th Fundamentals of Analytical Chemistry: 9 ed. Chapter 3 Chapter 3 3-1. (a) SQRT returns the square root of a number or result of a calculation. (b) AVERAGE returns the arithmetic mean of a series of numbers. (c) PI returns the value of pi accurate to 15 digits (d) FACT returns the factorial of a number, equal to 1 × 2 × 3 × … × number. (e) EXP returns e raised to the value of a given number. (f) LOG returns the logarithm of a number to a base specified by the user. 3-2. Count(value 1, value2,…) returns the number of cells that contain numbers and numbers within the list of arguments. As it appears in Figure 3-10, the COUNT function should return a value of 8 for the number of data values in each column of the spreadsheet. 3-3. One method is to add comments for the appropriate cells. The comment group is found under the Review tab. An example for finding the mean (average) of four numbers is shown below. Another method is to add a text box to the worksheet by clicking on Text Box in the Insert tab. An arrow could be added pointing to the appropriate cell by clicking on Shapes in the Illustrations group under the Insert tab. th Fundamentals of Analytical Chemistry: 9 ed. Chapter 3 3-4. Replacing these values gives the worksheet below. Obviously, Column C has a nonsensical result with the replacement. 3-5. The result is: 3-6. The answer is contained in the problem. 2 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 Chapter 4 4-1. (a) The millimole is an amount of a chemical species, such as an atom, an ion, a molecule or an electron. There are 23 particles −3 mole 20 particles 6.02×10 × 10 = 6.02×10 mole millimole millimole (b) The molar mass is the mass in grams of one mole of a chemical species. (c) The millimolar mass is the mass in grams of one millimole of a chemical species. (d) Parts per million, cppm, is a term expressing the concentration of dilute solutions. Thus, mass of solute 6 cppm= × 10 ppm mass of solution The units of mass in the numerator and the denominator must be the same. 4-2. The molar species concentration is number of moles of that species contained in one liter of solution. The molar analytical concentration is the total number of moles of a solute in 1 liter of the solution, regardless of the solute’s chemical state. 3 3 1000 mL 1 cm  1 m  −3 3 4-3. The liter: 1 L = × × = 10 m   1 L 1 mL 100 cm   1mol 1 L 1 mol Molar concentration: 1 M = × = −3 3 −3 3 1 L 10 m 10 m 8 1 MHz 4-4. (a) 3.2×10 Hz × = 320 MHz 6 10 Hz 9 −7 10 ng (b) 4.56×10 g × = 456 ng 1 g th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 7 1 mol (c) 8.43×10 μmol × = 84.3 mol 6 10 μmol 10 1 Gs (d) 6.5×10 s × = 65 Gs 9 10 s 6 1 mm (e) 8.96×10 nm × = 8.96 mm 6 10 nm 1 kg (f) 48,0 g × = 48 kg 1000 g 23 4-5. For oxygen, for example 15.999 u/atom = 15.999 g/6.022 ×10 atoms = 15.999 g/mol. So 1 u = 1 g/mol. Thus, 1g = 1mol u. 23 4-6. From Pb. 4-5, 1 g = 1 mol u = 6.022 × 10 u 12 1 u = 1/12 mass of C atom So 1 kg = 1000 g = 1000× Avogadro’s number of u = 12 1000 × Avogadro’s number × 1/12 mass of C atom = 12 1000/12 × Avogadro’s number × mass of C atom + 23 + 1mol Na 3PO4 3 mol Na 6.022×10 Na 22 + 4-7. 2.92 g Na PO × × × = 3.2 × 10 Na 3 4 + 163.94 g 1 mol Na PO 1 mol Na 3 4 + 23 + 2 mol K 6.022×10 K 24 + 4-8. 3.41 mol K HPO × × = 4.1×10 K 2 4 + 1 mol K HPO 1 mol K 2 4 2 mol B 1 mol B O 2 3 4-9. (a) 8.75 g B O × × = 0.251 mol B 2 3 1 mol B O 69.62 g B O 2 3 2 3 1 g 7 mol O 167.2 mg Na B O •10H O× × 2 4 7 2 1000 mg 1 mol Na B O •10H O 2 4 7 2 (b) 1 mol Na 2B4O7 •10H2O −3 × = 3.07 × 10 molO = 3.07 mmol 381.37 g 2 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 1 mol Mn 3O4 3 mol Mn −2 (c) 4.96 g Mn O × × = 6.50 × 10 mol Mn 3 4 228.81 g Mn O 1 mol Mn O 3 4 3 4 1 g mol CaC 2O4 2 mol C −3 333 mg CaC O × × × = 5.20 × 10 mol C 2 4 (d) 1000 mg 128.10 g CaC O 1 mol CaC O 2 4 2 4 = 5.20 mmol 1 g 1 mol P O 1000 mmol 2 mol P 2 5 850 mg P O × × × × 2 5 4-1 (a) 0 . 1000 mg 141.94 g P2O5 1 mol 1 mol P2O5 =11.98 mmol P 1 mol CO 1000 mmol 1 mol C 2 (b) 40.0 g CO × × × = 909 mmol C 2 44.01 g CO 1 mol 1 mol CO 2 2 1 mol NaHCO 1000 mmol 3 mol O 3 (c) 12.92 g NaHCO × × × = 461.4 mol O 3 84.01 g NaHCO 1 mol 1 mol NaHCO 3 3 1 g 1 mol MgNH PO 1000 mmol 4 4 57 mg MgNH PO × × × 4 4 1000 mg 137.32 g MgNH PO 1 mol 4 4 (d) 1 mol Mg × = 0.42 mmol Mg 1 mol MgNH PO 4 4 0.0555 mol KMnO 1000 mmol 4 4-11. (a) × × 2.00 L = 111 mmol KMnO 4 L 1 mol −3 3.25 × 10 M KSCN 1000 mmol L × × ×750 mL (b) L 1 mol 1000 mL = 2.44 mmol KSCN 3.33 mg CuSO 1 g 1 mol CuSO 1000 mmol 4 4 × × × × 3.50 L (c) 1 L 1000 mg 159.61 g CuSO4 1 mol −2 = 7.30×10 mmol CuSO 4 0.414 mol KCl 1000 mmol 1 L (d) × × × 250 mL = 103.5 mmol KCl 1 L 1 mol 1000 mL 3 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 0.320 mol HClO 1000 mmol 1 L 4 × × ×226 mL 4-12. (a) 1 L 1 mol 1000 mL = 72.3 mmol HClO 4 −3 8.05×10 mol K CrO 1000 mmol 2 4 × ×25.0 L (b) 1 L 1 mol = 201 mmol K CrO 2 4 6.75 mg AgNO 1 g 1 mol AgNO 1000 mmol 3 3 × × × ×6.00 L (c) 1 L 1000 mg 169.87 g AgNO3 1 mol = 0.238 mmol AgNO 3 0.0200 mol KOH 1000 mmol 1 L × × ×537 mL (d) 1 L 1 mol 1000 mL = 10.7 mmol KOH 63.01g HNO 3 1000 mg 4 4-13. (a) 0.367 mol HNO × × = 2.31 × 10 mg HNO 3 3 1 mol HNO 1 g 3 1 mol 40.30 g MgO 1000 mg 3 (b) 245 mmol MgO× × × = 9.87 × 10 mg MgO 1000 mmol 1 mol MgO 1 g 80.04 g NH 4NO3 1000 mg 6 (c) 12.5 mol NH NO × × = 1.00 × 10 mg NH NO 4 3 4 3 1 mol NH NO 1 g 4 3 548.23 g (NH ) Ce(NO ) 1000 mg 4 2 3 6 4.95 mol (NH ) Ce(NO ) × × 4 2 3 6 (d) 1 mol (NH4)2Ce(NO3)6 1 g 6 = 2.71 × 10 mg (NH ) Ce(NO ) 4 2 3 6 119.0 g KBr 4-14. (a) 3.20 mol KBr × = 381 g KBr 1 mol KBr 1 mol 223.20 g PbO (b) 18.9 mmol PbO × × = 4.22 g PbO 1000 mmol 1 mol PbO 4 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 120.37 g MgSO 4 (c) 6.02 mol MgSO × = 725 g MgSO 4 4 1 mol MgSO 4 1 mol 392.23 g Fe(NH ) (SO ) •6H O 4 2 4 2 2 10.9 mmol Fe(NH ) (SO ) •6H O × × 4 2 4 2 2 (d) 1000 mmol 1 mol Fe(NH4 )2 (SO4 )2•6H2O = 4.28 g Fe(NH ) (SO ) •6H O 4 2 4 2 2 0.350 mol sucrose 1 L 342 g sucrose 1000 mg × × × 4-15. (a) 1 L 1000 mL 1 mol sucrose 1 g 3 × 16.0 mL = 1.92×10 mg sucrose −3 3.76×10 mol H O 34.02 g H O 1000 mg 2 2 2 2 × × (b) 1 L 1 mol H O 1 g 2 2 × 1.92 L = 246 mg H O 2 2 2.96 mg Pb(NO ) 1 L 3 2 × × 356 mL (c) 1 L 1000 mL = 1.05 mg Pb(NO ) 3 2 0.0819 mol KNO 101.10 g KNO 1000 mg 1 L 3 3 × × × (d) 1 L 1 mol 1 g 1000 mL × 5.75 mL = 47.6 mg KNO 3 0.264 mol H O 1 L 34.02 g H O 2 2 2 2 × × × 250 mL 4-16. (a) 1 L 1000 mL 1 mol H2O2 = 2.25 g H O 2 2 5 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 −4 5.75×10 mol benzoicacid 1 L 122 g benzoicacid × × (b) 1 L 1000 mL 1 mol benzoicacid −3 × 37.0 mL = 2.60×10 g benzoicacid 31.7 mg SnCl 1 g 2 (c) × ×4.50 L = 0.143 g SnCl 2 1 L 1000 mg 0.0225 mol KBrO 1 L 167 g KBrO 3 3 × × ×11.7 mL (d) 1 L 1000 mL 1 mol KBrO3 −2 = 4.40 × 10 g KBrO 3 4-17. (a) pNa = − log(0.0635 + 0.0403) = − log(0.1038) = 0.9838 pCl = − log(0.0635) = 1.197 pOH = − log(0.0403) =1.395 −3 (b) pBa = − log(4.65×10 ) = 2.3 p = − M = −n l o g ( 2 . 5 4 ) 0 . 4 0 5 −3 p = − C× × l + × =l− o = − g ( 2 4 . 6 5 1 0 2 2 . 5 4 ) l o g ( 5 . 0 8 9 ) 0 . 7 0 7 (c) pH = −log(0.400) = 0.398 pCl = − log(0.400 + 2×0.100) = − log(0.600) = 0.222 p Z= −n l o =g ( 0 . 1 0 0 ) 1 . 0 0 (d) −2 pCu = − log(5.78×10 ) = 1.238 p = − Z =n l o g ( 0 . 2 0 4 ) 0 . 6 9 0 p = − N × +O× = − l = o g ( 2 0 . 0 5 7 8 2 0 . 2 0 4 ) l o g ( 0 . 5 2 3 6 ) 0 . 2 8 1 3 6 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 (e) −7 −7 −6 pK = −log(4×1.62×10 + 5.12×10 ) = − log(1.16×10 ) = 5.936 −7 p =O− H × l o g ( 5 . 1 2 1 0 ) = 6 . 2 9 1 −7 p F= − e × ( C N ) l o g ( 1 . 6 2 1 0 ) = 6 . 7 9 0 6 (f) −4 pH = −log(4.75×10 ) = 3.32 −4 p = − B × a l o g ( 2 . 3 5 1 0 ) = 3 . 6 3 −4 −4 −4 pClO = −log(2×2.35×10 + 4.75×10 ) = − log(9.45×10 ) = 3.02 4 + + –5 4-1 (8a) pH. = 4.31, log[H3O ] = −4.31, [H3O ] = 4.9 × 10 M as in part (a) + −5 (b) [H3O ] = 3.3 × 10 M + (c) [H3O ] = 0.26 M + −14 (d) [H3O ] = 1.3 × 10 M + −8 (e) [H3O ] = 2.4 × 10 M + −6 (f) [H3O ] = 4.8 × 10 M + (g) [H3O ] = 5.8 M + (h) [H3O ] = 2.6 M 4-19. (a) pNa = pBr = −log(0.0300) = 1.523 (b) pBa = −log(0.0200) = 1.699; pBr = −log(2 × 0.0200) = 1.398 –3 −3 (c) pBa = −log(5.5 × 10 ) = 2.26; pOH = −log(2 × 5.5 × 10 ) = 1.96 (d) pH = −log(0.020) = 1.70: pNa = −log(0.010) = 2.00 7 th Fundamentals of Analytical Chemistry: 9 ed. Chapter 4 pCl = −log(0.020 + 0.010 ) = −log(0.030) = 1.52 −3 –3 (e) pCa = −log(8.7 × 10 ) = 2.06; pBa = −log(6.6 × 10 ) = 2.18 −3 −3 pCl = −log(2 × 8.7 × 10 + 2 × 6.6 × 10 ) = −log(0.0306) = 1.51 −8 −7 (f) pZn = −log(2.8 × 10 ) = 7.55; pCd = −log(6.6 × 10 ) = 6.18 −8 −7 pNO3 = −log(2.8 × 10 + 2 × 6.6 × 10 ) = 5.87 + + 4-20. (a) pH = 1.020; log[H3O ] = −1.020; [H3O ] = 0.0955 M – – (b) pOH = 0.0025; log[OH ] = −0.0025; [OH ] = 0.99 M – −8 (c) pBr = 7.77; [Br ] = 1.70 × 10 M 2+ (d) pCa = −0.221; [Ca ] = 1.66 M + −13 (e) pLi = 12.35; [Li ] = 4.5 × 10 M – (f) pNO3 = 0.034; [NO3 ] = 0.92 M 2+ (g) pMn = 0.135; [Mn ] = 0.733 M – −10 (h) pCl = 9.67; [Cl ] = 2.14 × 10 M + 3 + 1 1.02 g 1000 mL 1 mol Na −2 + 4-21. (a) 1.08×10 ppm Na × × × × = 4.79×10 M Na 6 10 ppm 1 mL 1 L 22.99 g 3− 2− 1 1.02 g 1000 mL 1 mol SO4 −3 2− 270 ppm SO × × × × = 2.87×10 M SO 4 6 4 10 ppm 1 mL 1 L 96.06 g −2 (b) pNa = − log(4.79×10 ) =1.320 −3 p =S− O × =l o g ( 2 . 8 7 1 0 ) 2 . 5 4 2 4 −9 4-2 (a2) 300 n. mol/L = 300 × 10 mol/L or 300 nM in plasma −3 2.2 mmol/L = 2.2 × 10 mol/L or 2.2 mM in whole blood –9 (b) pHb in plasma = −log(300 × 10 ) = 6.52 −3 pHb in blood = −log(2.2 × 10 ) = 2.66 8