey (Summer 2016)
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Illustrations
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ous section.
Table of Contents
Chapter 1 ......................................................5
Chapter 2 ......................................................9
Chapter 3 ....................................................15
Chapter 4 ....................................................21
Chapter 5 ....................................................39
Chapter 6 ....................................................51
Chapter 7 ....................................................79
Chapter 8 ....................................................95
Chapter 9 ..................................................115
Chapter 10 ................................................157
Chapter 11 ................................................181
Chapter 12 ................................................201
Chapter 13 ................................................217
Chapter 14 ................................................227
Chapter 15 ................................................245
Appendix ...................................................249
iii

Chapter 1 Introduction to Analytical Chemistry 5
Chapter 1
1. (a) A qualitative and a quantitative analysis is the best choice because
we need to determine the identify of the possible contaminants and
determine if their concentrations are greater than the expected back-
ground levels.
(b) A forged work of art often contains compounds that are not pres-
ent in authentic materials or contains a distribution of compounds
that do not match the authentic materials. Either a qualitative anal-
ysis (to identify a compound that should not be present in authentic
materials) or a quantitative analysis (to determine if the concentra-
tions of compounds present do not match the distribution expected
in authentic materials) is appropriate.
(c) Because we are interested in detecting the presence of specific
compounds known to be present in explosive materials, a qualitative
analysis is the best choice.
(d) A compound’s structure is one of its characteristic properties; a
characterization analysis, therefore, is the best approach.
(e) In searching for a new acid–base indicator we are seeking to im-
prove the performance of an existing analytical method, which re-
quires a fundamental analysis of the method’s properties.
(f) A quantitative analysis is used to determine if an automobile emits
too much carbon monoxide.
2. Answers to this problem will vary, but here is a list of important
points that you might address:
The goal of this research is to develop a fast, automated, and real-time
instrumental method for determining a coffee’s sensory profile that
yields results similar to those from trained human sensory panels.
One challenge the authors have to address is that a human sensory
panelist reports results on a relative scale, typically 0–10, for charac-
teristics that are somewhat arbitrary: What does it mean, for example,
to say that a coffee is bitter? An instrumental method, on the other
hand, reports results on an absolute scale and for a clearly defined
signal; in this case, the signal is a raw count of the number of ions
with a particular mass–to-charge ratio. Much of the mathematical
processing described by the authors is used to transform the instru-
mental data into a relative form and to normalize the two sets of data
to the same relative scale.
The instrumental technique relies on gas chromatography equipped
See Chapter 12 for a discussion of gas
with a mass spectrometer as a detector. The specific details of the chromatography and for detection using
instrument are not important, but the characteristics the authors de- a mass spectrometer.
scribe—low fragmentation, high time resolution, broad linear dy-
6 Solutions Manual for Analytical Chemistry 2.1
namic range—are important. When a species enters a mass spectrom-
eter it is ionized (the PTR—proton transfer reaction—in PTR-MS
simply describes the method of ionization) and the individual ions,
being unstable, may decompose into smaller ions. As a roasted coffee
has more than 1000 volatile components, many of which do not con-
tribute to the sensory profile, the authors wish to limit the number
of ions produced in the mass spectrometer. In addition, they want to
ensure that the origin of each ion traces back to just a small number of
volatile compounds so that the signal for each ion carries information
about a small number of compounds. Table 1, for example, shows
that the 16 ions monitored in this study trace back to just 32 unique
volatile compounds, and that, on average, each ion traces back to 3–4
unique volatile compounds with a range of one to eight.
The authors need high time resolution so that they can monitor the
release of volatile species as a function of time, as seen in Figure 1,
and so that they can report the maximum signal for each ion during
the three-minute monitoring period. A rapid analysis also means they
can monitor the production of coffee in real time on the production
line instead of relying on a lengthy off-line analysis completed by a
For a discussion of quality control and
sensory panel. This is advantageous when it comes to quality control
quality assurance, see Chapter 15.
where time is important.
A broad linear dynamic range simply means there is a linear relation-
ship between the measured signal and the concentration of the com-
pounds contributing to that signal over a wide range of concentra-
tions. The assumption of a linear relationship between signal and con-
For a discussion of the relationship be-
tween signal and concentration, see Chap- centration is important because a relative change in concentration has
ter 5.
the same affect on the signal regardless of the original concentration.
A broad range is important because it means the signal is sensitive to
a very small concentration of a volatile compound and that the signal
does not become saturated, or constant, at higher concentrations of
the volatile compound; thus, the signal carries information about a
much wider range of concentrations.
To test their method, the authors divide their samples into two sets:
a training set and a validation set. The authors use the training set to
build a mathematical model that relates the normalized intensities
of the 16 ions measured by the instrument to the eight normalized
relative attributes evaluated by members of the sensory panel. The
specific details of how they created the mathematical model are not
important here, but the agreement between the panel’s sensory profile
and that predicted using the instrumental method generally is very
good (see Figure 3; note that the results for Espresso No. 5 and No.
11 show the least agreement).
Any attempt to create a model that relates one measurement (results
from the sensory panel) to a second measurement (results from the
Chapter 1 Introduction to Analytical Chemistry 7
instrumental analysis) is subject to a number of limitations, the most
important of which is that the model works well for the data set used
to build the model, but that it fails to work for other samples. To test
the more general applicability of their model—what they refer to as
See Chapter 14 for a discussion of robust-
a robust model—the authors use the model to evaluate the data in ness and other ways to characterize an an-
their validation set; the results, shown in Figure 4, suggest that the alytical method.
can apply their model both to coffees of the same type, but harvested
in a different year, and to coffees of a different type.
8 Solutions Manual for Analytical Chemistry 2.1
Chapter 2 Basic Tools of Analytical Chemistry 9
Chapter 2
1. (a) 3 significant figures; (b) 3 significant figures; (c) 5 significant fig-
ures; (d) 3 significant figures; (e) 4 significant figures; (f) 3 significant
figures
For (d) and for (e), the zero in the tenths place is not a significant digit
as its presence is needed only to indicate the position of the decimal
point. If you write these using scientific notation, they are 9.03×10–2
and 9.030×10–2, with three and four significant figures respectively.
2. (a) 0.894; (b) 0.893; (c) 0.894; (d) 0.900; (e) 0.0891
3. (a) 12.01; (b) 16.0; (c) 6.022×1023 mol–1; (d) 9.65×104 C/mol
4. a. 4.591 + 0.2309 + 67.1 = 71.9219 ≈ 71.9 In problem 4 we use a bold red font to
help us keep track of significant figures.
b. 313 – 273.15 = 39.85 ≈ 39.8
For example, in (a) we mark the last sig-
nificant digit common to the numbers we
Note that for (b) we retain an extra significant figure beyond that sug-
are adding together, and in (e), where we
gested by our simple rules so that the uncertainty in the final answer are multiplying and dividing, we identify
(1 part out of 398) is approximately the same as the most uncertain the number with the smallest number of
significant digits.
of our two measurements (1 part out of 313). Reporting the answer as
40, or 4.0×101, as suggested by our simple rules, gives an uncertainty
in the final result of 1 part out of 40, which is substantially worse than
either of our two measurements.
c. 712 × 8.6 = 6123.2 ≈ 6.1×103
d. 1.43/0.026 = 55.00 ≈ 55
e. (8.314 × 298)/96 485 = 0.0256783 ≈ 2.57×10–2
f. log(6.53 × 10–5) = –4.18509 = –4.185
Note that when we take the logarithm of a number, any digits before
the decimal point provide information on the original number’s pow-
er of 10; thus, the 4 in –4.185 is not counted as a significant digit.
g. 10–7.14 = 7.244×10–8 ≈ 7.2×10–8
Note that we take the antilog of a number, the digits before the dec-
imal point provide information on the power of 10 for the resulting
answer; thus, the 7 in –7.14 is not counted as a significant digit.
h. (6.51 × 10–5) × (8.14 × 10–9) = 5.29914×10–13 ≈ 5.30×10–13
5. To find the %w/w Ni, we first subtract the mass of Co from point B
from the combined mass of Ni and Co from point B, and then divide
by the mass of sample; thus
(0.2306-0.0813)g Ni
#100 =
12.1374g sample
0.1493g Ni
= 1.230%w/w Ni
12.1374g sample
10 Solutions Manual for Analytical Chemistry 2.1
For problems in this chapter, all formu- 6. Using the atomic weights from Appendix 18, we find that the formula
la weights are reported to the number of weight for Ni(C H N O ) is
4 14 4 4 2
significant figures allowed by the atomic
weights in Appendix 18. As a compound’s (58.693)+(8#12.011)++(14#1.008)
formula weight rarely limits the uncer- (4#14.007)+(4#15.999) = 288.917g/mole
tainty in a calculation, in later chapters
– –
usually we will round formula weights to 7. First we convert the mass of Cl to moles of Cl
a smaller number of significant figures, 1gCl-
chosen such that it does not limit the cal- 256mgCl-# #
1000mgCl-
culation’s uncertainty.
1molCl-
For problem 7 we include an extra signifi- 35.45gCl- = 7.221#10-3 molCl-
cant figure in each of the calculation’s first
two steps to avoid the possibility of intro- –
and then the moles of Cl to the moles of BaCl
ducing a small error in the final calcula- 2
tion as a result of rounding. If we need to 1molBaCl
7.221#10-3 molCl-# 2 = 3.610#10-3 molBaCl
report the result for an intermediate calcu- 2molCl- 2
lation, then we round that result appropri-
ately; thus, we need to isolate 3.61×10–3 and finally the moles of BaCl2 to the volume of our BaCl2 solution
mol of BaCl2. 3.610#10-3 molBaCl #
2
1L 1000mL
# = 16.6mL
0.217molBaCl 1L
2
8. We can express a part per million in several ways—this is why some
organizations recommend against using the abbreviation ppm—but
here we must assume that the density of the solution is 1.00 g/mL and
that ppm means mg/L or µg/mL. As molarity is expressed as mol/L,
we will use mg/L as our starting point; thus
0.28mg Pb 1g 1molPb
# # = 1.4#10-6 MPb
L 1000mg 207.2g Pb
9. (a) The molarity of 37.0% w/w HCl is
37.0g HCl 1.18g solution
# #
1.00#102 g solution mlsolution
1000mL 1molHCl
# = 12.0MHCl
L 36.46g HCl
(b) To calculate the mass and volume of solution we begin with the
molarity calculated in part (a). To avoid any errors due to rounding
the molarity down to three significant, we will return one additional
significant figure, taking the molarity as 11.97 M.
1L
0.315molHCl# #
11.97molHCl
1000mL 1.18g solution
# = 31.1g
L mL
1L 1000mL
0.315molHCl# # = 26.3mL
11.97molHCl L