CHAPTER 5: SUPPLEMENTARY SOLUTIONS 10 QUALITY ASSURANCE AND CALIBRATION METHODS

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1 CHAPTER 5: SUPPLEMENTARY SOLUTIONS 10 QUALITY ASSURANCE AND CALIBRATION METHODS S51. (a) y(± ) = (± )x (± ) S52. (b) Uncertainty in x = [ 1 + (3.89) (3.89) 9 24 ]1/2 = 0.65 (a) Subtract from each area, and plot (µg/ml) vs area: slope = and intercept = (b) Corrected area = = [Ag] [Ag] = 17.6 µg/ml. S53. First, note that s y = (± ) = (± ) [Ag] (± ) Putting the uncertainties into the equation for uncertainty in x gives [Ag] = 17.6 (±0.4) µg/ml. S54. (a) [dopamine] f = [dopamine] i (b) [S] f = ( M ) 2.00 ml ml = mm (c) [dopamine] i mm [dopamine] i = 34.6 na 58.4 na [dopamine] i = mm S55. A A [A] = F A B [B] 0.71 [80.0 nm] F = = F 1 [64.0 nm] A A [A] = F A B [B] 1.21 [A] = [930 nm] [A] = nm = 1.98 µm

2 Chapter 5: Supplementary Solutions 11 S56. (a)(b) Corrected signal (mv) = (±3.7) [C 2 H 2, vol%] (±0.7 3 ) The xintercept is and Equation 517 gives an uncertainty of [C 2 H 2, vol%] = (±0.011) vol% S57. [X] i [S] f + [X] f = I X I S+X [X] i I S+X = [S] f I X + [X] f I X I S+X = ( I X [X] i ) [S] f + [X] f [X] i I X slope intercept Setting I S+X = 0 gives I X [X] i [S] f = [ X ] f [ X ] i IX, which is true when [S] f = [X] f S58. (a) y(± 0.17) = (± )x (± ) (b) y(at 2 010) = ( )(2 010) = temperature = 10 y = K. Extrapolation of experimental progress is nonsense. There is no way to predict future progress.

3 CHAPTER 6: SUPPLEMENTARY SOLUTIONS 12 CHEMICAL EQUILIBRIUM S61. (a) K = [Cl ] [OCl ] / [OH ] 2 P Cl2 (b) K = 1/P I2 S62. Remain unchanged. S63. Cu + + N3 1 CuN 3 (s) K 1 = HN 3 H + + N3 K 2 = Cu + + HN 3 CuN 3 (s) + H + K 3 = K 1 K 2 = S64. S65. S66. S67. Q = [H + ][OH ] = > K reaction goes to the left. K 3 = e G 3 /RT = K 1. K 2 = e G 1 /RT e G 2 /RT e G 3 /RT = e ( G 1 + G 2 )/RT G 3 = G 1 + G 2 K = e G /RT ln K = G /RT G = RT ln K J (a) G = ( mol. K ) ( K) ln ( ) = 23.9 kj/mol (b) G = 63.6 kj/mol (a) Positive. (b) G = (+) = H T S. Since T S is negative, H must be positive. The reaction is endothermic. K sp S68. (a) Ag 2 CrO 4 (s) 2Ag + + CrO 2 4 FM x x (2x) 2 (x) = x = M (b) M = g/l = g/100 ml (c) [Ag + ] = M = g/l = mg/ml = 14.4 µg/ml = 14.4 ppm S69. Cu(s) + Cu 2+ 2 Cu + K 1 = Cu Cl 2 CuCl(s) K 2 = 1/K sp 2 = 1/( ) 2 Cu(s) + Cu Cl 2 CuCl(s) K = K 1 K 2 =

4 Chapter 6: Supplementary Solutions 13 S610. (a) PbI 2 (s) Pb I FM x 2x x (2x) 2 = x = M = g/l (b) [Pb 2+ ] ( ) 2 = x = M = g/l K sp = S611. Ca(C 2 O 4 )(s) Ca 2+ + C 2 O 2 4 We want to reduce [C 2 O 2 4 ] to M : [Ca 2+ ] [ ] = [Ca 2+ ] = M S612. Ca(OH) 2 : K sp = Mg(OH) 2 : K sp = We want to reduce [Mg 2+ ] to M [Mg 2+ ] [OH ] 2 = [ ] [OH ] 2 = [OH ] = M Will this precipitate 0.10 M Ca 2+? Q = [Ca 2+ ] [OH ] 2 = (0.10)( ) 2 = < K sp Ca(OH) 2 will not precipitate. S613. [Pb 2+ ] = K sp /[I ] 2 = ( ) / (0.050) 2 = M [PbI + ]= K 1 [Pb 2+ ] [I ] = M [PbI 2 (aq)] = β 2 [Pb 2+ ] [I ] 2 = M [PbI 3 ] = β 3 [Pb 2+ ] [I ] 3 = M [PbI 2 4 ] = β 4 [Pb 2+ ] [I ] 4 = M S614. [Ag + ] = K sp /[Cl ] [AgCl 2 3 ] = K 3 [AgCl 2 ] [Cl ] [AgCl 2 ] = K 2 [Cl ] [Ag] total = [Ag + ] + [AgCl 2 ] + [AgCl 2 3 ] (a) M Cl (b) 0.20 M Cl (c) 2.0 M Cl [Ag + ] M M M [AgCl2 ] M M M [AgCl 2 3 ] M M M [Ag] total M M M S615. HSO 3, H 2 O K sp

5 Chapter 6: Supplementary Solutions 14 S616. acid base H 2 O OH H 3 + N CH 2 CH 2 NH 2 H 2 NCH 2 CH 2 NH 2 S617. (a) [H + ] = M ph = log [H + ] = 3.00 (b) [H + ] = M ph = 1.30 (c) [OH ] = M [H + ] = K w / [OH ] = M ph = (d) [OH ] = 3.0 M [H + ] = M ph = (e) [OH ] = M [H + ] = M ph = K S618. HCO 2 H HCO2 + H + CH 3 NH + 3 CH 3 NH 2 + H + K b + S619. N H + H 2 O N H 2 + OH K b C O 2 + H 2 O CO 2 H + OH K a S620. K a 2 HPO 4 H + + PO 3 4 HPO H 2 O H 2 PO4 + O H K b S621. HN N H + H O 2 K b 1 HN + NH + OH K b 2 HN N H H 2 O H 2 N NH OH CO 2 C O 2 + H 2 O K b 1 CO 2 H CO 2 + OH CO 2 H C O 2 + H 2 O K b 2 CO 2 H CO 2 + OH

6 Chapter 6: Supplementary Solutions 15 S622. 4nitrophenol has the larger K a. OH K a O + H + NO 2 NO 2 O 2 N O H K a O 2 N O + H + S623. Cyclohexylamine has the larger K b. NH 2 + H 2 O K b NH 3 + OH + N H N + H 2 O + K b NH + OH N H S624. OCl + H 2 O HOCl + OH K b = K w /K a = S625. HSO H + + SO 2 4 O K a2 OH + O H 2 O OH K b 2 + OH HO O S626. K b1 = K w K a3 = K b2 = K w K a2 = K b3 = K w K a1 =

7 CHAPTER 7: SUPPLEMENTARY SOLUTIONS 16 LET THE TITRATIONS BEGIN S ml of M Ce 4+ = mmol of Ce 4+. This will react with half as many mol of oxalic acid = mmol of H 2 C 2 O 4. 2H 2 O = 1.72 mg. S72. Let x = mg FeCl 2 and (27.73 x) = mg KCl mmol Ag + = 2 mmol FeCl 2 + mmol KCl 2 x mg (18.49 ml)( M ) = mg / mmol ( x) mg mg / mmol x = mg FeCl 2 = 7.76 mg Fe = 7.76 mg mg 100 = 28.0 wt % Fe. S73. (a) mol Hg 2+ = 1 2 mol Cl = [Hg 2+ ] = g g / mol mol = M L = mol (b) ( L Hg(NO 3 ) 2 ) ( mol/l) = mmol Hg 2+ = mmol Cl = mg Cl in ml = mg Cl /ml S74. Br + Ag + AgBr(s) Equivalence point = (a) [Br ] = (20.00 ml)( M ) ( M ) = ml ( M) = M Initial Dilution Fraction remaining concentration factor [Ag + K sp ] = [Br = ] = M pag + = log ( ) = 9.31 (b) At the equivalence point, [Ag + ] = [Br ] = K sp = M. pag + = log [Ag + ] = 6.15 (c) [Ag + ] = ( M ) Initial concentration = M pag + = Dilution factor

8 Chapter 7: Supplementary Solutions 17 S75. Hg SCN Hg(SCN) 2 (s) K sp = Equivalence point = 2 (50.00 ml)( M) (0.104 M) = ml 0.25 V e = 5.91 ml : [Hg 2+ ] = 3 4 ( M ) = M phg 2+ = log [Hg 2+ ] = 1.78 Fraction Initial Dilution remaining concentration factor 0.50 V e = ml : [Hg 2+ ] = 1 2 ( M ) V e = ml : [Hg 2+ ] = 1 4 ( M ) = M phg 2+ = 2.00 = M phg 2+ = 2.34 V e = [Hg 2+ ][SCN ] 2 = (x) (2x) 2 = K sp x = [Hg 2+ ] = M phg 2+ = V e = ml : [SCN ] = (0.104 M ) = M [Hg 2+ ] = K sp [SCN ] 2 = Initial Dilution concentration factor ( ) 2 = M phg 2+ = V e = ml : [SCN ] = (0.104 M ) = M [Hg 2+ K sp ] = [SCN ] 2 = M phg 2+ = S76. First reaction: SO Ra 2+ RaSO 4 (s) K sp = V e1 = (100 ml) M M = ml Second reaction: SO Sr 2+ SrSO 4 (s) K sp = V e2 = = ml

9 Chapter 7: Supplementary Solutions 18 (a) ml: Half of the Ra 2+ has reacted. [Ra 2+ ] = 1 2 ( M ) = M Fraction Initial Dilution remaining concentration factor [SO 2 K sp (RaSO 4 ) 4 ] = [Ra 2+ = ] 9 M pso 2 4 (b) ml: 19/20 of the Ra 2+ has reacted. [Ra 2+ ] = ( M ) = M [SO 2 K sp (RaSO 4 ) 4 ] = [Ra 2+ = ] 8 M pso 2 4 (c) ml: 1/20 of the Sr 2+ has reacted. [Sr 2+ ] = ( M ) = M [SO 2 K sp (SrSO 4 ) 4 ] = [Sr 2+ = ] 6 M pso 2 4 (d) ml: Half of the Sr 2+ has reacted. [Sr 2+ ] = 1 2 ( M ) = M [SO 2 K sp (SrSO 4 ) 4 ] = [Sr 2+ = ] 5 M pso 2 4 = 8.72 = 7.69 = 5.09 = 4.78 (e) ml: Second equivalence point. [Sr 2+ ][SO 2 4 ] = x 2 = [SO 2 4 ] = pso 2 4 = 3.25 (f) ml: There is ml of excess SO 2 4. [SO 2 4 ] = (0.250 M ) = M pso 2 4 = 1.78 S77. (a) I estimate that the end point is near 25.9 ml. It is not very sharp and my guess could easily be off by ±0.3 ml (maybe even a little worse). Based on this end point, mol Ag + delivered at the end point = ( mm)( ml) = mmol. This equals mol Cl in ml of stream water, so [Cl ] = mmol/100.0 ml = mm = 37.9 mg Cl /L = 37.9 µg Cl /ml

10 Chapter 7: Supplementary Solutions End point 25.9 ± 0.3 ml mv ml (b) 25.9 ± 0.3 ml = 25.9 ± %. An uncertainty of % in the resulting molarity will be ( )( mm) = mm. A reasonable expression of the molarity is ± mm or 1.07 ± 0.01 mm. (c) The average difference between the two methods is ( )/7 = The standard deviation of the differences is s d = 2 ( d i d ) = n 1 d t calculated = n = s = 1.24 < t table = d t calculated < t table for 6 degrees of freedom at 95% confidence The results are not different from each other.

11 CHAPTER 8: SUPPLEMENTARY SOLUTIONS 20 ACTIVITY AND THE SYSTEMATIC TREATMENT OF EQUILIBRIUM S81. (a) (b) (c) 0.10 (d) S82. The ionic strength is half way between 0.01 and 0.05 M γ = 1 2 ( ) = S83. (a) log γ = ( / 305) = γ = = (b) γ = ( ) = S84. [Ag + ] 2 γ 2 Ag + [CrO 4 2 ]γ CrO 2 4 = (a) For 0.05 M KClO 4, µ = 0.05, γ Ag + = 0.80, γ CrO 2 = (2x) 2 (0.80) 2 (x) (0.445) = x = M [Ag + ] = 2x = M (b) For M AgNO 3, µ = M, γ Ag + = 0.924, γ CrO 2 = [ ] 2 (0.924) 2 [CrO 2 4 ] (0.740) = [CrO 2 4 ] = M S85. Since we don't know the ionic strength, we begin by neglecting activity coefficients: [Tl + ][Br ] = x 2 = K sp = x = M µ = M γ Tl + = and γ Br = For a second approximation, we add activity coefficients from the first calculation: [Tl + ]γ Tl + [Br ]γ Br = (x)(0.955)(x)(0.955) = K sp x = M µ = M γ Tl + = and γ Br = The third approximation is (x)(0.954)(x)(0.954) = K sp x = M S86. (a) µ = M γ H + = 0.86 A H + = (0.050)(0.86) = ph = log A H + = 1.37 (b) µ = 0.10 M γ H + = 0.83 A H + = (0.10)(0.83) = ph = log A H += 1.08

12 Chapter 8: Supplementary Solutions 21 S87. (a) To find the activity coefficient for µ = 0.001, we write log γ = (0.001) = γ = = In a similar manner, we calculate the results below: µ γ γ ( values from table) (b) log γ = 3.23(±0.32) (±0.32) (±0.012). 0.1 = (± ) γ = (± ) = ± e. Table 31 tells us that for y = 10 x, e y = y e x = (0.286)(2.303)( ) = γ = ±