(Solution Manual) Analytical Chemistry

Harvey, 2

Chemistry

Solution Manual

2nd Edition

• $15.00 $50.00

---

Description

Solution Manual for Analytical Chemistry, The downloadable file consists of 394 pages, and the total size of the file is 4.82 MB. Purchase now only for highly discounted price.

Author(s): Harvey

---

---

Study Note Preview

(a) (b) Phase 2 Phase 1 2.95 3.00 3.05 3.10 3.15 3.20 3.25 Mass of Pennies (g) Analytical Chemistry 2.1 Solutions Manual 1.0 stoichiometric mixture pH 0.8 0.6 0.4 0.2 0.0 0.0 0.2 0.4 XL 0.6 0.8 1.0 In– pH = pKa,HIn HIn indicator is color of In– indicator’s color transition range indicator is color of HIn Print Version Production History Solutions Manual to Modern Analytical Chemistry by David Harvey ISBN 0–697–39760 –3 Copyright © 2000 by McGraw -Hill Companies Copyright transferred to David Harvey, February 15, 2008 Electronic Version Solutions Manual to Analytical Chemistry 2.1 by David Harvey (Summer 2016) Copyright This work is licensed under the Creative Commons Attribution -NonCommercial -ShareAlike 4.0 Unported License. To view a copy of this license, visit http://creativecommons.org/licenses/by -nc-sa/4.0/. Under the conditions of this copyright you are free to share this work with others in any medium or format. You also are free to remix, transform, and bulid upon this material provided that you attribute the original work and author and that you distribute your work under the same licence. You may not use this work for commercial purposes. Illustrations All illustrations are original to this solutions manual and are covered by the copyright described in the previ - ous section. Table of Contents Chapter 1 ................................ ...................... 5 Chapter 2 ................................ ...................... 9 Chapter 3 ................................ .................... 15 Chapter 4 ................................ .................... 21 Chapter 5 ................................ .................... 39 Chapter 6 ................................ .................... 51 Chapter 7 ................................ .................... 79 Chapter 8 ................................ .................... 95 Chapter 9 ................................ .................. 115 Chapter 10 ................................ ................ 157 Chapter 11 ................................ ................ 181 Chapter 12 ................................ ................ 201 Chapter 13 ................................ ................ 217 Chapter 14 ................................ ................ 227 Chapter 15 ................................ ................ 245 Appendix ................................ .................. 249 iii Chapter 1 Introduction to Analytical Chemistry 5 Chapter 1 1. (a) A qualitative and a quantitative analysis is the best choice because we need to determine the identify of the possible contaminants and determine if their concentrations are greater than the expected back - ground levels. (b) A forged work of art often contains compounds that are not pres - ent in authentic materials or contains a distribution of compounds that do not match the authentic materia ls. Either a qualitative anal- ysis (to identify a compound that should not be present in authentic materials) or a quantitative analysis (to determine if the concentra - tions of compounds present do not match the distribution expected in authentic materials) is appropriate. (c) Because we are interested in detecting the presence of specific compounds known to be present in explosive materials, a qualitative analysis is the best choice. (d) A compound’s structure is one of its characteristic properties; a characterization analysis, therefore, is the best approach. (e) In searching for a new acid –base indicator we are seeking to im - prove the performance of an existing analytical method, which re- quires a fundamental analysis of the method’s properties. (f) ) A quantitative analysis is used to determine if an automobile emits too much carbon monoxide. 2. Answers to this problem will vary, but here is a list of important points that you might address: The goal of this research is to develop a fast, automated, and real-time instrumental method for determining a coffee’s sensory profile that yields results similar to those from trained human sensory panels. One challenge the authors have to address is that a human sensory panelist reports results on a relative scale, typically 0 –10, for charac - teristics that are somewhat arbitrary: What does it mean, for example, to say that a coffee is bitter? An instrumental method, on the other hand, reports results on an absolute scale and for a clearly defined signal; in this case, the signal is a raw count of the number of ions with a particular mass –to-charge ratio. Much of the mathematical processing described by the authors is used to transform the instru - mental data into a relative form and to normalize the two sets of data to the same relative scale. The instrumental technique relies on gas chromatography equipped with a mass spectrometer as a detector. The specific details of the instrument are not important, but the characteristics the authors de - scribe —low fragmentation, high time resolution, broad linear dy - See Chapter 12 for a discussion of gas chromatography and for detection using a mass spectrometer. Solutions Manual for Analytical Chemistry 2.1 6 For a discussion of quality control and quality assurance, see Chapter 15. For a discussion of the relationship be - tween signal and concentration, see Chap - ter 5. namic range —are important. When a species enters a mass spectrom - eter it is ionized (the PTR—proton transfer reaction —in PTR-MS simply describes the method of ionization) and the individual ions, being unstable, may decompose into smaller ions. As a roasted coffee has more than 1000 volatile components, many of which do not con - tribute to the sensory profile, the authors wish to limit the number of ions produced in the mass spectrometer. In addition, they want to ensure that the origin of each ion traces back to just a small number of volatile compounds so that the signal for each ion carries information about a small number of compounds. Table 1, for example, shows that the 16 ions monitored in this study trace back to just 32 unique volatile compounds, and that, on average, each ion traces back to 3–4 unique volatile compounds with a range of one to eight. The authors need high time resolution so that they can monitor the release of volatile species as a function of time, as seen in Figure 1, and so that they can report the maximum signal for each ion during the three -minute monitoring period. A rapid analysis also means they can monitor the production of coffee in real time on the production line instead of relying on a lengthy off -line analysis completed by a sensory panel. This is advantageous when it comes to quality control where time is important. A broad linear dynamic range simply means there is a linear relation - ship between the measured signal and the concentration of the com - pounds contributing to that signal over a wide range of concentra - tions. The assumption of a linear relationship between signal and con- centration is important because a relative change in concentration has the same affect on the signal regardless of the original concentration . A broad range is important because it means the signal is sensitive to a very small concentration of a volatile compound and that the signal does not become saturated, or constant, at higher concentrations of the volatile compound; thus, the signal carries information about a much wider range of concentrations. To test their method, the authors divide their samples into two sets: a training set and a validation set. The authors use the training set to build a mathematical model that relates the normalized intensities of the 16 ions measured by the instrument to the eight normalized relative attributes evaluated by members of the sensory panel. The specific details of how they created the mathematical model are not important here, but the agreemen t between the panel’s sensory profile and that predicted using the instrumental method generally is very good (see Figure 3; note that the results for Espresso No. 5 and No. 11 show the least agreement). Any attempt to create a model that relates one measurement (results from the sensory panel) to a second measurement (results from the Chapter 1 Introduction to Analytical Chemistry 7 instrumental analysis) is subject to a number of limitations, the most important of which is that the model works well for the data set used to build the model, but that it fails to work for other samples. To test the more general applicability of their model—what they refer to as a robust model —the authors use the model to evaluate the data in their validation set; the results, shown in Figure 4, suggest that the can apply their model both to coffees of the same type, but harvested in a different year, and to coffees of a different type. See Chapter 14 for a discussion of robust - ness and other ways to characterize an an- alytical method. Solutions Manual for Analytical Chemistry 2.1 8 Chapter 2 Basic Tools of Analytical Chemistry 9 Chapter 2 1. (a) 3 significant figures; (b) 3 significant figures; (c) 5 significant fig- ures; (d) 3 significant figures; (e) 4 significant figures; (f ) 3 significant figures For (d) and for (e), the zero in the tenths place is not a significant digit as its presence is needed only to indicate the position of the decimal point. If you write these using scientific notation, they are 9.03 ×10–2 and 9.030 ×10–2, with three and four significant figures respectively. 2. (a) 0.894; (b) 0.893; (c) 0.894; (d) 0.900; (e) 0.0891 3. (a) 12.01; (b) 16.0; (c) 6.022 ×1023 mol–1; (d) 9.65 ×104 C/mol 4. a. 4.591 + 0.2309 + 67.1 = 71.9219 ≈ 71.9 b. 313 – 273.15 = 39.85 ≈ 39.8 Note that for (b) we retain an extra significant figure beyond that sug- gested by our simple rules so that the uncertainty in the final answer (1 part out of 398) is approximately the same as the most uncertain of our two measurements (1 part out of 313). Reporting the answer as 40, or 4.0×101, as suggested by our simple rules, gives an uncertainty in the final result of 1 part out of 40, which is substantially worse than either of our two measurements. c. 712 × 8.6 = 6123.2 ≈ 6.1×103 d. 1.43/0.0 26 = 55.00 ≈ 55 e. (8.314 × 298)/96 485 = 0.0256 783 ≈ 2.57 ×10–2 f. log(6.53 × 10–5) = –4.18509 = –4.185 Note that when we take the logarithm of a number, any digits before the decimal point provide information on the original number’s pow - er of 10; thus, the 4 in –4.185 is not counted as a significant digit. g. 10–7.14 = 7.244×10–8 ≈ 7.2×10–8 Note that we take the antilog of a number, the digits before the dec - imal point provide information on the power of 10 for the resulting answer; thus, the 7 in –7.14 is not counted as a significant digit. h. (6.51 × 10–5) × (8.14 × 10–9) = 5.29914×10–13 ≈ 5.30 ×10–13 5. To find the %w/w Ni, we first subtract the mass of Co from point B from the combined mass of Ni and Co from point B, and then divide by the mass of sample; thus (0.2306 - 0.0813) g Ni 12.1374 g sample 0.1493 g Ni 12.1374 g sample In problem 4 we use a bold red font to help us keep track of significant figures. For example, in (a) we mark the last sig- nificant digit common to the numbers we are adding together, and in (e), where we are multiplying and dividing, we identify the number with the smallest number of significant digits. # 100 = = 1.230 %w/w Ni Solutions Manual for Analytical Chemistry 2.1 10 2 mol Cl- 11.97 mol HCl For problems in this chapter, all formu - la weights are reported to the number of significant figures allowed by the atomic weights in Appendix 18. As a compound’s formula weight rarely limits the uncer - tainty in a calculation, in later chapters usually we will round formula weights to a smaller number of significant figures, chosen such that it does not limit the cal- culation’s uncertainty. 6. Using the atomic weights from Appendix 18, we find that the formula weight for Ni(C 4H14N4O4)2 is (58.693) + (8 # 12.011) + (14 # 1.008) + (4 # 14.007) + (4 # 15.999) = 288.917 g/mole 7. First we convert the mass of Cl– to moles of Cl– 1 g Cl- 256 mg Cl- # 1000 mg Cl- # For problem 7 we include an extra signifi - cant figure in each of the calculation’s first 1 mol Cl- 35.45 g Cl- = 7.221 # 10-3 mol Cl- two steps to avoid the possibility of intro - ducing a small error in the final calcula - tion as a result of rounding. If we need to report the result for an intermediate calcu - lation, then we round that result appropri - and then the moles of Cl– to the moles of BaCl 2 7.221 # 10-3 mol Cl- # 1 mol BaCl2 = 3.610 # 10-3 mol BaCl2 ately; thus, we need to isolate 3.61 ×10–3 mol of BaCl2. and finally the moles of BaCl 2 to the volume of our BaCl 2 solution 3.610 # 10-3 mol BaCl 2 # 1 L # 1000 mL = 16.6 mL 0.217 mol BaCl 2 1 L 8. We can express a part per million in several ways —this is why some organizations recommend against using the abbreviation ppm —but here we must assume that the density of the solution is 1.00 g/mL and that ppm means mg/L or µg/mL. As molarity is expressed as mol/L, we will use mg/L as our starting point; thus 0.28 mg Pb # 1 g # 1 mol Pb = 1.4 # 10-6 M Pb L 1000 mg 207.2 g Pb 9. (a) The molarity of 37.0% w/w HCl is 37.0 g HCl # 1.18 g solution # 1.00 # 102 g solution ml solution 1000 mL # 1 mol HCl = 12.0 M HCl L 36.46 g HCl (b) To calculate the mass and volume of solution we begin with the molarity calculated in part (a). To avoid any errors due to rounding the molarity down to three significant, we will return one additional significant figure, taking the molarity as 11.97 M. 0.315 mol HCl # 1 L # 1000 mL # 1.18 g solution = 31.1 g L mL 0.315 mol HCl # 1 L # 1000 mL = 26.3 mL 11.97 mol HCl L Chapter 2 Basic Tools of Analytical Chemistry 11 250.0 ml 15 mL 15 mL 0.1000 L 294.181 g K Cr O 10. A volume of 1.0×103 mL is equivalent to 1.0 L; thus 1.0 L # 0.036 mol NH 3 # 17.031 g NH 3 # L 1.00 # 102 g solution 28.0 g NH 3 mol NH 3 # 1.00 mL 0.899 g solution = 2.4 mL 11. As we have information about the solution’s volume and no informa - tion about its density, we will assume that ppm and ppb are expressed as a mass of analyte per unit volume; thus, 45.1µg # 1 g # 100 = 1.80 # 10-5 % w/w 250.0 ml 1 # 106 µg 45.1µg = 0.180 ppm 45.1µg # 1000 mL = 1.80 # 102 ppb 250.0 ml L 12. To obtain a total concentration of 1.6 ppm F– we must add sufficient NaF such that we increase the concentration of F– by 1.4 ppm; thus 1 gal # 3.785 L gal 1.4 mg F- # L # Note that “per gallon” implies that 1 gal is an exact number that does not limit the 41.988 mg NaF 18.998 mg F- = 12 mg NaF number of significant figures in our final answer. 13. pH = –log[H+] = –log(6.92 ×10–6) = 5.160 [H3O+] = 10–pH=10–8.923 = 1.19 ×10–9 M 14. When using a 25 -mL graduated cylinder to measure 15 mL, the ab - solute uncertainty is ±1% of 25 mL, or ±0.25 mL and the relative uncertainty is !0.25 mL # 100 =!1.7% When using a 50 -mL graduated cylinder to measure 15 mL, the ab - solute uncertainty is ±1% of 50 mL, or ±0.50 mL and the relative uncertainty is !0.50 mL # 100 =!3.3% 15. First, we calculate the moles of K2Cr2O7 9.67 g K2 Cr 2 O7 # 1 mol K2 Cr 2 O7 = 0.03287 mol K2 Cr 2 O7 2 2 7 and then we calculate the solution’s molarity 0.03287 mol K2 Cr 2 O7 = 0.329 M K 2 Cr 2 O7 Solutions Manual for Analytical Chemistry 2.1 12 16. Given that the uncertainty in the volume and in the concentration is 1% (1.0 L ± 0.1 L or 0.10 M ± 0.01 M), we can prepare these Chapter 2 Basic Tools of Analytical Chemistry 13 Given the uncertainty in the volume and in the concentration, there is no advan - tage to taking the extra time needed to measure the solid’s mass to three or four decimal places, to quantitatively transfer solution by weighing out the appropriate amount of solid to two significant figures, place it in a 1 -L beaker or bottle, and dissolve in 1000 mL of water. What remains is to calculate the amount of reagent for each solution; thus: (a) for KCl we have the solid to a volumetric flask, and dilute to volume. 1.0 L # 0.10 mol K+ L # 1 mol KCl mol K+ 74.55 g KCl # mol KCl = 7.5 g KCl 1.0 L # 1.0 # 102 mg K+ L 1g K+ # 1000 mg K+ 1000 mL # 74.55 g KCl 39.098 g K+ 1.0 g K+ = 0.19 g KCl 1.0 L # L # 1.0 # 102 mL and (b) for K2SO 4 we have # 74.55 g KCl 39.098 g K+ = 19 g KCl 1.0 L # 0.10 mol K+ # 1 mol K2 SO 4 L 2 mol K+ # 174.25 g K2 SO 4 = 8.7 g K SO mol K2 SO 4 2 4 1.0 L # 1.0 # 102 mg K+ L 1g K+ # 1000 mg K+ # 174.25 g K2 SO 4 = 0.22 g K SO 1.0 L # 1000 mL L 78.196 g K+ 2 4 1.0 g K+ # 1.0 # 102 mL # 174.25 g K2 SO 4 = 22 g K SO 78.196 g K+ 2 4 and (c) for K3Fe(CN) 6 we have 1.0 L # 0.10 mol K+ # 1 mol K 3 Fe(CN) 6 L 3 mol K+ # 329.247 g K3 Fe(CN) 6 = 11 g K Fe(CN) mol K3 Fe(CN) 6 1.0 L # 1.0 # 102 mg K+ L 1g K+ # 1000 mg K+ # 329.247 g K3 Fe(CN) 6 = 0.28 g K Fe(CN) 3 6 Solutions Manual for Analytical Chemistry 2.1 14 117.294 g K+ 3 6 1.0 L # 1000 mL L 1.0 g K+ # 1.0 # 102 mL # 329.247 g K3 Fe(CN) 6 = 28 g K Fe(CN) 117.294 g K+ 3 6 Chapter 2 Basic Tools of Analytical Chemistry 15 250.0 mL 100.0 mL 500.0 mL 3 0.050 L 0.2500 L 0.0500 L 0.050 mol NO# 3 3 3 3 3 3 3 3 3 3 17. For a serial dilution, we need the concentration of solution A to cal - culate the concentration of solution B, and the concentration of solu - tion B to calculate the concentration of solution A; thus Solution A: 0.100 M # 10.00 mL = 4.00 # 10-3 M Solution B: 4.00 # 10-3 M # 25.00 mL Solution C: 1.00 # 10-3 M # 20.00 mL = 1.00 # 10-3 M = 4.00 # 10-5 M 18. When we dissolve 1.917 g NaCl in 50 mL of water measured using a graduated cylinder, the reported concentration of NaCl is limited to just two significant figures because the uncertainty for the graduated cylinder is approximately 1%, or 1 part in 50; thus 1.917 g NaCl # 1 mol NaCl 58.44 g NaCl = 0.66 M NaCl When we quantitatively transfer this solution to a 250.0 mL volu - metric flask and dilute to volume, we can report the concentration of NaCl to four significant figures because the uncertainty in the volumetric flask is ±0.01 mL, or 1 part in 2500; thus 1.917 g NaCl # 1 mol NaCl 58.44 g NaCl = 0.1312 M NaCl Note that the second calculation does not begin with the concen - tration from the previous calculation, as we did in problem 17 for a serial dilution. A quantitative transfer is not a serial dilution; instead, all 1.917 g of NaCl added to the 50-mL beaker is transfered to the 250.0 mL volumetric flask, so we begin our calculation with this mass of NaCl. 19. First, we calculate the moles of NO- in 50.0 mL of 0.050 M KNO 3 and in 40.0 mL of 0.075 M KNO 3. As we are interested only in the concentra - tion of NaCl in our final solution, there is no particular reason for us to complete the intermediate calculation; we did so here simply to make this point: The uncertain - ty in a calculated result is determined by the measurements that contribute to the calculation only, and is not affected by other measurements that we happen to make. What matters in this case is that 1.917 g of NaCl are dissolved in 250.0 mL of water. If we fail to complete a quan - titative transfer, then our calculated con- centration is in error, but this is an error in the accuracy of our work, not an un - certainty in the inherent precision of the balance or volumetric pipet. We will have more to say about accuracy and precision in Chapter 4. – 3 L 0.075 mol NO– = 2.50 # 10-3 mol NO- Here, again, we keep an extra significant figure through the intermediate steps of our calculation. 0.0400 L # L 3 = 3.00 # 10-3 mol NO- Next, we add together these results to obtain the total moles of NO- in the combined solutions, and then divide by the total volume to find the concentration of NO- and pNO- . 2.50 # 10-3 mol NO- + 3.00 # 10-3 mol NO- 0.0500 L+0.0400 L = 0.061 M NO- pNO- =-log (NO-) =-log(0.061) = 1.21 Solutions Manual for Analytical Chemistry 2.1 16 20. First, we calculate the moles of Cl– in 25.0 mL of 0.025 M NaCl and in 35.0 mL of 0.050 M BaCl 2. Chapter 2 Basic Tools of Analytical Chemistry 17 L 0.0050 L Here, again, we keep an extra significant figure throught the intermediate steps of 0.0250 L # 0.025 mol NaCl # 0.050 mol BaCl 1 mol Cl- mol NaCl 2 mol Cl- = 6.25 # 10-4 mol Cl- our calculation. 0.0350 L # L 2 # mol BaCl = 3.50 # 10-3 mol Cl- Next, we add together these results to obtain the total moles of Cl– in the combined solutions, and then divide by the total volume to find the concentration of Cl– and pCl. 6.25 # 10-4 mol Cl- + 3.50 # 10-3 mol Cl- 0.0250 L+0.0350L = 0.069 M Cl- pCl- =-log(Cl-) =-log (0.069 ) = 1.16 21. The concentration is 0.0844 M ethanol # 0.500 L = 8.44 M ethanol 2 Chapter 3 The Vocabulary of Analytical Chemistry 15 mL Chapter 3 1. In a total analysis technique the signal is proportional to the absolute amount of analyte, in grams or in moles, in the original sample. If we double the amount of sample, then the signal also doubles. For this reason, an accurate analysis requires that w e recover all analyte present in the original sample, which is accomplished here in two key ways: (a) the beaker in which the digestion is carried out is rinsed several times and the rinsings are passed through the filter paper, and (b) the filter paper itself is rinsed several times. The volume of solvent used in the digestion and the volumes used to rinse the beaker and the filter paper are not critical because they do not affect the mass or moles of analyte in the filtrate. In a concentration technique the signal is proportional to the relative amount, or concentration of analyte, which means our treatment of the sample must not change the analyte’s concentration or it must allow us to do so in a precise way. Having completed the digestion, we need to ensure that the concentration of analyte in the beaker and the concentration of analyte in the filtrate are the same, which is ac - complished here by not washing the beaker or the filter paper, which would dilute the analyte’s concentration, and by taking a precisely measured volume of the filtrate to the next step in the procedure. Because we know precisely the original volume of sample (25.00 mL) and the volume of filtrate taken (5.00 mL), we can work back from the concentration of analyte in the filtrate to the absolute amount of analyte in the original sample. 2. (a) Here we must assume that a part per billion is expressed as a mass per unit volume, which, in this case, is best expressed as ng/mL; thus 10 ng # 0.5 mL = 5 ng (b) A concentration of 10% w/v is equivalent to 10 g of analyte per 100 mL of sample or 108 ng/mL. Because the final concentration is 10 ng/mL, we must dilute the sample by a factor of 107, which we can accomplish, for example, by diluting 0.1 µL of sample to a final volume of 1 L. (c) A concentration of 10% w/w is equivalent to 10 g of analyte per 100 g of sample. To prepare the solution we need to take 1000 mL # 10 ng analyte # 1 g analyte # mL 109 ng analyte or 0.1 mg of sample. 100 g sample 10 g analyte = 1 # 10-4 g sample Solutions Manual for Analytical Chemistry 2.1 16 (d) This method is not particularly suited for a major analyte because we must dissolve a very small amount of sample (0.1 µL or 0.1 mg) in a large volume of solution (1000 mL), which is difficult to do with precision and with accuracy. We might consider a serial dilution from an initial solution that is more concentrated; however, multiple dilutions increase the opportunity for introducing error. 3. (a) The analyte’s sensitivity, kA, is kA = SA = 23.3 = 1.55 ppm-1 . 1.6 ppm-1 CA 15 ppm (b) The interferent’s sensitivity, kI, is kI = SI = 13.7 = 0.548 ppm-1 . 0.55 ppm-1 CI 25 ppm (c) The selectivity coefficient, KA,I, is KA,I = kI kA 0.548 ppm-1 1.55 ppm-1 = 0.354 . 0.35 (d) Because kA is greater than kI, which makes KA,I less than one, we know that the method is more selective for the analyte than for the interferent. (e) To achieve an error of less than 1% we know that KA,I # CI < 0.01 # CA Rearranging for the ratio CI/CA and solving gives CI < 0.01 = 0.01 = 0.028 . 0.03 CA KA,I 0.354 Thus, the interferent can be present with a concentration that is no more than 3% of the analyte’s concentration. 4. We know that Stotal = SA + Sreag = kA CA + Sreag . Making appropri - ate substitutions 35.2 = (17.2 ppm-1) # CA + 0.06 and solving for CA gives the analyte’s concentration as 2.01 ppm. 5. A relative error of –2.0% means that KCa,Zn # CZn =- 0.020 # CCa We know that the concentrations of Ca and Zn are in a 50:1 ratio, so it is convenient to assign a concentration of 50 to Ca and a concen - tration of 1 to Zn; making appropriate substitutions KCa,Zn # 1 = –0.02 # 50 and solving for KCa,Zn gives its value as –1.0. Note that an absolute value for KCa,Zn of one implies the method is equally sensitive to the analyte, Ca, and the interferent, Zn, and that the negative sign for = Chapter 3 The Vocabulary of Analytical Chemistry 17 KCa,Zn implies the interferent, Zn, decreases the signal. A sample for which CCa = CZn will have Ssamp = 0! 6. In the absence of ascorbic acid the signal is S1 = kGL # CGL = kGL # (10.0 ppb) where GL represents glutathione. In the presence of ascorbic acid, AA, the signal is S2 = kGL (CGL + KGL,AA # CAA) = kGL (10.0 ppb + KGL,AA # 1.5 ppb) We know that the signal in the presence of ascorbic acid, S1, is 5.43 × the signal in the absence of ascorbic acid, S2; thus S2 = 5.43 = kGL (10.0 ppb + KGL,AA # 1.5 ppb) S1 kGL # (10.0 ppb) 5.43 = 10.0 ppb + KGL,AA # 1.5 ppb 10.0 ppb 54.3 ppb = 10.0 ppb + KGL,AA # (1.5 ppb) Solving for KGL,AA gives its value as 3.0×101. When the interferent is methionine, which we abbreviate as ME, we have S kGL (10.0 ppb + KGL,ME # 3.5 # 102 ppb) 2 = 0.906 = 1 kGL # (10.0 ppb) S2 S1 = 0.906 = 10.0 ppb + KGL,ME # 3.5 # 102 ppb 10.0 ppb 9.06 ppb= 10.0 ppb+ KGL,ME # (3.50 # 102 ppb) which gives KGL,ME as –0.0027. There are two important differences between these two interferents. First, although the method is more sensitive for that analyte glutathione than it is for the interferent me - thionine (the absolute value for KGL,ME is less than one), it is more sensitive for the interferent ascorbic acid than it is for the analyte glutathione (KGL,AA is greater than one). Second, the positive value for KGL,AA indicates that ascorbic acid increases the total signal and the negative value for KGL,ME indicates that methionine decreases the total signal. 7. (a) In the absence of ascorbic acid the signal is S1 = kGA CGA where GA represents glycolic acid, and in the presence of ascorbic acid, AA, the signal is S2 = kGA (CGA + KGA,AA # CAA) We know that the signal in the presence of ascorbic acid, S1, is 1.44 × the signal in the absence of ascorbic acid, S2; thus S Solutions Manual for Analytical Chemistry 2.1 18 k C C 1.44 = kGA (CGA + KGA,AA # CAA) GA GA = CGA + KGA,AA # CAA CGA 1.44 = (1 # 10-4 M) + KGA,AA # (1 # 10-5 M) 1 # 10-4 M Solving for KGA,AA gives its value as 4.4. (b) The method is more selective for the interferent, ascorbic acid, than it is for the analyte, glycolic acid, because KGA,AA is greater than one. (c) To avoid an error of more than 1%, we require that KGA,AA # CAA < 0.01 # CGA which requires that C KGA,AA # CAA 4.4 # (1.0 # 10-5 M) GA > 0.01 = 0.01 = 4.4 # 10 M 8. (a) To determine the sensitivity for the analyte, we begin with the equation Ssamp = kACA and solve for kA; thus kA = Ssamp = A 7.45 # 10-5 A 1.12 # 10-6 M = 66.5 A/M (b) In the presence of an interferent, the signal is Ssamp = kA (CA + KA,I # CI ) Rearranging to solve for KA,I and making appropriate substitutions KA,I = Ssamp - kA CA kA CI 4.04 # 10-5 A - (66.5 A/M) # (1.12 # 10-6 M) = (66.5 A/M) # (6.5 # 10-5 M) gives –7.9×10–3 as the value for KA,I. (c) The method is more selective for the analyte, hypoxanthine, than for the interferent, ascorbic acid, because the absolute value of KA,I is less than one. (d) To avoid an error of 1.0% requires that KA,I # CI < –0.01 # CA , where we use a relative error of –0.010 because the interferent de- creases the signal (note that KA,I is negative). Rearranging and making appropriate substitutions gives CI < -0.010 CA = KA,I -0.010 # (1.12 # 10-6 M) –7.9 # 10-3 or a concentration of ascorbic acid less than 1.4×10-6 M. 9. Answers will vary with the selected procedure, but what follows is an -3 Chapter 3 The Vocabulary of Analytical Chemistry 19 example of a typical response. Solutions Manual for Analytical Chemistry 2.1 20 3 Surfactants are compounds that decrease the surface tension between normally immiscble compounds, allowing them to mix together. Common examples of surfactants, which have many practical appli - cations, include detergents and emulsifiers . Many surfactants consist of a long non -polar hydrocarbon chain of 10 –20 carbon atoms with a polar functional group on one end that either carries a charge (anionic or cationic) or is neutral. Although surfactants themselves generally are not a health hazard, their presence in the environment may help solubilize other, more harmful compounds. One method for deter - mining the concentration of anionic surfactants in water is the Meth - ylene Blue Method for Methylene -Blue-Active Substances (MBAS), which is Method 5540 C in Standard Methods for the Examination of Water and Wastewater . This method relies on the reaction of methylene blue (MB), which is a cationic dye, with anionic surfactants to form a neutral complex. The aqueous sample is made slightly basic, a strongly acidic solution of MB is added, and the resulting complex extracted into chloroform. When the extraction is complete, the chlorofor m layer is isolated and then washed with an acidic solution of water to remove interferents. The intensity of the complex’s color in chloroform is proportional to the concentration of anionic surfactants in the original sample. The absorbance of the surfactant -MB complex is measured in a spec- trophotometer using a cell with a 1-cm pathlength at a wavelength of 652 nm. A blank consisting of chloroform is used to calibrate the spectrophotometer. Although a sample will contain a variety of different anionic surfac - tants, the method is standardized using a single, standard reference material of linear alkylbenzene sulfonates (LAS). A stock standard solution is prepared that is 1.00 g LAS/L, which is used to prepare a working standard solution that is 10.0 µg LAS/mL. At least five calibration standards are prepared from the working standard with concentrations of LAS in the range of 0.10 µg/mL to 2.0 µg/L. The method is sensitive to a variety of interferents. If cationic sur- factants are present, they will compete with methylene blue for the anionic surfactants, decreasing the reported concentration of MBAS; when present, their concentration is minimized by first passing the sample through a cation -exchange column. Some organic anions, such as chloride ions and organic sulfates, form complexes with meth - ylene blue that also extract into chloroform, increasing the reported concentration of MBAS; these interferences are minimized by the acidic wash that follows the extraction step. The volume of sample taken for the analysis is based on the expected MBAS concentration as follows: 400 mL if the expected concentra - Not all anionic surfactants react with MB, which is why the procedure’s name includes the qualifying statement “meth - ylene -blue-active substances. (MBAS)” The most important class of MBAS surfactants are linear alkylbenzene sul- fonates (LAS) with the general formula R–C6 H4 SO– where R is an linear alkyl chain of 10–14 carbons. See Chapter 7 for more details on solvent extractions. See Chapter 10 for more details about ab - sorption spectrophotometry. See Chapter 5 for more details about methods of standardization, including calibration curves. See Chapter 12 for more details about ion-exchange. Chapter 3 The Vocabulary of Analytical Chemistry 21 tion is between 0.025 –0.080 mg/L; 250 mL if the expected concen - tration is between 0.08 –0.40 mg/L; and 100 mL if the expected con - centration is between 0.4 –2.0 mg/L. If the expected concentration is greater than 2 mg/L, a sample that contains between 40–200 µg is diluted to 100 mL with distilled water.