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Assigned Problems, Chapter 4 Example Problems, Chapter 5

Assigned Problems, Chapter 4 Example Problems, Chapter 5. 2. Al 3+ + F - = AlF 2+ d[Al 3+ ] / dt= k f [Al 3+ ][F - ] – k r [AlF 2+ ] Assume reverse is negligible for early stage of reaction and given that [Al 3+ ] o = [F - ] o , d[Al 3+ ] / dt= kf[Al 3+ ] 2

elvis-mccormick
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Assigned Problems, Chapter 4 Example Problems, Chapter 5

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  1. Assigned Problems, Chapter 4 Example Problems, Chapter 5

  2. 2. • Al3+ + F- = AlF2+ • d[Al3+] / dt= kf[Al3+][F-] – kr[AlF2+] • Assume reverse is negligible for early stage of reaction and given • that [Al3+]o = [F-]o, • d[Al3+] / dt= kf[Al3+]2 • 1 / [Al3+] – 1/ [Al3+]o = kft and ½-life is from • 1 / [Al3+]o = kft1/2 or t1/2 = 1 / kf [Al3+]o • t1/2 @ pH = 3.9, 1 / (110 L mol-1 s-1 x10-5 mol L-1) = 909 s • t1/2 @ pH = 4.9, 1 / (726 L mol-1 s-1 x10-5 mol L-1) = 138 s

  3. 6. Distribution coefficients, αi, obtained by expressing concentrations of all other species in terms of i and [H+]. Thus, for H2CO3, [HCO3-] = K2[H+][CO32-] [CO32-] = [H2CO3] / K1[H+]2 [HCO3-] = K2[H+][H2CO3] / K1[H+] = (K2 / K1) [H2CO3] / [H+] [CO3]T = [H2CO3] x [1 + (K2 / K1) 10pH + (1 / K1) 102pH] αH2CO3 = [H2CO3] / [CO3]T= 1/ [1 + (K2 / K1) 10pH + (1 / K1) 102pH] αHCO3- = (K1 / K2) 10-pH + 1 + (1 / K2) 10pH αCO3- = K110pH + K210pH + 1

  4. When is HCO3- dominant, i.e., αHCO3- > 0.5? Bounds would be (K1 / K2) 10-pH + 1 + (1 / K2) 10pH = 2 Substituting for Ki 106.4-pH + 10pH-10.3 = 1 @ pH = 6.4 and pH = 10.3, αHCO3- = 0.5. Therefore, in this range. 6.4 0.50 6.7 0.67 7.0 0.80 7.3 0.89 7.6 0.94 7.9 0.97 8.2 0.98 8.5 0.98 8.8 0.97 9.1 0.94 9.4 0.89 9.7 0.80 10.0 0.67 10.3 0.50

  5. 13. [Al3+] = (Al3+) / γ3+ = 10-6.23 / γ3+ I γ3+ Davies model was used for activity coefficient. [Al3+] increases with increasing I.

  6. 8. Chapter 5 Kdis = (Ca2+)(SO42-)(H2O)2 / (gypsum) = 10-4.62 log [(gypsum) / (Ca2+)] = -log Kdis + log (SO42-) + 2log (H2O) ARgypsum = 4.62 – 3.00 + 2 (0.00) = 1.62 Kdis = (Ca2+)(CO2)(H2O) / (calcite)(H+)2 = 109.75 log [(calcite) / (Ca2+)] = -Kdis + log PCO2 + log (H2O) + 2pH ARcalcite = -9.75 + log PCO2 + 0.00 + 2 (8) = 6.25 + log PCO2

  7. 10. CaSO4 2H2O = Ca2+ + SO42- + 2H2O (CaSO4 2H2O) / (Ca2+) = (SO42-)(H2O)2 / Kdis ARgypsum = -log Kdis + log (SO42-) + 2log (H2O) = 4.62 – log (SO42-) CaCO3 + 2H+ = Ca2+ + CO2 + H2O (CaCO3) / (Ca2+) = (CO2)(H2O) / Kdis (H+)2 ARcalcite = -log Kdis + log PCO2 + log (H2O) + 2pH = -9.75 + log PCO2 + 2pH

  8. CaF2 = Ca2+ + 2F- (CaF2) / (Ca2+) = (F-)2 / Kdis ARfluorite = -log Kdis + 2log (F-) = 9.80 + 2log (F-) Substituting for (SO42-) = 0.003, (H+) = 10-8 and (F-) = 0.00003 ARgypsum = 2.10 ARcalcite = 6.25 + log PCO2 (= 2.73 @ PCO2 = 0.0003; 4.73 @ PCO2 = 0.03 ARfluorite = 0.76 ARcalcite > ARgypsum > ARfluorite

  9. 13. ARTPbOP = -log Kdis + 2/3 log (H2PO4-) + 4/3 pH ARchloro- = -log Kdis + 3/5 log (H2PO4-) + 1/5 (Cl-) + 6/5 pH For (H2PO4-) = 10-6 and (Cl-) = 10-3 ARTPbOP = -2.20 + 4/3 pH ARchloro- = 0.86 + 6/5 pH and ARchloro- = 0.46 + 6/5 pH @ (Cl-) = 10-5 Where intersect? pH = meaninglessly large values, irrespective of (Cl-) so chloropyromorphite controls Pb2+ solubility throughout pH range.

  10. Time Parent Metabolites CO2 d P / P0 M / P0 CO2 / P0 0 1.000 0.000 0.000 2 0.819 0.178 0.004 4 0.670 0.316 0.014 6 0.549 0.423 0.029 8 0.449 0.504 0.047 12 0.301 0.607 0.092 16 0.202 0.655 0.143 20 0.135 0.669 0.196 24 0.091 0.660 0.249 32 0.041 0.608 0.351 40 0.018 0.539 0.443 48 0.008 0.468 0.523 56 0.004 0.403 0.593 72 0.001 0.295 0.704 88 0.000 0.215 0.785 104 0.000 0.156 0.844 120 0.000 0.113 0.887 dX / dt = -k1X X / X0 = exp(-k1t) dM / dt = k1X – k2M = k1X0 exp(-k1t) –k2M dCO2 / dt = k2M M = 0 @ t = 0 and t = infinite CO2 = 0 @ t = 0 and CO2 = X0 @ t = infinite

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