1. Problems Chapter 5

2. 7.
Kdis = (Cd2+)(CO32-) / (otavite) =
For the co-precipitated form assuming (otavite*) = mole fraction = 1 / 20
Kso = Kdis (otavite*) = /20 =
The assumption of same soil solution composition sets (CO32-), thus,
(Cd2+)* / (Cd2+) = Kso* / Kso = 1 / 20

3. 10.
CaSO4 2H2O = Ca2+ + SO H2O
(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
= log PCO2 + 2pH

4. CaF2 = Ca2+ + 2F-
(CaF2) / (Ca2+) = (F-)2 / Kdis
ARfluorite = -log Kdis + 2log (F-)
= log (F-)
Substituting for (SO42-) = 0.003, (H+) = 10-8 and (F-) =
ARgypsum = 2.10
ARcalcite = log PCO2 (= PCO2 = ; PCO2 = 0.03
ARfluorite = 0.76
ARcalcite > ARgypsum > ARfluorite

6. 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 = /3 pH
ARchloro- = /5 pH and
ARchloro- = /5 (Cl-) = 10-5
Where intersect? pH = meaninglessly large values, irrespective of (Cl-) so chloropyromorphite controls Pb2+ solubility throughout pH range.