Solutions Manual for Quantitative Chemical Analysis (10th Ed)

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Solutions Manual for Quantitative Chemical Analysis Tenth Edition Daniel C. Harris Michelson Laboratory, China Lake, California Charles A. Lucy University of Alberta, Edmonton, Alberta Macmillan Learning Austin I Boston I New York Plymouth 2020, 2016, 2010, 2007 W. H. Freeman and Company All rights reserved. Printed in the United States of America. Macmillan Learning One New York Plaza Suite 4600 New York, NY CHAPTER 0 THE ANALYTICAL PROCESS Qualitative analysis finds out what is in a sample. Quantitative analysis measures how much is in a sample. Steps in a chemical analysis: (1) Formulate the question: Convert a general question into a specific one that can be answered 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. Masking converts an interfering species to a noninterfering species. 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. (a) A homogeneous material has the same composition everywhere. In a 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 portion, we will capture each of the different compositions that are present. 1 CHAPTER 1 CHEMICAL MEASUREMENTS A note from Dan and Chuck: worry if your numerical answers are slightly different from ours. You or we may have rounded intermediate results. In general, retain many extra digits for intermediate answers and save your roundoff until the end. study this process in Chapter 3. (a) meter (m), kilogram (kg), second (s), ampere (A), kelvin (K), mole (mol) (b) hertz (Hz), newton (N), pascal (Pa), joule (J), watt (W) See Table Abbreviations above kilo are capitalized: M (mega, 106), G (giga, 109), T (tera, 1012), P (peta, 1015), E (exa, 1018), Z (zetta, 1021), and Y (yotta, 1024). (a) mW milliwatt watt (b) picometer meter (c) kO kiloohm 103 ohm (d) uF microfarad farad (e) TJ terajoule 1012 joule (f) ns nanosecond second (g) fg femtogram gram (h) dPa decipascal pascal (a) 100 fJ or 0 pJ (d) 0 nm or 100 pm (b) 43 nF (e) 21 TW (c) 299 THz or 0 79 PHz (f) 0 amol or 483 zmol (a) 8 C (b) The formula mass of CO2 is 12 44 1012 (c) 2 2 X 1010 tons of CO2 2 tons CO2 4 tons CO2 per person 7 x 109 people 3 4 Chapter 1 We will convert ounces of tuna to grams of tuna and find out how many ug of mercury are in one can (6 oz) of tuna. For a body mass of 68 kg, we will compute how many days are allowed between eating tuna SO that the average dose does not exceed 0 ug body weight per day. Table tells us that 1 lb 0 kg 6 oz 1b kg) 0 kg 170 g One part per million means 1 ug Hg per gram of tuna There is 0 ppm of mercury in chunk white tuna 0 ug tuna. A can contains (170 Hg A dose of 0 ug body weight per day for a 68 kg person is If I eat 102 ug Hg in one day from one can of tuna, I have eaten the amount of mercury allowed in (102 Hg) I should wait 15 days before consuming my next can of tuna SO that my average intake does not exceed 6 ug Chunk light tuna contains 0 ppm Hg 0 ug tuna. Substituting this number for 0 ug tuna in the sequence of calculations gives a period of 3 days. I could eat 2 cans of chunk light tuna per week. Table tells us that 1 horsepower 745 W 745 100 horsepower (100 horsepower) 104 7.457x104 4 (a) 2 (120 6 Chapter 1 (b) A nanogram is g. molecules 0 day (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 of solution or mixture) (e) volume percent 100 X (volume of of solution or mixture) (f) parts per million 106 X (grams of of sample) (g) parts per billion 109 X (grams of of sample) 6 Chapter 1 (b) A nanogram is g. molecules 0 day (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 of solution or mixture) (e) volume percent 100 X (volume of of solution or mixture) (f) parts per million 106 X (grams of of sample) (g) parts per billion 109 X (grams of of sample) (h) formal concentration moles of of solution Acetic acid (CH3CO2H) is a weak electrolyte that is partially dissociated. When we dissolve 0 mol in a liter, the concentrations of CH3CO2H plus CH3CO 2 add to 0 M. The concentration of CH3CO2H alone is less than 0 M. 0 mol NaCl 0 0 L 1 0 mol CH3OH ( 0. 171 (a) If atmospheric pressure is 1 bar, then a concentration of 1 ppb is bar. A concentration of 39 ppb is 39 X bar. There are exactly 105 Pa in a bar. So, first convert bar to Pa: Chemical Measurements 7 (39 Pa Then convert Pa to mPa: (39 X The O3 concentration of 3 mPa is about of the stratospheric concentration of 19 mPa. (b) The pressure of the atmosphere at 16 km altitude is 9 kPa. The pressure of O3 is 19 mPa. From the definition of ppb, the O3 pressure in ppb is O3 pressure (Pa) X atmospheric pressure (Pa) 2 X 103 ppb (a) 19 19 x Pa. 19 X bar mPa X (b) T (K) 273 70 K bar 1 X M 11 nM X 203 (a) PV nRT (5 X n 2,11 x mol 2 M (b) Ar: means 0 L of Ar per L of air PV nRT: n 3 x mol IN 3 x 1 M Kr: 1 ppm 1 uL Kr per L of air PV nRT bar)(1 x K n 4 x Xe: 87 ppb 87 nL Xe per L of air Chemical Measurements 9 (a) 2 cell vessicle vessicle (b) 106 molecules (c) Volume (200 m)³ 3 L (d) 0 M 4 Similarly, 120 L 6 x M. (a) Mass of L 1 1046 Grams of C2H6O2 per liter (b) L contains 376 g of C2H6O2 and 1046 376 669 of H2O 0 kg Molality C2H6O2 H2O C2H6O2 9 m Shredded wheat: g contains 0 g protein 0 g carbohydrate X Doughnut: contains 0 protein 0 g carbohydrate 0 g fat 0 0 Cal In a similar manner, we find 2 hamburger and 0 for apple. There are 16 ounces in 1 pound, which Table says is equal to 453 g 453 g 16 ounce 28 g . ounce To convert to multiply times (28 10 Chapter 1 Shredded Wheat Doughnut Hamburger Apple 3 3 2 0 102 79 14 2 0 6 g in a 2 L volumetric flask Weigh out 2 X 0 mol 0 mol 6 g B(OH)3 and dissolve in 2 kg H2O. We need 1 X X 0 mol 0 mol NaOH 4 g NaOH 0 NaOH 4 8 g solution g solution (a) Vcon 55 mL (b) One liter of H2SO4 contains (18 mol ) (98 mol 1 X 103 g of H2SO4. Since the solution contains 98 H2SO4, and the mass of H2SO4 per mL is 1 g, the mass of solution per milliliter (the density) is 1 1 2 L of 0 M NaOH 0 mol NaOH 13 g NaOH solution solution 13 1 mL (16 mL solution)