# Solution Manual For Principles of Instrumental Analysis, 7th Edition

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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 โ 1.0 V โ 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 ฮฉ , R2 = 250 ร 4.0 = 1.0 kฮฉ, and R3 = 1.0 kฮฉ ร 5.0/4.0 = 1.25 kฮฉ . Use a 1.0 kฮฉ resistor and a 250 kฮฉ resistor in series. The 500 kฮฉ resistor is not used. (b) V3 = IR3 = 10.0 V โ 1.0 V โ 4.0 V = 5.0 V (c) I = V/( R1 + R2 + R3) = 10.0 V/(250 ฮฉ + 1000 ฮฉ + 1250 ฮฉ) = 0.004 A (4.0 mA) (d) P = IV = 0.004 A ร 10.0 V = 0.04 W 2-2. (Equation 2-2) (a) From Equation 2-10, V2 = 15 ร 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) ร 100 = 15% 2-3. V2,4 = 24.0 ร [(2.0 + 4.0) ร 103]/[(1.0 + 2.0 + 4.0) ร 103] = 20.6 V With the meter in parallel across contacts 2 and 4, R + 6.0 kฮฉ 1 1 1 = + = M R2,4 (2.0 + 4.0) kฮฉ RM RM ร 6.0 kฮฉ R2,4 = (RM ร 6.0 kฮฉ)/(RM + 6.0 kฮฉ) (a) R2,4 = (4.0 kฮฉ ร 6.0 kฮฉ)/(4.0 kฮฉ + 6.0 kฮฉ) = 2.40 kฮฉ VM = (24.0 V ร 2.40 kฮฉ)/(1.00 kฮฉ + 2.40 kฮฉ) = 16.9 V rel error = 16.9 V โ 20.6 V ร 100% = โ 18% 20.6 V Proceeding in the same way, we obtain (b) โ1.2% and (c) โ0.2% 2-4. Applying Equation 2-19, we can write (a) โ1.0% = โ 1000 ฮฉ ร100% (RM โ 1000 ฮฉ) RM = (1000 ร 100 โ 1000) ฮฉ = 99000 ฮฉ or 99 k ฮฉ (b) โ0.1% = โ 1000 ฮฉ ร100% (RM โ 1000 ฮฉ) RM = 999 k ฮฉ 2-5. Resistors R2 and R3 are in parallel, the parallel combination Rp is given by Equation 2-17 Rp = (500 ร 250)/(500 + 250) = 166.67 ฮฉ (a) This 166.67 ฮฉ Rp is in series with R1 and R4. Thus, the voltage across R1 is V1 = (15.0 ร 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 ร 166.67/1416.67 = 1.76 V V4 = 15.0 V ร 1000/1416.67 = 10.59 V (b) I1R1 = V1 = 2.647 V I1 = 2.647/250 = 0.01059A (1.06 ร 10โ2 A) I2 = 1.76 V/500 ฮฉ= 3.5 ร 10โ3 A I3 = 1.76 V /250 ฮฉ = 7.0 ร 10โ3 A I4 = 10.59 V/ 1000 ฮฉ = 0.01059 ฮ (1.06 ร 10โ2 A) (c) P = IV = 1.76 V ร 7.0 ร 10โ3 A = 1.2 ร 10โ2 W (d) Since point 3 is at the same potential as point 2, the voltage between points 3 and 4 (Vโฒ) is the sum of the drops across the 166.67 ฮฉ and the 1000 ฮฉ resistors. Or, Vโฒ = 1.76 V + 10.59 V = 12.35 V. It is also the source voltage minus the V1 Vโฒ = 15.0 โ 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 kฮฉ ร 4.0 kฮฉ/(3.0 kฮฉ + 4.0 kฮฉ) = 1.71 kฮฉ Similarly the resistance between points 2 and 3 is R2,3 = 2.0 kฮฉ ร 1.0 kฮฉ/(2.0 kฮฉ + 1.0 kฮฉ) = 0.667 kฮฉ These two resistors are in series with RA for a total series resistance RT of RT = 1.71 kฮฉ + 0.667 kฮฉ + 2.0 kฮฉ = 4.38 kฮฉ I = 24/(4380 ฮฉ) = 5.5 ร 10โ3 A (a) P1,2 = I2R1,2 = (5.5 ร 10โ3)2 ร 1.71 ร 103 = 0.052 W (b) As above I = 5.5 ร 10โ3 A (5.5 mA) (c) VA = IRA = 5.5ร 10โ3 A ร 2.0 ร 103 ฮฉ = 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 ร R2,3/RT = 24 ร 0.667/4.38 = 3.65 V (e) V5,4 = 24 โ VA = 24 โ 11.0 = 13 V 2-7. With the standard cell in the circuit, Vstd = Vb ร AC/AB where Vb is the battery voltage 1.018 = Vb ร 84.3/AB With the unknown voltage Vx in the circuit, Vx = Vb ร 44.2/AB Dividing the third equation by the second gives, 1.018 V 84.3 cm = Vx 44.3 cm Vx = 1.018 ร 44.3 cm/84.3 cm = 0.535 V 2-8. Er = โ RS ร 100% RM + RS For RS = 20 ฮฉ and RM = 10 ฮฉ, Er = โ Similarly, for RM = 50 ฮฉ, Er = โ 20 ร 100% = โ 67% 10 + 20 20 ร 100% = โ 29% 50 + 20 The other values are shown in a similar manner. 2-9. Equation 2-20 is Er = โ Rstd ร 100% RL + Rstd For Rstd = 1 ฮฉ and RL = 1 ฮฉ, Er = โ 1ฮฉ ร 100% = โ 50% 1 ฮฉ +1 ฮฉ 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 ฮฉ, Er = โ 1ฮฉ ร 100% = โ 9.1% 10 ฮฉ + 1 ฮฉ The other values are shown in a similar manner. 2-10. (a) Rs = V/I = 1.00 V/20 ร 10โ6 A = 50000 ฮฉ or 50 kฮฉ (b) Using Equation 2-19 โ1% = โ 50 kฮฉ ร 100% RM + 50 kฮฉ RM = 50 kฮฉ ร 100 โ 50 kฮฉ = 4950 kฮฉ or โ 5 Mฮฉ 2โ11. ฮ1 = 90/(25 + 5000) = 1.791 ร 10โ2 A I2 = 90/(45 + 5000) = 1.784 ร 10โ2 A % change = [(1.784 ร 10โ2 A โ 1.791 ร 10โ2 A)/ 1.791 ร 10โ2 A] ร 100% = โ 0.4% 2-12. I1 = 12.5/420 = 2.976 ร 10โ2 A I2 = 12.5/440 = 2.841ร 10โ2 A % change = [(2.841ร 10โ2 โ 2.976 ร 10โ2)/ 2.976 ร 10โ2] ร 100% = โ4.5% 2-13. i = Iinit eโt/RC (Equation 2-35) RC = 25 ร 106 ฮฉ ร 0.2 ร 10โ6 F = 5.00 s Iinit = 24V/(25 ร 106 ฮฉ) = 9.6 ร 10โ7 A i = 9.6 ร 10โ7 eโt/5.00 A or 0.96 ร eโt/5.0 ยตA t, s 0.00 0.010 0.10 i, ยตA 0.96 0.958 0.941 t, s 1.0 10 i, ยตA 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 eโt/RC Chapter 2 (Equation 2-40) vC/VC = 1.00/100 for discharge to 1% โ6 0.0100 = eโt/RC = eโt /( R ร 0.025 ร 10 ) ln 0.0100 = โ4.61 = โt/(2.5 ร 10โ8R) t = 4.61 ร 2.5 ร 10โ8R = 1.15 ร 10โ7R (a) When R = 10 Mฮฉ or 10 ร 106 ฮฉ, t = 1.15 s (b) Similarly, when R = 1 Mฮฉ, t = 0.115 s (c) When R = 1 kฮฉ, t = 1.15 ร 10โ4 s 2-15. (a) When R = 10 Mฮฉ, RC = 10 ร 106 ฮฉ ร 0.025 ร 10โ6 F = 0.25 s (b) RC = 1 ร 106 ร 0.025 ร 10โ6 = 0.025 s (c) RC = 1 ร 103 ร 0.025 ร 10โ6 = 2.5 ร 10โ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 โ vR For part (d) we calculate the quantities from i= โvC โt / RC , vR = iR, and vC = โ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 ฯ from XC = 2/2ฯfC, Z = R 2 + X C2 , and ฯ = arc tan(XC/R) 2-19. Let us rewrite Equation 2-54 in the form y= (V p )o (V p )i = 1 (2ฯfRC ) 2 + 1 y2(2ฯfRC)2 + y2 = 1 f = 1 2ฯRC 1 1 1 โ y2 โ 1 = y2 2ฯRC 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/ 2ฯfRC ) 2 Squaring this equation yields y2 + y2/(2ฯfRC)2 = 1 2ฯfRC = f = y2 1โ y2 1 y2 2ฯRC 1 โ 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.

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