Problems Chapter 5.

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Presentation transcript:

Problems Chapter 5

7. Kdis = (Cd2+)(CO32-) / (otavite) = 10-11.2 For the co-precipitated form assuming (otavite*) = mole fraction = 1 / 20 Kso = Kdis (otavite*) = 10-11.2 /20 = 10-12.5 The assumption of same soil solution composition sets (CO32-), thus, (Cd2+)* / (Cd2+) = Kso* / Kso = 1 / 20

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.