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Solution Manual For Quantitative Chemical Analysis Eighth Edition

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    i guy 7 I, 04 : AG of WY ED TO i Ares PN | af {0 WD 1) y= pa |: URC Waa a | | | eighth Edition / i) SOLUTIONS | | | | MANUAL FOR | 49 rr a RATED Quantitative _¥ or i Chemical ANGE | , en ST. aT ON Analysis ¥en REAR " Z \ FSS ) =a a A Daniel C. Harris yy BR N p $5 f \ + StudyXy - = : E 3 § J = )

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    Downloaded from StudyXY.com ® + StudyXY Sd Ye. o> \ | iF ’ pr E \ 3 S Stu dy Anything This ContentHas been Posted On StudyXY.com as supplementary learning material. StudyXY does not endrose any university, college or publisher. Allmaterials posted are under the liability of the contributors. wv 8) www.studyxy.com

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    Solutions Manual for Harris’ Quantitative Chemical Analysis Eighth Edition Daniel C. Harris Michelson Laboratory

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    Contents Chapter 0 The Analytical Process 1 Chapter 1 Measurements 3 Chapter 2 Tools of the Trade 12 Chapter 3 Experimental Error 17 Chapter 4 Statistics 24 Chapter 5 Quality Assurance and Calibration Methods 38 Chapter 6 Chemical Equilibrium 53 Chapter 7 Activity and Systematic Treatment of Equilibrium 61 Chapter 8 Monoprotic Acid-Base Equilibria 71 Chapter 9 Polyprotic Acid-Base Equilibria 82 Chapter 10 Acid-Base Titrations 97 Chapter 11 EDTA Titrations 126 Chapter 12 Advanced Topics in Equilibrium 142 Chapter 13 Fundamentals of Electrochemistry 175 Chapter 14 Electrodes and Potentiometry 189 Chapter 15 Redox Titrations 202 Chapter 16 ~~ Electroanalytical Techniques 217 Chapter 17 Fundamentals of Spectrophotometry 230 Chapter 18 Applications of Spectrophotometry 238 Chapter 19 Spectrophotometers 251 Chapter 20 Atomic Spectroscopy 260 Chapter 21 ~~ Mass Spectrometry 269 Chapter 22 Introduction to Analytical Separations 284 Chapter 23 ~~ Gas Chromatography 298 Chapter 24 High-Performance Liquid Chromatography 311 Chapter 25 Chromatographic Methods and Capillary Electrophoresis 327 Chapter 26 ~~ Gravimetric Analysis, Precipitation Titrations, 345 and Combustion Analysis Chapter 27 ~~ Sample Preparation 360

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    CHAPTER 0 THE ANALYTICAL PROCESS 0-1. Qualitative analysis finds out what is in a sample. Quantitative analysis measures how much is in a sample. 0-2. Steps in a chemical analysis: (1) Formulate the question: Convert a general question into a specific one that can be answered by a chemical measurement. (2) Select the appropriate analytical procedure. (3) Obtain a representative sample. (4) Sample preparation: Convert the representative sample into a sample suitable for analysis. If necessary, concentrate the analyte and remove or mask interfering species. (5) Analysis: Measure the unknown concentration in replicate analyses. (6) Produce a clear report of results, including estimates of uncertainty. (7) Draw conclusions: Based on the analytical results, decide what actions to take. 0-3. Masking converts an interfering species to a noninterfering species. 0-4. A calibration curve shows the response of an analytical method as a function of the known concentration of analyte in standard solutions. Once the calibration curve is known, then the concentration of an unknown can be deduced from a measured response. 0-5. (a) A homogeneous material has the same composition everywhere. Ina heterogeneous material, the composition is not the same everywhere. (b) In a segregated heterogeneous material, the composition varies on a large scale. There could be large patches with one composition and large patches with another composition. The differences are segregated into different regions. In a random heterogeneous material, the differences occur on a fine scale. If we collect a “reasonable-size” portion, we will capture each of the different compositions that are present. (c) To sample a segregated heterogeneous material, we take representative amounts from each of the obviously different regions. In panel b in Box 0-1, 66% of the area has composition A, 14% is B, and 20% is C. To construct a 1

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    2 Chapter 0 ’ representative bulk sample, we could take 66 randomly selected samples from region A, 14 from region B, and 20 from region C. To sample a random heterogeneous material, we divide the material into imaginary segments and collect random segments with the help of a table of random numbers. 0-6. We are apparently observing interference by Mn2* in the I- analysis by method A. The result of the I analysis is affected by the presence of Mn2*. The greater the concentration of Mn2* in the mineral water, the greater is the apparent concentration of I- found by method A. Method B is not subject to the same interference, so the concentration of I- is low and independent of addition of Mn2*, There must be some Mn2+ in the original mineral water, which causes method A to give a higher result than method B even when no Mn2* is deliberately added.

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    CHAPTER 1 MEASUREMENTS A note from Dan: Don’t worry if your numerical answers are slightly different from those in the Solutions Manual. You or I may have rounded intermediate results. In general, retain many extra digits for intermediate answers and save our roundoff until the end. We'll study this process in Chapter 3. 1-1. (a) meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol) (b) hertz (Hz), newton (N), pascal (Pa), joule (J), watt (W) 1-2. Abbreviations above kilo are capitalized: M (mega, 106), G (giga, 109), T (tera, 1012), P (peta, 1015), E (exa, 1018), Z (zetta, 102!) and Y (yotta, 1024). 1-3. (@ mW = milliwatt = 10-3 watt (b) pm = picometer ~~ = 10-12 meter (c) kQ = kiloohm = 103 ohm (d) uF = microfarad ~~ = 10-6 farad (e) TJ = terajoule = 1012 joule (f) ns = nanosecond = 10-9 second (g) fg = femtogram ~~ = 10-15 gram (h) dPa = decipascal = 10-! pascal 1-4. (a) 100 {Jor 0.1 pJ (d) 0.1 nm or 100 pm (b) 43.1728 nF (e) 21 TW (c) 299.79 THz or 0.299 79 PHz (f) 0.483 amol or 483 zmol 1-5. (a) 5.4Pg=54x1015g 54x1015 £x i =5.4x1012kgof C (b) The formula mass of CO is 12.010 7 + 2(15.999 4) = 44.009 5 .009 5 kg 5.4% 1012 kge x HI a = 2.01013 kg CO, 1 ton 2.0 x 1013 COy x ———— =2.0x1010¢ f CO (c) K 2) To00 74 ons 0 2) 2.0x10" tons ~———5——— "= 4 tons per person 5x10" people 3 Study

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    4 Chapter 1 1-6. Table 1-4 tells us that | horsepower = 745.700 W = 745.700 J/s. 100.0 horsepower = (100.0 horsepower (raze) = 7.457 x 104 J/s. 0) er 7457 x 10% 4 { —— i x 3600 £ — 6416x107 4184 4 h h cal 1 2.2 x 10° ol 4.184 1 dag ny day wal )\ 24 | 3600s 7, @) 2 ATT 20(skg) (120 pound [oss 6 5] = 2.0 Wikg Similarly, 3.4 x 103 rel 3.0 J/(s'kg) = 3.0 W/kg. ay (b) The office worker’s power output is 1 2210624 (4180 1) \(_K “1x12 — ix 102w dag wl J 24K )(3600s 5 The person’s power output is greater than that of the 100 W light bulb. 1-8. 5.00 x 103 BE 1055.06 1K =1.47 x 103 La 1.47 x 103 W x Bt )( 3600s s m\( ljnch | 1 fost 1 mile mile 1-9. a) 11000 —|| —— | —— || ———— | = 0.621377 — ® ( i) re | me) km ®) 100 kn (re 37 miles )(3.785 410) _ | miles 4.68 km gallon gallon a (c) The diesel engine produces 223 g CO,/km, which we will convert into g/mile: (23 B22 =) Un) a ECD krfi J 0.621 37 mile mie In 15 000 miles, CO, = (15 000 mites)(359 g/mite) = 5.38 x 10° g or 5.38 x 10° kg = 5.38 metric tons. The gasoline engine produces 266 g CO,/km,

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    Measurements 5 which we convert into 428 g/mile or 6.42 metric tons in 15 000 miles. 1-10. Newton = force = mass x acceleration = (3) s m m? Joule = energy = force x distance = kg! — | m= kg — s s m kg Pascal = pressure = force/area = kg = /m2 = —= s ms 2 1 1-11. 003 FE |[ 10002 (535 Jari) 1), #2. ga Jatt 1000 pg 57 1 ton day ton —= | "OR 3652 >| = 6 —— 1000 £ 1000 74 year year 1-12. (a) molarity = moles of solute / liter of solution (b) molality = moles of solute / kilogram of solvent (c) density = grams of substance / milliliter of substance (d) weight percent = 100 x (mass of substance/mass of solution or mixture) (e) volume percent = 100 x (volume of substance/volume of solution or mixture) (f) parts per million = 106 x (grams of substance/grams of sample) (2) parts per billion = 109 x (grams of substance/grams of sample) (h) formal concentration = moles of formula/liter of solution 1-13. Acetic acid (CH3CO,H) is a weak electrolyte that is partially dissociated. When we dissolve 0.01 mol in a liter, the concentrations of CH3CO2H plus CH3CO; add to 0.01 M. The concentration of CH3CO2H alone is less than 0.01 M. 1-14. 32.0 g/[(22.990 + 35.453) g/mol] = 0.548 mol NaCl 0.548 mol /0.500L = 1.10M . 1-15, [1m CHO oo Lsokution) = 0.171 mol CH30H on Study

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    6 Chapter 1 0171 cine) 20) = 548g mi 1bar 1-16. (2) 19mPa=19x103Pa. 19x 103 P§ x —— =1.9x 107 bar 10° p§ (b) T7(K)=273.15+°C=273.15-70=203 K 7 rE Coed = L1x108M = 110M 0.08314—=° x 203K mol. X 1 g solute . . . 1-17. 1 ppm = 766 2 solution’ Since 1 L of dilute solution = 103 g, 1 ppm = 10-3 g solute/L. (= 10-3 g solute / 103 g solution). ’ Since 103 g=103 pg, 1 ppm=103 pg/L or 1 pg/mL. Since 10-3 g = 1 mg, 1 ppm = 1 mg/L. 1-18. 0.2 ppb means 0.2 x 10-9 g of Cy9Hyz per g of rainwater BN _ 8CyHy 0.2 x 106 g CyoHgp = 02x10 1000 g rainwater ~ L rainwater ~~ ° -6 02x10°4/L _ 7x 10-10 ] 282.55 ¢ / mol 1-19. [os 88901 |r gsolition) =26.5 g HCIO4 on 37.6 g solution — 26.5 g HCIO4 = 11.1 g Hy0 i g solution ol . 1-20. (a) |1.67=———|1000— | = 1.67 x 103 lution © [to om) <1 g HCIO4 3 b) | 0.705=——— |(1.67 x 10 0] = 1.18 x 103 g HCIO, (b) polation (1.67 x 10° g solution ) gHCIO4 (©) (1.18% 103 ¢)/(100.46 ¢ /mol) = 11.7 mol ._ _molKI 1-21. molality = Kg solvent solvent _ __200gKI _ 200gKI 20.0 wt% KI = 7600 g solution ~ 800 g H0 To find the grams of KI in 1 kg of H20, we set up a proportion: Study

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    Measurements 7 200g KI x gKI _ 800g H,0 ~ 1000 gH0 =X ~ 250gKl But 250 g KI = 1.51 mol KI, so the molality is 1.51 m. 150 x 10°15 mol/eett amol 1-22. @ 25107 vesicles/cett ~ 0 Vessicle : (b) (6.0x107'8 i) 5.2 «10% — = 3.6 x 106 molecules a6 (c) Volume = 2x (200 x 109 m)3 = 3.35 x 10-20 m3; 20 3 odd he A 335k 1017L 107° 6° /L ~18 @ 10 x 10 ol ~ 030M 335x107 'L 3 4 80 x 10 1-23. 80x 10% = 4.4 x 104 mol; 44x10" mol _ 44% 103M; 180.2 ¢ / mol 100x107 L Similarly, 120 mg/100 L = 6.7 x 10-3 M. a ol” 1-24. (a) Mass of 1.000 L = 1.046 x 1000x1000 ¥ = 1046 ¢ sal )4 aol g 2 Grams of CaHgO3 per liter = 6.067 —— x 62.07—— = 376.6 2Hg02 p L prs L (b) 1.000 L contains 376.6 g of CoHgO2 and 1046 — 376.6 = 669 g of H,0 = 0.669 kg 6.067 mol CoHgO2 Bn mol CoHgO2 B Molality = T0660 ke HhO kg 0 9.07 TkgH0 0 9.07 m 1-25. Shredded wheat: 1.000 g contains 0.099 g protein + 0.799 g carbohydrate 0.099 ¢ x w +0799 ¢ x wg = 3.6Cal Doughnut: 1.000 g contains 0.046 g protein + 0.514 g carbohydrate + 0.186 g fat 0.046 £ x 0 +0514 4 x wg + 0.186 ¢ x 207 =3.9Cal C In a similar manner, we find 2.8 - for hamburger and 0.48 or apple. There are 16 ounces in | pound, which Table 1-4 says is equal to 453.59237 g Study

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    8 Chapter 1 8 = 28.35 pr To convert Cal/g to Cal/ounce, multiply by 28.35: Shredded Wheat ~~ Doughnut Hamburger Apple Cal/g 3.6 39 2.8 0.48 Cal/ounce 102 111 79 14 1-26. Mass of water = 7 (225 pi)2 (10.0 A) 5 a ) = 1.59 x 109 kg _16x103gF L6ppm = “0 H0 Mass of F~ required = [1e- 1073 Sos aw kgH70) = 2.5% 106g F~. (If we retain three digits for the next calculation, this last number is 2.54 x 106.) The atomic mass of F is 18.998 and the formula mass of HySiFg is 144.09. One mole of HySiFg contains 6 moles of F. massof © 6x 18.998 2.54x106gF _ 5 . mass of HoSiFg ~ 144.00 ~ xgHoSiFg = % = 3:2 100g HaSiFe ’ 1-27. (a) PV = nRT (1.000 bar)(5.24 x 10-6) = n | 0.083 jg Leshan (298.15K) mol - K =n =211%107mol = 2.11 x 107M (b) Ar: 0.934% means 0.009 34 L of Ar per L of air (1.000 bar)(0.00934 L) = n [08s ater) (298.15 K) mol + K| = n=3.77x104mol = 3.77 x 104M Kr: 1.14 ppm = 1.14 pL Kr per L of air = 4.60 x 10-8 M Xe: 87 ppb => 87 nL Xe per L of air = 3.5x 109M 1-28. 2.00 ¥ 0.050 02, 6153-8 — 618 gina?2 L volumetric flask x pat 1-29. Weigh out 2 x 0.050 0 mol = 0.100 mol = 6.18 g B(OH)3 and dissolve in 2.00 ’ kg Hy0. Study

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    Measurements 9 1-30. Mcon* Yeon = Mail Vail 0.80% 1.00 ¥) = (0252) Vai = Vail = 32L }4 L 1-31. We need 1.00 ¥ x 010% = 0.10 mol NaOH = 4.0 g NaOH 40g NaOH Son = 8.0 g solution 0.50 =——— g solution Mail 1.00 M 132. (@) Veon = Vail jg, = 1000 mL br) = 55.6mL (b) One liter of 98.0% H,SO4 contains (18.0 yw6l )(98.079 / mel ) = 1.77 x 103 g of HpSO4. Since the solution contains 98.0 wt% HS04, and the mass of H,S04 per mL is 1.77 g, the mass of solution per milliliter (the density) is 1.77 g Hp804 /mL _ TERRA TE 80 g solution/mL 0.980 gH25807 /g solution 1-33. 2.00 L of 0.169 M NaOH = 0.338 mol NaOH = 13.5 g NaOH . _ _gsolution density = {solution 135g NaOTl = EE a. i1nE (16.7 mL ion) 0524 E00 g solution 1-34. FM of Ba(NO3); = 261.34 4.35 g of solid with 23.2 wt% Ba(NO3), contains (0.232)(4.35 g) = 1.01 g Ba(NO3)2 - (1.01 g BagNO3)2) mol Ba2t = ————=— ~~~ = 3.86 x 10-3 mol (261.34 g BagNO73); / mol) mol HSO4 = mol Ba2t = 3.86 x 10-3 mol (3.86 x 10-3 mol) volume of HySO4 = (3.00 mol/L) = 1.29 mL 1-35. 25.0 mL of 0.023 6 M Th** contains (0.0250 L)(0.0236 M) = 5.90 x 10-4 mol Th4+ mol HF required for stoichiometric reaction = 4 x mol Th#* = 2.36 x 10-3 mol 50% excess = 1.50(2.36 x 10-3 mol) = 3.54 x 10-3 mol HF - Study

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    10 Chapter | Required mass of pure HF = (3.54 x 10-3 mol)(20.01 g/mol) = 0.0708 g 0.0708 g HF Mass of 0.491 wt% HF solution = — CC OS HE) (0.00491 g HF / g solution) 1-36. Concentrations of reagents used in an analysis are determined either by weighing out supposedly pure primary standards or by reaction with such standards. If the standards are not pure, none of the concentrations will be correct. 1-37. The equivalence point occurs when the exact stoichiometric quantities of reagents have been mixed. The end point, which comes near the equivalence point, is marked by a sudden change in a physical property brought about by the disappearance of a reactant or appearance of a product. 1-38. Ina blank titration, the quantity of titrant required to reach the end point in the absence of analyte is measured. By subtracting this quantity from the amount of titrant needed in the presence of analyte, we reduce the systematic error. 1-39. Ina direct titration, titrant reacts directly with analyte. In a back titration, a known excess of reagent that reacts with analyte is used. The excess is then measured with a second titrant. 1-40. Primary standards are purer than reagent-grade chemicals. The assay of a primary standard must be very close to the nominal value (such as 99.95 100.05%), whereas the assay on a reagent chemical might be only 99%. Primary standards must have very long shelf lives. 1-41. Since a relatively large amount of acid might be required to dissolve a small amount of sample, we cannot tolerate even modest amounts of impurities in the acid for trace analysis. Otherwise, the quantity of impurity could be greater than quantity of analyte in the sample. 1-42. 40.0 mL of 0.040 0 M Hga(NO3)> = 1.60 mmol of Hg" , which will require 3.20 n PY 3.20 1 mmol of KI. This is contained in volume = 5100 mmol = 32.0 mL. 1-43. 108.0 mL of 0.1650 M oxalic acid = 17.82 mmol, which requires 2 mol MnO, i S mol HhCa0u H2C105 (17.82 mol HyC204) = 7.128 mmol of MnO, . Study

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    Measurements 11 7.128 mmol / (0.165 0 mmol/mL) = 43.20 mL of KMnOy4. Another way to see this is to note that the reagents are both 0.1650 M. Therefore, volume of MnO}; = (volume of oxalic acid) . For second question, volume of oxalic acid = 3(volume of MnO) = 270.0 mL. 1-44. 1.69 mg of NH3 = 0.099 2 mmol of NH3. This will react with 20.099 2) = 0.149 mmol of OBr™. The molarity of OBr™ is 0.149 mmol/1.00 mL = 0.149 M. 1-45. mol sulfamic acid = Sa = 3.4369 mmol . _ 3.4369 mmol molarity of NaOH = “3426mL 0.1003 M 1-46. HCl added to powder = (10.00 mL)(1.396 M) = 13.96 mmol NaOH required = (39.96 mL)(0.1004 M) = 4.012 mmol HCI consumed by carbonate = 13.96 — 4.012 = 9.94g mmol mol CaCO3 = 3 mol HCI consumed = 4.974 mmol = 0.497g g CaCO3 N _ 0.497g g CaCO3 B 5 wt% CaCO3 = (75413 g limestone g limestone x 100 = 92.0 wt% Study

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    CHAPTER 2 - TOOLS OF THE TRADE 2-1. The primary rule is to familiarize yourself with the hazards of what you are about to do and not to do something you consider to be dangerous. 2-3. Dichromate (Cr,0%) is soluble in water and contains carcinogenic Cr(VI). Reducing Cr(VI) to Cr(IlI) decreases the toxicity of the metal. Converting aqueous Cr(III) to solid Cr(OH); decreases the solubility of the metal and therefore decreases its ability to be spread by water. Evaporation produces the minimum volume of waste. 2-4. The upper “0” means that the reagent has no fire hazard. The right hand “0” indicates that the reagent is stable. The “3” tells us that the reagent is corrosive or toxic and we should avoid skin contact or inhalation. 2-5. The lab notebook must: (1) state what was done; (2) state what was observed; and (3) be understandable to a stranger. i 2-6. See Section 2.3. 2-7. The buoyancy correction is 1 when the substance being weighed has the same density as the weights used to calibrate the balance. 0.0012 g/mL (14.82 g) (1 ~~ 8.0 g/mL ) 2-8. m= ( 0.0012 ul} =1485¢g ~ 0.626 g/mL 2-9. The smallest correction will be for PbO», whose density is closest to 8.0 g/mL. The largest correction will be for the least dense substance, lithium. 42366 e(1 0.0012 Sal _ 8.0g/mL ) 2-10. m= TC (-AEAmn 0.0012 BE) =42391g “71.636 g/mL Without correcting for buoyancy, we would think the mass of primary standard is less than the actual mass and we would think the molarity of base reacting with the standard is also less than the actual molarity. The percentage error would be true mass — measured mass 4.2391-4.2366 true mass = 100 = 4.2391 * 100 = 0.06%. 12 ! Study

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