Principles of Instrumental Analysis 6th Edition Instructor's Manual

235 Pages·1987·5.92 MB·English

by  Douglas A. Skoog

Skoog/Holler/Crouch Chapter 1
Principles of Instrumental Analysis, 6th ed. Instructor’s Manual

CHAPTER 1
1-1. A transducer is a device that converts chemical or physical information into an electrical
signal or the reverse. The most common input transducers convert chemical or physical
information into a current, voltage, or charge, and the most common output transducers
convert electrical signals into some numerical form.
1-2. The information processor in a visual color measuring system is the human brain.
1-3. The detector in a spectrograph is a photographic film or plate.
1-4. Smoke detectors are of two types: photodetectors and ionization detectors. The
photodetectors consist of a light source, such as a light-emitting diode (LED) and a
photodiode to produce a current proportional to the intensity of light from the LED.
When smoke enters the space between the LED and the photodiode, the photocurrent
decreases, which sets off an alarm. In this case the photodiode is the transducer.
In ionization detectors, which are the typical battery-powered detectors found in homes, a
small radioactive source (usually Americium) ionizes the air between a pair of electrodes.
When smoke enters the space between the electrodes, the conductivity of the ionized air
changes, which causes the alarm to sound. The transducer in this type of smoke detector
is the pair of electrodes and the air between them.
1-5. A data domain is one of the modes in which data may be encoded. Examples of data
domain classes are the analog, digital and time domains. Examples of data domains are
voltage, current, charge, frequency, period, number.
1
Principles of Instrumental Analysis, 6th ed. Chapter 1

1-6. Analog signals include voltage, current, charge, and power. The information is encoded
in the amplitude of the signal.
1-7.
Output Transducer Use
LCD display Alphanumeric information
Computer monitor Alphanumeric information, text,
graphics
Laser printer Alphanumeric and graphical
information
Motor Rotates to change position of
attached elements

1-8. A figure of merit is a number that provides quantitative information about some
performance criterion for an instrument or method.
1-9. Let c = molar concentration of Cu2+ in standard = 0.0287 M
s
c = unknown Cu2+ concentration
x
V = volume of standard = 0.500 mL
s
V = volume of unknown = 25.0 mL
x
S = signal for unknown = 23.6
1
S = signal for unknown plus standard = 37.9
2
Assuming the signal is proportional to c and c , we can write
x s
S = Kc or K = S /c
1 x 1 x
After adding the standard
⎛V c +V c ⎞
S = K⎜ x x s s ⎟
2 V +V
⎝ ⎠
x s
Substituting for K and rearranging gives,
SV c
c = 1 s s
x S (V +V )−SV
2 x s 1 x
2
Principles of Instrumental Analysis, 6th ed. Chapter 1

23.6×0.500 mL×0.0287 M
c = = 9.00×10−4 M
x
37.9(0.500 mL + 25.0 mL)−(23.6×25.0 mL)

1-10. The results are shown in the spreadsheet below.

(a) Slope, m = 0.0701, intercept, b = 0.0083
(b) From LINEST results, SD slope, s = 0.0007, SD intercept, s = 0.0040
m b
(c) 95% CI for slope m is m ± ts where t is the Student t value for 95% probability and N –
m
2 = 4 degrees of freedom = 2.78
95% CI for m = 0.0701 ± 2.78 × 0.0007 = 0.0701 ± 0.0019 or 0.070 ± 0.002
For intercept, 95% CI = b ± ts = 0.0083 ± 2.78 × 0.004 = 0.0083 ± 0.011 or 0.08 ± 0.01
b
(d) c = 4.87 ± 0.086 mM or 4.87 ± 0.09 mM
u
3
Principles of Instrumental Analysis, 6th ed. Chapter 1

1-11. The spreadsheet below gives the results

(a) See plot in spreadsheet.
(b) c = 0.410 μg/mL
u
(c) S = 3.16V + 3.25
s
bc 3.246×2.000 μg/mL
(d) c = s = =0.410 μg/mL
u mV 3.164 mL−1×5.00 mL
u
(e) From the spreadsheet s = 0.002496 or 0.002 μg/mL
c
4
Principles of Instrumental Analysis, 6th ed. Chapter 2

2-3. V = 12.0 × [(2.5 + 4.0) × 103]/[(1.0 + 2.5 + 4.0) × 103] = 10.4 V
2.4
With meter in parallel across contacts 2 and 4,
1 1 1 R + 6.5 kΩ
= + = M
R (2.5 + 4.0) kΩ R R ×6.5 kΩ
2,4 M M
R = (R × 6.5 kΩ)/(R + 6.5 kΩ)
2,4 M M
(a) R = (5.0 kΩ × 6.5 kΩ)/(5.0 kΩ + 6.5 kΩ) = 2.83 kΩ
2,4
V = (12.0 V × 2.83 kΩ)/(1.00 kΩ + 2.83 kΩ) = 8.87 V
M
8.87 V−10.4 V
rel error = × 100% = −15%
10.4 V
Proceeding in the same way, we obtain (b) –1.7% and (c) –0.17%
2-4. Applying Equation 2-19, we can write
750 Ω
(a) −1.0% = ×100%
(R −750 Ω)
M
R = (750 × 100 – 750) Ω = 74250 Ω or 74 kΩ
M
750 Ω
(b) −0.1% = ×100%
(R −750 Ω)
M
R = 740 kΩ
M
2-5. Resistors R and R are in parallel, the parallel combination R is given by Equation 2-17
2 3 p
R = (500 × 200)/(500 + 200) = 143 Ω
p
(a) This 143 Ω R is in series with R and R . Thus, the voltage across R is
p 1 4 1
V = (15.0 × 100)/(100 + 143 + 1000) = 1.21 V
1
V = V = 15.0 V × 143/1243 = 1.73 V
2 3
V = 15.0 V × 1000/1243 = 12.1 V
4
2
Principles of Instrumental Analysis, 6th ed. Chapter 2

(b) I = I = 15.0/(100 + 143 + 1000) = 1.21 × 10–2 A
1 5
I = 1.73 V/500 Ω= 3.5 × 10–3 A
2
I = I = 1.73 V /200 Ω = 8.6 × 10–3 A
3 4
(c) P = IV = 1.73 V× 8.6 × 10–3 A = 1.5 × 10–2 W
(d) Since point 3 is at the same potential as point 2, the voltage between points 3 and
4 V′ is the sum of the drops across the 143 W and the 1000 W resistors. Or,
V ′ = 1.73 V + 12.1 V = 13.8 V. It is also the source voltage minus the V
1
V ′ = 15.0 – 1.21 = 13.8 V
2-6. The resistance between points 1 and 2 is the parallel combination or R and R
B C
R = 2.0 kΩ × 4.0 kΩ/(2.0 kΩ + 4.0 kΩ) = 1.33 kΩ
1,2
Similarly the resistance between points 2 and 3 is
R = 2.0 kΩ × 1.0 kΩ/(2.0 kΩ + 1.0 kΩ) = 0.667 kΩ
2,3
These two resistors are in series with R for a total series resistance R of
A T
R = 1.33 kΩ + 0.667 kΩ + 1.0 kΩ = 3.0 kΩ
T
I = 24/(3000 Ω) = 8.0 × 10–3 A
(a) P = I2R = (8.0 × 10–3)2 × 1.33 × 103 = 0.085 W
1,2 1,2
(b) As above I = 8.0 × 10–3 A
(c) V = IR = 8.0× 10–3 A × 1.0 × 103 Ω = 8.0 V
A A
(d) V = 14 × R /R = 24 × 0.667/3.0 = 5.3 V
D 2,3 T
(e) V = 24 – V = 24 -8.0 = 16 V
5,2 A
2-7. With the standard cell in the circuit,
V = V × AC/AB where V is the battery voltage
std b b
3
Principles of Instrumental Analysis, 6th ed. Chapter 2

1.018 = V × 84.3/AB
b
With the unknown voltage V in the circuit,
x
V = V × 44.2/AB
x b
Dividing the third equation by the second gives,
1.018 V 84.3 cm
=
V 44.3 cm
x
V = 1.018 × 44.3 cm/84.3 cm = 0.535 V
x
R
2-8. E = − S ×100%
r R +R
M S
20
For R = 20 Ω and R = 10 Ω, E = − ×100% = −67%
S M r 10+20
20
Similarly, for R = 50 Ω, E = − ×100% = −29%
M r 50+20
The other values are shown in a similar manner.
R
2-9. Equation 2-20 is E = − std ×100%
r R +R
L std
1 Ω
For R = 1 Ω and R = 1 Ω, E = − ×100% = −50%
std L r 1 Ω+1 Ω
1 Ω
Similarly for R = 10 Ω, E = − ×100% = −9.1%
L r 10 Ω+1 Ω
The other values are shown in a similar manner.
2-10. (a) R = V/I = 1.00 V/50 × 10–6 A = 20000 Ω or 20 kΩ
s
(b) Using Equation 2-19
20 kΩ
−1% = − ×100%
R +20 kΩ
M
4
Principles of Instrumental Analysis, 6th ed. Chapter 2

R = 20 kΩ × 100 – 20 kΩ = 1980 kΩ or ≈ 2 MΩ
M
2−11. Ι = 90/(20 + 5000) = 1.793 × 10–2 A
1
I = 90/(40 + 5000) = 1.786 × 10–2 A
2
% change = [(1.786 × 10–2 – 1.793 × 10–2 A)/ 1.793 × 10–2 A] × 100% = – 0.4%
2-12. I = 9.0/520 = 1.731 × 10–2 A
1
I = 9.0/540 = 1.667 × 10–2 A
2
% change = [(1.667 × 10–2 – 1.731 × 10–2 A)/ 1.731 × 10–2 A] × 100% = – 3.7%
2-13. i = I e–t/RC (Equation 2-35)
init
RC = 10 × 106 Ω × 0.2 × 10–6 F = 2.00 s I = 24V/(10 × 106 Ω) = 2.4 × 10–6 A
init
i = 2.4 × 10–6 e–t/2.00 A or 2.4 e–t/2.0 μA

t, s i, μA t, s i, μA
0.00 2.40 1.0 1.46
0.010 2.39 10 0.0162
0.10 2.28

2-14. v = V e–t/RC (Equation 2-40)
C C
v /V = 1.00/100 for discharge to 1%
C C
0.0100 = e–t/RC = e−t/R×0.015×10−6
ln 0.0100 = –4.61 = –t/1.5 × 10–8R
t = 4.61 × 1.5 × 10–8R = 6.90 × 10–8R
(a) When R = 10 MΩ or 10 × 106 Ω, t = 0.69 s
(b) Similalry, when R = 1 MΩ, t = 0.069 s
5
Principles of Instrumental Analysis, 6th ed. Chapter 2

(c) When R = 1 kΩ, t = 6.9 × 10–5 s
2-15. (a) When R = 10 MΩ, RC = 10 × 106 Ω × 0.015 × 10–6 F = 0.15 s
(b) RC = 1 × 106 × 0.015 × 10–6 = 0.015 s
(c) RC = 1 × 103 × 0.015 × 10–6 = 1.5 × 10–5 s
2-16. Parts (a) and (b) are given in the spreadsheet below. For part (c), we calculate the
quantities from
i = I e-t/RC, v = iR, and v = 25 – v
init R C R
For part (d) we calculate the quantities from
−v
i = C e−t/RC, v = iR, and v = –v
R C R
R
The results are given in the spreadsheet.
6
PPrriinncciipplleess ooff IInnssttrruummeennttaall AAnnaallyyssiiss,, 66tthh eedd.. CChhaapptteerr 22

77