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Principles of Instrumental Analysis, 6th ed. ...

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Skoog/Holler/Crouch Principles of Instrumental Analysis, 6th ed.

Chapter 1 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. 1-6.

Chapter 1

Analog signals include voltage, current, charge, and power. The information is encoded in the amplitude of the signal.

1-7. Output Transducer LCD display Computer monitor

Use Alphanumeric information Alphanumeric information, text, graphics Alphanumeric and graphical information Rotates to change position of attached elements

Laser printer Motor

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 cs= molar concentration of Cu2+ in standard = 0.0287 M cx = unknown Cu2+ concentration Vs = volume of standard = 0.500 mL Vx = volume of unknown = 25.0 mL S1 = signal for unknown = 23.6 S2 = signal for unknown plus standard = 37.9 Assuming the signal is proportional to cx and cs , we can write S1 = Kcx or

K = S1/cx

After adding the standard

⎛ V c +V c ⎞ S2 = K ⎜ x x s s ⎟ ⎝ V x + Vs ⎠ Substituting for K and rearranging gives, cx =

S1Vs cs S2 (Vx + Vs ) − S1Vx

2

Principles of Instrumental Analysis, 6th ed.

cx =

Chapter 1

23.6 × 0.500 mL × 0.0287 M = 9.00 × 10−4 M 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, sm = 0.0007, SD intercept, sb = 0.0040

(c)

95% CI for slope m is m ± tsm where t is the Student t value for 95% probability and N – 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 ± tsb = 0.0083 ± 2.78 × 0.004 = 0.0083 ± 0.011 or 0.08 ± 0.01

(d)

cu = 4.87 ± 0.086 mM or 4.87 ± 0.09 mM 3

Principles of Instrumental Analysis, 6th ed.

Chapter 1

1-11. The spreadsheet below gives the results

(a)

See plot in spreadsheet.

(b)

cu = 0.410 μg/mL

(c)

S = 3.16Vs + 3.25

(d)

cu =

(e)

From the spreadsheet sc = 0.002496 or 0.002 μg/mL

3.246 × 2.000 μg/mL bcs = = 0.410 μg/mL mVu 3.164 mL−1 × 5.00 mL

4

Principles of Instrumental Analysis, 6th ed. 2-3.

Chapter 2

V2.4 = 12.0 × [(2.5 + 4.0) × 103]/[(1.0 + 2.5 + 4.0) × 103] = 10.4 V With meter in parallel across contacts 2 and 4, 1 1 1 R + 6.5 kΩ = + = M R 2,4 (2.5 + 4.0) k Ω RM RM ×6.5 k Ω

R2,4 = (RM × 6.5 kΩ)/(RM + 6.5 kΩ) (a) R2,4 = (5.0 kΩ × 6.5 kΩ)/(5.0 kΩ + 6.5 kΩ) = 2.83 kΩ VM = (12.0 V × 2.83 kΩ)/(1.00 kΩ + 2.83 kΩ) = 8.87 V rel error =

8.87 V − 10.4 V × 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 (a)

− 1.0% =

750 Ω × 100% (RM − 750 Ω )

RM = (750 × 100 – 750) Ω = 74250 Ω or 74 k Ω (b)

− 0.1% =

750 Ω × 100% (RM − 750 Ω )

RM = 740 k Ω 2-5.

Resistors R2 and R3 are in parallel, the parallel combination Rp is given by Equation 2-17 Rp = (500 × 200)/(500 + 200) = 143 Ω (a) This 143 Ω Rp is in series with R1 and R4. Thus, the voltage across R1 is V1 = (15.0 × 100)/(100 + 143 + 1000) = 1.21 V V2 = V3 = 15.0 V × 143/1243 = 1.73 V V4 = 15.0 V × 1000/1243 = 12.1 V 2

Principles of Instrumental Analysis, 6th ed. (b)

Chapter 2

I1 = I5 = 15.0/(100 + 143 + 1000) = 1.21 × 10–2 A I2 = 1.73 V/500 Ω= 3.5 × 10–3 A I3 = I4 = 1.73 V /200 Ω = 8.6 × 10–3 A

(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 V1 V′ = 15.0 – 1.21 = 13.8 V

2-6.

The resistance between points 1 and 2 is the parallel combination or RB and RC R1,2 = 2.0 kΩ × 4.0 kΩ/(2.0 kΩ + 4.0 kΩ) = 1.33 kΩ Similarly the resistance between points 2 and 3 is R2,3 = 2.0 kΩ × 1.0 kΩ/(2.0 kΩ + 1.0 kΩ) = 0.667 kΩ These two resistors are in series with RA for a total series resistance RT of RT = 1.33 kΩ + 0.667 kΩ + 1.0 kΩ = 3.0 kΩ I = 24/(3000 Ω) = 8.0 × 10–3 A (a) P1,2 = I2R1,2 = (8.0 × 10–3)2 × 1.33 × 103 = 0.085 W (b) As above I = 8.0 × 10–3 A (c) VA = IRA = 8.0× 10–3 A × 1.0 × 103 Ω = 8.0 V (d) VD = 14 × R2,3/RT = 24 × 0.667/3.0 = 5.3 V (e) V5,2 = 24 – VA = 24 -8.0 = 16 V

2-7.

With the standard cell in the circuit, Vstd = Vb × AC/AB where Vb is the battery voltage 3

Principles of Instrumental Analysis, 6th ed.

Chapter 2

1.018 = Vb × 84.3/AB With the unknown voltage Vx in the circuit, Vx = Vb × 44.2/AB Dividing the third equation by the second gives, 1.018 V 84.3 cm = Vx 44.3 cm

Vx = 1.018 × 44.3 cm/84.3 cm = 0.535 V 2-8.

Er = −

RS × 100% RM + RS

For RS = 20 Ω and RM = 10 Ω, E r = − Similarly, for RM = 50 Ω, Er = −

20 × 100% = − 67% 10 + 20

20 × 100% = − 29% 50 + 20

The other values are shown in a similar manner. 2-9.

Equation 2-20 is Er = −

Rstd ×100% RL + Rstd

For Rstd = 1 Ω and RL = 1 Ω, Er = − Similarly for RL = 10 Ω, Er = −

1Ω × 100% = − 50% 1 Ω +1 Ω

1Ω × 100% = − 9.1% 10 Ω + 1 Ω

The other values are shown in a similar manner. 2-10. (a) Rs = V/I = 1.00 V/50 × 10–6 A = 20000 Ω or 20 kΩ (b) Using Equation 2-19 −1% = −

20 kΩ × 100% R M + 20 kΩ

4

Principles of Instrumental Analysis, 6th ed.

Chapter 2

RM = 20 kΩ × 100 – 20 kΩ = 1980 kΩ or ≈ 2 MΩ 2−11.

Ι1 = 90/(20 + 5000) = 1.793 × 10–2 A I2 = 90/(40 + 5000) = 1.786 × 10–2 A % change = [(1.786 × 10–2 – 1.793 × 10–2 A)/ 1.793 × 10–2 A] × 100% = – 0.4%

2-12.

I1 = 9.0/520 = 1.731 × 10–2 A I2 = 9.0/540 = 1.667 × 10–2 A % change = [(1.667 × 10–2 – 1.731 × 10–2 A)/ 1.731 × 10–2 A] × 100% = – 3.7%

2-13.

i = Iinit e–t/RC (Equation 2-35) RC = 10 × 106 Ω × 0.2 × 10–6 F = 2.00 s

Iinit = 24V/(10 × 106 Ω) = 2.4 × 10–6 A

i = 2.4 × 10–6 e–t/2.00 A or 2.4 e–t/2.0 μA

t, s 0.00 0.010 0.10 2-14. vC = VC e–t/RC

i, μA 2.40 2.39 2.28

t, s 1.0 10

(Equation 2-40)

vC/VC = 1.00/100 for discharge to 1% −6

0.0100 = e–t/RC = e− t / R × 0.015× 10

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

i, μA 1.46 0.0162

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 = Iinit e-t/RC, vR = iR, and vC = 25 – vR For part (d) we calculate the quantities from i=

− vC −t / RC , vR = iR, and vC = –vR e R

The results are given in the spreadsheet.

6

Principles of Instrumental Analysis, 6th ed.

Chapter 2

7

Principles of Instrumental Analysis, 6th ed.

Chapter 2

2-17. Proceeding as in Problem 2-16, the results are in the spreadsheet

8

Principles of Instrumental Analysis, 6th ed.

Chapter 2

2-18. In the spreadsheet we calculate XC, Z, and φ from XC = 2/2πfC, Z =

R 2 + XC2 , and φ = arc tan(XC/R)

2-19. Let us rewrite Equation 2-54 in the form

y=

(V p )o = (V p )i

1 (2πfRC ) 2 + 1

y2(2πfRC)2 + y2 = 1 f =

1 2πRC

1 1 1− y 2 − 1 = y2 y2 2πRC

The spreadsheet follows

9

Principles of Instrumental Analysis, 6th ed.

Chapter 2

2-20. By dividing the numerator and denominator of the right side of Equation 2-53 by R, we obtain y=

(V p )o (V p )i

=

1 1 + (1/ 2π fRC) 2

Squaring this equation yields y2 + y2/(2πfRC)2 = 1 2πfRC =

f =

y2 1− y 2

y2 1 2πRC 1 − y 2

The results are shown in the spreadsheet that follows.

10

Principles of Instrumental Analysis, 6th ed.

Chapter 2

11

Skoog/Holler/Crouch Principles of Instrumental Analysis, 6th ed.

Chapter 3 Instructor’s Manual

CHAPTER 3 3-1.

vo = Avs = A(v+ – v–) For + limit, (a).

+13 V =A(v+ – v–) For – limit, –14 V =A(v+ – v–)

If A = 200,000, for + limit, (v+ – v–) = 13 V/200, 000 = 65 μV. So v+ must exceed v– by 65 μV for + limit to be reached. For – limit, (v+ – v–) = – 14 V/200,000 = – 75 μV, so v– must exceed v+ by 75 μV.

(b).

For + limit (v+ – v–) = 13 V/500,000 = 26 μV. For – limit (v+ – v–) = –14 V/500,000 = – 28 μV.

(c).

+ limit, (v+ – v–) = 13 V/1.5 × 106 = 8.7 μV; – limit, (v+ – v–) = –14 V/1.5 × 106 = –9.3 μV

3-2.

CMRR =

Ad Acm

The difference gain, Ad = 10 V/500 μV = 2.0 × 104

The common mode gain, Acm = 1 V/500 mV = 2.0 CMRR = 2.0 × 104/ 2.0 = 1.0 × 104 or in dB, CMRR = 20 log(1.0 × 104) = 80 dB 3-3.

For vs = 5.0 μV, A = vo/vs = 13/5 × 10–6 = 2.6 × 106.

3-4.

⎛ A ⎞ 5 5 (a). From Equation 3-2, vo = vi ⎜ ⎟ = 2.0 × 1.0 × 10 /( 1 + 1.0 × 10 ) = 1.99998 V ⎝ 1+ A ⎠ Error = 2.0 – 1.99998 V = 0.00002 V or 20 μV Rel Error = (20 × 10–6/ 2.0) × 100%= 0.001% (b).

Current drawn from the source, i, is i =

2.0 V − = 1.99999998 × 10 12 A 12 (10 + 10 ) Ω 4

The IR drop across the source resistance is 1.99999998 × 10–12 A × 104 Ω = 20 nV 1

Principles of Instrumental Analysis, 6th ed.

Chapter 3

Rel Error = (20 nV/2.0 V) × 100% = 1.0 × 10–6% 3-5.

⎛ R1 ⎞ Resistors R1 and R2 form a voltage divider. A fraction of the output voltage ⎜ ⎟ vo ⎝ R1 + R 2 ⎠

is feedback to the inverting input. The amplifier maintains the voltage at the ⎛ R noninverting input, vi equal to the voltage at the inverting input ⎜ 1 ⎝ R1 + R2

⎞ ⎟ vo . Hence, if ⎠

⎛ R1 + R2 ⎞ ⎛ R1 ⎞ vi = ⎜ ⎟ v i . This is a voltage follower ⎟ vo , it follows that vo = ⎜ ⎝ R1 ⎠ ⎝ R1 + R2 ⎠

configuration, but since (R1 + R2) > R1, there will be gain. 3-6.

R1 + R2 = 3.5 × R1 = 10.0 × 103 Ω. So, R1 = 10.0 × 103 Ω/3.5 = 2.86 × 103 Ω (2.9 kΩ). R2 = 10.0 × 103 Ω – R1 = 10.0 × 103 – 2.9 × 103 = 7.1 k Ω.

3-7.

vo = –ix and v+ = v– = 0 vi = i(R – x) vo x −ix = = − vi i (R − x) R− x

⎛ x ⎞ vo = − ⎜ ⎟ vi ⎝ R− x ⎠ 3-8.

(a).

The output voltage vo = –iRf. So Rf = –vo/I = 1.0 V/10.0 × 10–6 A = 1.0 × 105 Ω. Use the circuit of Figure 3-6 with Rf = 100 kΩ.

(b).

From Equation 3-5, Ri = Rf/A = 100 kΩ/2 × 105 = 0.5 Ω.

(c).

⎛ A ⎞ Equation 3-6 states vo = − Rf (ii − ib ) ⎜ ⎟ ⎝ 1+ A ⎠

2

Principles of Instrumental Analysis, 6th ed.

Chapter 3

⎛ 2 × 105 ⎞ v o = − 100 kΩ ( 25 μA − 2.5 nA) ⎜ 5 ⎟ = 2.4997 V ⎝ 1 + 2 × 10 ⎠

2.5000 − 2.4997 ⎞ Rel error = ⎛⎜ ⎟ ×100% = 0.01% 2.5000 ⎝ ⎠ 3-9.

(a).

Gain = Rf/Ri so if Ri = 10 kΩ and the gain = 25, Rf= 25 × 10 kΩ = 250 kΩ. Use the circuit of Figure 3-7, with Rf = 250 kΩ and Ri = 10 kΩ.

(b).

⎛R ⎞ ± 10 V = ± 0.40 V v o = − ⎜ f ⎟ vi so if vo = ± 10 V, and gain = 25, vi = 25 ⎝ Ri ⎠

(c).

Ri is determined by the input resistor, so Ri = 10 kΩ.

(d).

Insert a voltage follower between the input voltage source and Ri as in Figure 311.

3

Principles of Instrumental Analysis, 6th ed.

Chapter 3

3-10.

4

Principles of Instrumental Analysis, 6th ed.

Chapter 3

3-11. The rise time tr is given by Equation 3-9

tr =

1 1 = = 6.7 × 10 − 9 s = 6.7 ns 6 3Δ f 3 × 50 × 10

The slew rate is given by

Δv 10 V 9 = = 1.5 × 10 V/s = 1500 V/μs 6.7 ns Δf

3-12. Several resistor combinations will work including the following

⎛V V V′ ⎞ Vo = − Rf ⎜ 1 + 2 + 3 ⎟ ⎝ R1 R2 R3 ⎠

V3′ = −

1000 V3 = − V3 1000

V V′ ⎞ V V ⎞ ⎛ V ⎛ V Vo = − 3000 ⎜ 1 + 2 + 3 ⎟ = − 3000 ⎜ 1 + 2 − 3 ⎟ ⎝ 1000 600 500 ⎠ ⎝ 1000 600 500 ⎠ Vo = –3V1 –5V2 + 6V3

5

Principles of Instrumental Analysis, 6th ed.

Chapter 3

3-13. Several combinations will work, including the following

⎛V V V ⎞ Vo ′ = −1 × ⎜ 1 + 2 + 3 ⎟ ⎝3 3 3 ⎠ Vo = −Vo ′ ×

1000 ⎛ V +V +V ⎞ =1000 ⎜ 1 2 3 ⎟ 1 3 ⎝ ⎠

3-14. Several resistor combinations will work including

⎛V V ⎞ V ⎛ V Vo = − Rf ⎜ 1 + 2 ⎟ = − 3000 ⎜ 1 + 2 ⎞⎟ = –(0.500 V1 + 0.300 V2) ⎝ 6000 10000 ⎠ ⎝ R1 R2 ⎠

−Vo =

1 ( 5V1 + 3V2 ) 10

6

Principles of Instrumental Analysis, 6th ed.

Chapter 3

3-15. The circuit is shown below. Resistances values that work include Rf = 1.00 kΩ and R1 = 250 Ω

I1 + I2 = If

Note I2 = Ii

V V1 + Ii = − o R1 Rf −Vo =

V1Rf + I i Rf R1

or V o = −

V1 Rf − I iRf R1

Substituting the vales of Rf and Ri gives

Vo = − 3-16. (a).

V1 × 1000 − 1000 Ii = −4V1 −1000 I i 250

Let vx be the output from the first operational amplifier,

vx = −

v1Rf1 vR − 2 f1 R1 R2

vo = −

v x Rf2 vR R vR R vR = 1 f1 f2 + 2 f1 f2 − 3 f2 R4 R1 R4 R2 R4 R3

and

(b).

vo =

200 × 400 200 × 400 400 v1 + v2 − v 3 = v1 + 4v 2 − 40v3 200 × 400 50 × 400 10

7

Principles of Instrumental Analysis, 6th ed.

Chapter 3

3-17. Let vx be the output of the first operational amplifer,

vx = −

15 15 v1 − v2 = − 5 v1 − 3 v2 3 5

vo = −

12 12 12 vx − v3 − v4 = − 2vx − 3v3 − 2 v4 = 10v1 + 6 v 2 − 3v3 − 2 v4 6 4 6

3-18. Operational amplifer 2 is an integrator whose output voltage is given by Equation 3-22. Thus, ( vo ) 1 = −

t ⎛ −v 1 vi dt = ⎜ i RiC f 0 ⎝ R iC f

∫

⎞ ⎟t ⎠

Operational amplifier 2 is in the differentiator configuration where Equation 3-23 applies ( vo ) 2 = − Rf Ci

( vo ) 2 =

d ( vo )1 d ⎛ − vi = − Rf Ci ⎜ dt dt ⎝ Ri Cf

⎞ t⎟ ⎠

Rf Ci vi Ri Cf

In the sketch below, it is assumed that RfCi/RiCf > 1

8

Principles of Instrumental Analysis, 6th ed.

Chapter 3

3-19. i1 + i2 = if

V1 V2 dv dv + = − C o = − 0.010 ×10 6 o 6 6 20 ×10 5 ×10 dt dt dv o V ⎞ ⎛ V = −⎜ 1 + 2 ⎟ dt ⎝ 0.20 0.05 ⎠ vo = − (5v1 + 20 v2 )

t

∫0 dt

3-20. This circuit is analogous to that shown in Figure 3-13 with the added stipulation that R1 = Rk1 = R2 = Rk2 Equation 3-15 then applies and we can write

vo =

Rk ( v2 − v1 ) = v2 − v1 Ri

3-21. 1.02 = Vo × BC / AB = 3.00 × BC /100 BC = 100 cm × 1.02/3.00 = 34.0 cm

3-22. Here we combine the outputs from two integrators (Figure 3-16c) with a summing amplifier (Figure 3-16b). If both integrators have input resistances of 10 MΩ and feedback capacitances of 0.1 μF, the outputs are

(v o )1 = − Similarly,

t

1 − 6 10× 10 × 0.1× 10 6

∫0

t

∫0

v1dt = − 1.0 v 1dt

t

∫0

(vo )2 = − 1.0 v2dt

If the summing amplifier has input resistances of 5.0 kΩ for integrator 1 and 4.0 kΩ for integrator 2 and a feedback resistance of 20 kΩ,

9

Principles of Instrumental Analysis, 6th ed.

Chapter 3

(v ) (v ) vo = −20 ⎡⎢ o 1 + o 2 ⎤⎥ = − 4.0( vo )1 − 5.0( vo ) 2 4.0 ⎦ ⎣ 5.0 vo = +4.0

t

∫0

t

∫0

v1dt + 5.0 v2 dt

3-23. The circuit is shown below with one possible resistance-capacitance combination

Here,

( vo )1 = −

1 10 ×10 ×0.1 ×10 −6 6

t

t

∫0 v dt = − 1.0 ∫0 v dt 1

1

⎡ (v ) v v ⎤ v o = − 12.0 ⎢ o 1 + 2 + 3 ⎥ = − 2(v o ) 1 − 6( v 2 + v 3) 2 2⎦ ⎣ 6 t

∫0

vo = +2.0 v1dt − 6(v2 + v3 )

10

Principles of Instrumental Analysis, 6th ed. 3-24.

vo = −

1 Ri Cf

Chapter 3

t

t

1

∫0v dt = − 2× 10 × 0.25× 10 ∫0 4.0 ×10 1

−6

6

t

∫0

v o = − 2 4.0 ×10 −3dt = − 8.0 mV × t t, s 1 3 5 7

vo, mV –8.0 –24 –40 –56

11

−3

dt

Skoog/Holler/Crouch Principles of Instrumental Analysis, 6th ed.

Chapter 4 Instructor’s Manual

CHAPTER 4 4-1.

(a) For 24 24 = 16 24 – 16 = 8 23 = 8

binary

1 1 × 24

8–8=0 1 + 1 × 23

+

0 0 × 22

+

0 0 × 21

+