Assigned Problems, Chapter 4

Example Problems, Chapter 5

2.

Al3+ + F- = AlF2+

-d[Al3+] / dt= kf[Al3+][F-] – kr[AlF2+]

Assume reverse is negligible for early stage of reaction and giventhat [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

6.

Distribution coefficients, αi, obtained by expressing concentrations ofall other species in terms of i and [H+]. Thus, for H2CO3,

[HCO3-] = K2[H+][CO3

2-]

[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

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 0.8010.0 0.6710.3 0.50

0.0001 0.9005640.0003 0.8355270.0010 0.7246550.0030 0.5819190.0050 0.5041920.0100 0.3934730.0300 0.2297070.1000 0.1074370.3000 0.060813

13.

[Al3+] = (Al3+) / γ3+ = 10-6.23 / γ3+

I γ3+

Davies model was used for activity coefficient.

[Al3+] increases with increasing I.

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

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

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

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.

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