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

2. 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 1 / [Al 3+ ] – 1/ [Al 3+ ]o = k f t and ½-life is from 1 / [Al 3+ ] o = k f t1/2 or t 1/2 = 1 / k f [Al 3+ ] o t 1/2 @ pH = 3.9, 1 / (110 L mol -1 s -1 x10 -5 mol L -1 ) = 909 s t 1/2 @ pH = 4.9, 1 / (726 L mol -1 s -1 x10 -5 mol L -1 ) = 138 s

3. 6. Distribution coefficients, α i, obtained by expressing concentrations of all other species in terms of i and [H + ]. Thus, for H 2 CO 3, [HCO 3 - ] = K 2 [H + ][CO 3 2- ] [CO 3 2- ] = [H 2 CO 3 ] / K 1 [H + ] 2 [HCO 3- ] = K 2 [H + ][H 2 CO 3 ] / K 1 [H + ] = (K 2 / K 1 ) [H 2 CO 3 ] / [H + ] [CO 3 ] T = [H 2 CO 3 ] x [1 + (K 2 / K 1 ) 10 pH + (1 / K 1 ) 10 2pH ] αH 2 CO 3 = [H 2 CO 3 ] / [CO 3 ] T = 1/ [1 + (K 2 / K 1 ) 10 pH + (1 / K 1 ) 10 2pH ] αHCO 3 - = (K 1 / K 2 ) 10 -pH + 1 + (1 / K 2 ) 10 pH αCO 3 - = K 1 10 pH + K 2 10 pH + 1

4. When is HCO 3 - dominant, i.e., αHCO3 - > 0.5? Bounds would be (K 1 / K 2 ) 10 -pH + 1 + (1 / K 2 ) 10 pH = 2 Substituting for K i 10 6.4-pH + 10 pH-10.3 = 1 @ pH = 6.4 and pH = 10.3, αHCO 3 - = 0.5. Therefore, in this range. 6.40.50 6.70.67 7.00.80 7.30.89 7.60.94 7.90.97 8.20.98 8.50.98 8.80.97 9.10.94 9.40.89 9.70.80 10.00.67 10.30.50

5. 0.00010.900564 0.00030.835527 0.00100.724655 0.00300.581919 0.00500.504192 0.01000.393473 0.03000.229707 0.10000.107437 0.30000.060813 13. [Al 3+ ] = (Al 3+ ) / γ 3+ = 10 -6.23 / γ 3+ I γ 3+ Davies model was used for activity coefficient. [Al 3+ ] increases with increasing I.

6. 8. Chapter 5 K dis = (Ca 2+ )(SO 4 2- )(H 2 O) 2 / (gypsum) = 10 -4.62 log [(gypsum) / (Ca 2+ )] = -log K dis + log (SO 4 2- ) + 2log (H 2 O) AR gypsum = 4.62 – 3.00 + 2 (0.00) = 1.62 K dis = (Ca 2+ )(CO 2 )(H 2 O) / (calcite)(H + ) 2 = 10 9.75 log [(calcite) / (Ca 2+ )] = -K dis + log P CO2 + log (H 2 O) + 2pH AR calcite = -9.75 + log P CO2 + 0.00 + 2 (8) = 6.25 + log P CO2

7. 10. CaSO 4 2H 2 O = Ca 2+ + SO 4 2- + 2H 2 O (CaSO 4 2H 2 O) / (Ca 2+ ) = (SO 4 2- )(H 2 O) 2 / K dis AR gypsum = -log K dis + log (SO 4 2- ) + 2log (H 2 O) = 4.62 – log (SO 4 2- ) CaCO 3 + 2H + = Ca 2+ + CO 2 + H2O (CaCO 3 ) / (Ca 2+ ) = (CO 2 )(H 2 O) / K dis (H + ) 2 AR calcite = -log K dis + log P CO2 + log (H 2 O) + 2pH = -9.75 + log P CO2 + 2pH

8. CaF 2 = Ca 2+ + 2F - (CaF 2 ) / (Ca 2+ ) = (F - ) 2 / K dis ARfluorite = -log K dis + 2log (F - ) = 9.80 + 2log (F - ) Substituting for (SO 4 2- ) = 0.003, (H + ) = 10 -8 and (F - ) = 0.00003 AR gypsum = 2.10 AR calcite = 6.25 + log P CO2 (= 2.73 @ P CO2 = 0.0003; 4.73 @ P CO2 = 0.03 AR fluorite = 0.76 ARcalcite > ARgypsum > ARfluorite

9. 13. AR TPbOP = -log K dis + 2/3 log (H 2 PO 4- ) + 4/3 pH AR chloro - = -log K dis + 3/5 log (H 2 PO 4- ) + 1/5 (Cl - ) + 6/5 pH For (H 2 PO 4- ) = 10 -6 and (Cl-) = 10 -3 AR TPbOP = -2.20 + 4/3 pH AR chloro - = 0.86 + 6/5 pH and AR chloro - = 0.46 + 6/5 pH @ (Cl - ) = 10 -5 Where intersect? pH = meaninglessly large values, irrespective of (Cl - ) so chloropyromorphite controls Pb 2+ 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 = -k 1 X X / X 0 = exp(-k 1 t) dM / dt = k 1 X – k 2 M = k 1 X 0 exp(-k 1 t) –k 2 M d CO2 / dt = k 2 M M = 0 @ t = 0 and t = infinite CO 2 = 0 @ t = 0 and CO 2 = X 0 @ t = infinite