Solution Manual For Principles of Instrumental Analysis, 7th Edition

Skoog/Holler/Crouch Principles of Instrumental Analysis, 7th ed. Chapter 2 CHAPTER 2 2-1. (a) Applying the voltage divider equation (2-10) R1 1.0 = 10 R1 + R2 + R3 R2 4.0 = 10 R1 + R2 + R3 V3 = 10.0V – 1.0 V – 4.0 V = 5.0 V R3 5.0 = 10 R1 + R2 + R3 Dividing the first equation by the second, gives R1/R2 = 1.0/4.0 Similarly, R2/R3 = 4.0/5.0 Letting R1 = 250 Ω , R2 = 250 × 4.0 = 1.0 kΩ, and R3 = 1.0 kΩ × 5.0/4.0 = 1.25 kΩ . Use a 1.0 kΩ resistor and a 250 kΩ resistor in series. The 500 kΩ resistor is not used. (b) V3 = IR3 = 10.0 V – 1.0 V – 4.0 V = 5.0 V (c) I = V/( R1 + R2 + R3) = 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 2-2. (Equation 2-2) (a) From Equation 2-10, V2 = 15 × 400/(200 + 400 + 2000) = 2.31 V (b) P = V22 /R2 = (2.31)2/400 = 0.013 W (c) Total P = V2/Rs = (15)2/2600 = 0.087 W 1 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 Percentage loss in R2 = (0.013/0.087) × 100 = 15% 2-3. V2,4 = 24.0 × [(2.0 + 4.0) × 103]/[(1.0 + 2.0 + 4.0) × 103] = 20.6 V With the meter in parallel across contacts 2 and 4, R + 6.0 kΩ 1 1 1 = + = M R2,4 (2.0 + 4.0) kΩ RM RM × 6.0 kΩ R2,4 = (RM × 6.0 kΩ)/(RM + 6.0 kΩ) (a) R2,4 = (4.0 kΩ × 6.0 kΩ)/(4.0 kΩ + 6.0 kΩ) = 2.40 kΩ VM = (24.0 V × 2.40 kΩ)/(1.00 kΩ + 2.40 kΩ) = 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) −1.0% = − 1000 Ω ×100% (RM − 1000 Ω) RM = (1000 × 100 – 1000) Ω = 99000 Ω or 99 k Ω (b) −0.1% = − 1000 Ω ×100% (RM − 1000 Ω) RM = 999 k Ω 2-5. Resistors R2 and R3 are in parallel, the parallel combination Rp is given by Equation 2-17 Rp = (500 × 250)/(500 + 250) = 166.67 Ω (a) This 166.67 Ω Rp is in series with R1 and R4. Thus, the voltage across R1 is V1 = (15.0 × 250)/(250 + 166.67 + 1000) = 2.65 V 2 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 V2 = V3 = 15.0 V × 166.67/1416.67 = 1.76 V V4 = 15.0 V × 1000/1416.67 = 10.59 V (b) I1R1 = V1 = 2.647 V I1 = 2.647/250 = 0.01059A (1.06 × 10–2 A) I2 = 1.76 V/500 Ω= 3.5 × 10–3 A I3 = 1.76 V /250 Ω = 7.0 × 10–3 A I4 = 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 V1 V′ = 15.0 – 2.65 = 12.35 V 2-6. The resistance between points 1 and 2 is the parallel combination or RB and RC R1,2 = 3.0 kΩ × 4.0 kΩ/(3.0 kΩ + 4.0 kΩ) = 1.71 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 = 1.71 kΩ + 0.667 kΩ + 2.0 kΩ = 4.38 kΩ I = 24/(4380 Ω) = 5.5 × 10–3 A (a) P1,2 = I2R1,2 = (5.5 × 10–3)2 × 1.71 × 103 = 0.052 W (b) As above I = 5.5 × 10–3 A (5.5 mA) (c) VA = IRA = 5.5× 10–3 A × 2.0 × 103 Ω = 11.0 V 3 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 (d) VD = 24 × R2,3/RT = 24 × 0.667/4.38 = 3.65 V (e) V5,4 = 24 – VA = 24 – 11.0 = 13 V 2-7. With the standard cell in the circuit, Vstd = Vb × AC/AB where Vb is the battery voltage 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, Er = − 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 = − 1Ω × 100% = − 50% 1 Ω +1 Ω 4 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 Similarly for RL = 10 Ω, Er = − 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/20 × 10–6 A = 50000 Ω or 50 kΩ (b) Using Equation 2-19 −1% = − 50 kΩ × 100% RM + 50 kΩ RM = 50 kΩ × 100 – 50 kΩ = 4950 kΩ or ≈ 5 MΩ 2−11. Ι1 = 90/(25 + 5000) = 1.791 × 10–2 A I2 = 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. I1 = 12.5/420 = 2.976 × 10–2 A I2 = 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 = Iinit e–t/RC (Equation 2-35) RC = 25 × 106 Ω × 0.2 × 10–6 F = 5.00 s Iinit = 24V/(25 × 106 Ω) = 9.6 × 10–7 A i = 9.6 × 10–7 e–t/5.00 A or 0.96 × e–t/5.0 µA t, s 0.00 0.010 0.10 i, µA 0.96 0.958 0.941 t, s 1.0 10 i, µA 0.786 0.130 5 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. 2-14. vC = VC e–t/RC Chapter 2 (Equation 2-40) vC/VC = 1.00/100 for discharge to 1% −6 0.0100 = e–t/RC = e−t /( R × 0.025 × 10 ) ln 0.0100 = –4.61 = –t/(2.5 × 10–8R) t = 4.61 × 2.5 × 10–8R = 1.15 × 10–7R (a) When R = 10 MΩ or 10 × 106 Ω, t = 1.15 s (b) Similarly, when R = 1 MΩ, t = 0.115 s (c) When R = 1 kΩ, t = 1.15 × 10–4 s 2-15. (a) When R = 10 MΩ, RC = 10 × 106 Ω × 0.025 × 10–6 F = 0.25 s (b) RC = 1 × 106 × 0.025 × 10–6 = 0.025 s (c) RC = 1 × 103 × 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 = 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 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 2-17. Proceeding as in Problem 2-16, the results are in the spreadsheet 7 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 8 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 2-18. In the spreadsheet we calculate XC, Z, and φ from XC = 2/2πfC, Z = R 2 + X C2 , 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 − y2 − 1 = y2 2πRC y2 The spreadsheet follows 9 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th 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/ 2πfRC ) 2 Squaring this equation yields y2 + y2/(2πfRC)2 = 1 2πfRC = f = y2 1− y2 1 y2 2πRC 1 − y 2 The results are shown in the spreadsheet that follows. 10 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Principles of Instrumental Analysis, 7th ed. Chapter 2 11 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.