Page 61 - Soil and water contamination, 2nd edition

48 Soil and Water Contamination

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distribution coefficient of 70 l kg for sediment–water partitioning and a concentration
factor of 65 for fish–water partitioning.
6. The adsorption of caesium (Cs) to illite (a clay mineral) was studied by means of a batch
experiment. During this experiment an amount of 8 mg of illite was added to 80 ml
solution with a known initial Cs concentration. The solution was buffered at a pH of 8.0.
After shaking for 12 days, the dissolved Cs concentration was measured. The results are
shown in the table below

Experiment no. Initial Cs concentration Cs concentration after 12 days
µg l -1 µg l -1
1. 1 0.5
2. 2 1.1
3. 5 3.1
4. 10 7.0
5. 20 15.5
-1
a. Calculate the adsorbed concentration in μg g for each of the five experiments after
12 days.
b. Draw the isotherm of Cs.
c. Is this a Freundlich isotherm or a Langmuir isotherm? Explain your answer.

7. The precipitation reaction of barite (barium sulphate; BaSO ):
4
2-
2+
Ba + SO ↔ BaSO (s)
4 4
0
-1
is characterised by the following thermodynamic properties: ΔG = -56.91 kJ mol ;
-1
0
ΔH = -6.35 kJ mol .
a. Is the dissolution of barium sulphate endothermic or exothermic?
b. Calculate the solubility product for barium sulphate at 25 °C.
2+
c. Calculate the Ba concentration in a solution saturated with BaSO at 25 °C.
4
2+
d. Calculate the Ba concentration in a solution saturated with BaSO at 10 °C.
4
e. Explain what happens if a small amount of sulphuric acid (H SO ) is added to the
2 4
saturated solution.
2-
-1
2+
8. A water sample has a calcium (Ca ) activity of 0.001 mol l and a sulphate (SO )
4
-1
activity of 0.02 mol l . Evaluate the saturation index of gypsum (CaSO ∙2H O) given the
4
2
solubility product of gypsum = 10 -4.61 . What is your conclusion?
0
9. A solution in equilibrium with gypsum also contains a certain amount of the CaSO
4
0
complex. Discuss what the presence of aqueous CaSO complexes implies for the total
4
amount of calcium in solution.
10. Phosphoric acid (H PO ) dissociates in three steps at different pHs and with different K
3 4 a
values:
+
-
First dissociation step: H PO ↔ H PO + H K = 6.92∙10 -3
3 4 2 4 a
+
2-
-
Second dissociation step: H PO ↔ HPO + H K = 6.17∙10 -8
2 4 4 a
2-
+
3-
Third dissociation step: HPO ↔ PO + H K = 2.09∙10 -12
4 4 a
3-
2-
-
What is the dominant phosphate species (i.e. H PO , H PO , HPO , or PO ) at
3 4 2 4 4 4
pH = 8, pH = 6, and pH= 4?
+
-
11. Consider the reaction: NH + H O ↔ OH + NH K = 1.8∙10 -5
3 2 4 b
-1
Calculate the pH of a solution of 0.15 mol l of ammonium chloride (NH Cl)
4
(Hint: Note that pK + pK = pK = 14).
a b w
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