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Solution Manual For Principles of Instrumental Analysis, 7th Edition

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    Skoog/Holler/Crouch Chapter 1 Principles of Instrumental Analysis, 7 th ed. 1 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.

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    Principles of Instrumental Analysis, 7 th ed. Chapter 1 2 1-6. Time-domain signals include period, frequency, and pulse width. The information is encoded in the time relationship of signal fluctuations. 1-7. Input Transducer Use Phototube Convert light intensity to an analog current Glass electrode Convert electrode potential to a voltage related to pH Electron multiplier Convert ion intensity to a corresponding electric current Thermal conductivity Converts sample thermal conductivity to related voltage 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 s = molar concentration of Cu 2+ in standard = 0.0275 M c x = unknown Cu 2+ concentration V s = volume of standard = 0.500 mL V x = volume of unknown = 25.0 mL S 1 = signal for unknown = 25.2 S 2 = signal for unknown plus standard = 45.1 Assuming the signal is proportional to c x and c s , we can write S 1 = Kc x or K = S 1 / c x After adding the standard 2 x x s s x s V c V c S K V V ⎛ ⎞ + = ⎜ ⎟ + ⎝ ⎠ Substituting for K and rearranging gives,

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    Principles of Instrumental Analysis, 7 th ed. Chapter 1 3 1 2 1 ( ) s s x x s x S V c c S V V S V = + − c x 4 25.2 0.500 mL 0.0275 M = 6.66 10 M 45.1(0.500 mL + 25.0 mL) (25.2 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 m = 0.0007, SD intercept, s b = 0.0040 (c) 95% CI for slope m is m ± t s m For 95% probability and N – 2 = 4 degrees of freedom, t = 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 b = 0.0083 ± 2.78 × 0.004 = 0.0083 ± 0.011 or 0.08 ± 0.01 (d) c u = 4.87 ± 0.086 mM or 4.87 ± 0.09 mM

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    Principles of Instrumental Analysis, 7 th ed. Chapter 1 4 1-11. The spreadsheet below gives the results (a) See plot in spreadsheet. (b) c u = 0.410 μ g/mL (c) S = 3.16 V s + 3.25 (d) 1 3.246 2.000 μg/mL 0.410 μg/mL 3.164 mL 5.00 mL s u u bc c mV − × = = = × (e) From the spreadsheet s c = 0.002496 or 0.002 μ g/mL

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    Skoog/Holler/Crouch Chapter 2 Principles of Instrumental Analysis, 7th ed. 1 CHAPTER 2 2-1. (a) Applying the voltage divider equation (2-10) 1 1 2 3 1.0 = 10 + + R R R R 2 1 2 3 4.0 = 10 + + R R R R V 3 = 10.0V – 1.0 V – 4.0 V = 5.0 V 3 1 2 3 5.0 = 10 + + R R R R Dividing the first equation by the second, gives R 1 / R 2 = 1.0/4.0 Similarly, R 2 / R 3 = 4.0/5.0 Letting R 1 = 250 Ω , R 2 = 250 × 4.0 = 1.0 k Ω , and R 3 = 1.0 k Ω × 5.0/4.0 = 1.25 k Ω . Use a 1.0 k Ω resistor and a 250 k Ω resistor in series. The 500 k Ω resistor is not used. (b) V 3 = IR 3 = 10.0 V – 1.0 V – 4.0 V = 5.0 V (c) I = V /( R 1 + R 2 + R 3 ) = 10.0 V/(250 Ω + 1000 Ω + 1250 Ω ) = 0.004 A (4.0 mA) (d) P = IV = 0.004 A × 10.0 V = 0.04 W (Equation 2-2) 2-2. (a) From Equation 2-10, V 2 = 15 × 400/(200 + 400 + 2000) = 2.31 V (b) P = 2 2 V / R 2 = (2.31) 2 /400 = 0.013 W (c) Total P = V 2 / R s = (15) 2 /2600 = 0.087 W

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 2 Percentage loss in R 2 = (0.013/0.087) × 100 = 15% 2-3. V 2,4 = 24.0 × [(2.0 + 4.0) × 10 3 ]/[(1.0 + 2.0 + 4.0) × 10 3 ] = 20.6 V With the meter in parallel across contacts 2 and 4, 2,4 + 6.0 k 1 1 1 = + = (2.0 + 4.0) k 6.0 k M M M R R R R Ω Ω × Ω R 2,4 = ( R M × 6.0 k Ω )/( R M + 6.0 k Ω ) (a) R 2,4 = (4.0 k Ω × 6.0 k Ω )/(4.0 k Ω + 6.0 k Ω ) = 2.40 k Ω V M = (24.0 V × 2.40 k Ω )/(1.00 k Ω + 2.40 k Ω ) = 16.9 V rel error = 16.9 V 20.6 V 100% = 18% 20.6 V − × − Proceeding in the same way, we obtain (b) –1.2% and (c) –0.2% 2-4. Applying Equation 2-19, we can write (a) 1000 1.0% = 100% ( 1000 ) M R Ω − − × − Ω R M = (1000 × 100 – 1000) Ω = 99000 Ω or 99 k Ω (b) 1000 0.1% = 100% ( 1000 ) M R Ω − − × − Ω R M = 999 k Ω 2-5. Resistors R 2 and R 3 are in parallel, the parallel combination R p is given by Equation 2-17 R p = (500 × 250)/(500 + 250) = 166.67 Ω (a) This 166.67 Ω R p is in series with R 1 and R 4 . Thus, the voltage across R 1 is V 1 = (15.0 × 250)/(250 + 166.67 + 1000) = 2.65 V

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 3 V 2 = V 3 = 15.0 V × 166.67/1416.67 = 1.76 V V 4 = 15.0 V × 1000/1416.67 = 10.59 V (b) I 1 R 1 = V 1 = 2.647 V I 1 = 2.647/250 = 0.01059A (1.06 × 10 –2 A) I 2 = 1.76 V/500 Ω = 3.5 × 10 –3 A I 3 = 1.76 V /250 Ω = 7.0 × 10 –3 A I 4 = 10.59 V/ 1000 Ω = 0.01059 Α (1.06 × 10 –2 A) (c) P = IV = 1.76 V × 7.0 × 10 –3 A = 1.2 × 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 166.67 Ω and the 1000 Ω resistors. Or, V ′ = 1.76 V + 10.59 V = 12.35 V. It is also the source voltage minus the V 1 V ′ = 15.0 – 2.65 = 12.35 V 2-6. The resistance between points 1 and 2 is the parallel combination or R B and R C R 1,2 = 3.0 k Ω × 4.0 k Ω /(3.0 k Ω + 4.0 k Ω ) = 1.71 k Ω Similarly the resistance between points 2 and 3 is R 2,3 = 2.0 k Ω × 1.0 k Ω /(2.0 k Ω + 1.0 k Ω ) = 0.667 k Ω These two resistors are in series with R A for a total series resistance R T of R T = 1.71 k Ω + 0.667 k Ω + 2.0 k Ω = 4.38 k Ω I = 24/(4380 Ω ) = 5.5 × 10 –3 A (a) P 1,2 = I 2 R 1,2 = (5.5 × 10 –3 ) 2 × 1.71 × 10 3 = 0.052 W (b) As above I = 5.5 × 10 –3 A (5.5 mA) (c) V A = IR A = 5.5 × 10 –3 A × 2.0 × 10 3 Ω = 11.0 V

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 4 (d) V D = 24 × R 2,3 / R T = 24 × 0.667/4.38 = 3.65 V (e) V 5,4 = 24 – V A = 24 – 11.0 = 13 V 2-7. With the standard cell in the circuit, V std = V b × AC / AB where V b is the battery voltage 1.018 = V b × 84.3/ AB With the unknown voltage V x in the circuit, V x = V b × 44.2/ AB Dividing the third equation by the second gives, 1.018 V 84.3 cm = 44.3 cm x V V x = 1.018 × 44.3 cm/84.3 cm = 0.535 V 2-8. = 100% S r M S R E R R − × + For R S = 20 Ω and R M = 10 Ω , 20 = 100% = 67% 10 20 r E − × − + Similarly, for R M = 50 Ω , 20 = 100% = 29% 50 20 r E − × − + The other values are shown in a similar manner. 2-9. Equation 2-20 is std std = 100% r L R E R R − × + For R std = 1 Ω and R L = 1 Ω , 1 = 100% = 50% 1 1 r E Ω − × − Ω + Ω

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 5 Similarly for R L = 10 Ω , 1 = 100% = 9.1% 10 1 r E Ω − × − Ω + Ω The other values are shown in a similar manner. 2-10. (a) R s = V / I = 1.00 V/20 × 10 –6 A = 50000 Ω or 50 k Ω (b) Using Equation 2-19 50 k 1% = 100% 50 k M R Ω − − × + Ω R M = 50 k Ω × 100 – 50 k Ω = 4950 k Ω or ≈ 5 M Ω 2 − 11. Ι 1 = 90/(25 + 5000) = 1.791 × 10 –2 A I 2 = 90/(45 + 5000) = 1.784 × 10 –2 A % change = [(1.784 × 10 –2 A – 1.791 × 10 –2 A)/ 1.791 × 10 –2 A] × 100% = – 0.4% 2-12. I 1 = 12.5/420 = 2.976 × 10 –2 A I 2 = 12.5/440 = 2.841 × 10 –2 A % change = [(2.841 × 10 –2 – 2.976 × 10 –2 )/ 2.976 × 10 –2 ] × 100% = –4.5% 2-13. i = I init e – t / RC (Equation 2-35) RC = 25 × 10 6 Ω × 0.2 × 10 –6 F = 5.00 s I init = 24V/(25 × 10 6 Ω ) = 9.6 × 10 –7 A i = 9.6 × 10 –7 e – t /5.00 A or 0.96 × e –t/5.0 μ A t , s i , μ A t , s i , μ A 0.00 0.96 1.0 0.786 0.010 0.958 10 0.130 0.10 0.941

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 6 2-14. v C = V C e –t/RC (Equation 2-40) v C / V C = 1.00/100 for discharge to 1% 0.0100 = e –t/RC = 6 /( 0.025 10 ) t R e − − × × ln 0.0100 = –4.61 = – t /(2.5 × 10 –8 R ) t = 4.61 × 2.5 × 10 –8 R = 1.15 × 10 –7 R (a) When R = 10 M Ω or 10 × 10 6 Ω , t = 1.15 s (b) Similarly, when R = 1 M Ω , t = 0.115 s (c) When R = 1 k Ω , t = 1.15 × 10 –4 s 2-15. (a) When R = 10 M Ω , RC = 10 × 10 6 Ω × 0.025 × 10 –6 F = 0.25 s (b) RC = 1 × 10 6 × 0.025 × 10 –6 = 0.025 s (c) RC = 1 × 10 3 × 0.025 × 10 –6 = 2.5 × 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 init e -t/RC , v R = iR , and v C = 25 – v R For part (d) we calculate the quantities from / = t RC C v i e R − − , v R = iR , and v C = – v R The results are given in the spreadsheet.

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 7 2-17. Proceeding as in Problem 2-16, the results are in the spreadsheet

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 8

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 9 2-18. In the spreadsheet we calculate X C , Z , and φ from X C = 2/2 π fC , 2 2 = + C Z R X , and φ = arc tan( X C / R ) 2-19. Let us rewrite Equation 2-54 in the form o 2 ( ) 1 = = ( ) (2 π ) 1 p p i V y V fRC + y 2 (2 π fRC ) 2 + y 2 = 1 2 2 2 1 1 1 1 = 1 = 2 π RC 2 π RC y f y y − − The spreadsheet follows

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 10 2-20. By dividing the numerator and denominator of the right side of Equation 2-53 by R , we obtain o 2 ( ) 1 = = ( ) 1 + (1/ 2 π ) p p i V y V fRC Squaring this equation yields y 2 + y 2 /(2 π fRC ) 2 = 1 2 2 2 π = 1 y fRC y − 2 2 1 = 2 π 1 y f RC y − The results are shown in the spreadsheet that follows.

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    Principles of Instrumental Analysis, 7 th ed. Chapter 2 11

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