Page 46 - Soil and water contamination, 2nd edition

Basic environmental chemistry 33

–0.062 t = ln(0.1) = -2.303
t = 37.1 s

So, it takes 37.1 s to decrease the N O concentration to 10 percent of its initial value.
2 5

2.7 DISSOLUTION–PRECIPITATION REACTIONS

Dissolution and precipitation reactions are an important control on the transition of
chemicals between the solid and dissolved phases . As noted earlier, some compounds dissolve
in water as molecules, whereas electrolyte minerals dissociate and dissolve as ions. Consider
the following dissociation reaction:

M A xM m yA a (2.43)
x y
where M = the cation with charge m+ and A = the anion with charge a-, and x and y refer
to the stoichiometric constants. In an aqueous solution in contact with an excess of an
undissolved, solid, mineral, the mineral will dissolve until the solution is saturated and no
more solid can dissolve. The concentration of the saturated solution at a given temperature
and pressure is termed the solubility of the substance. In such a saturated solution,
equilibrium is established, which means that the forward reaction in Equation (2.43) has the
same rate as the backward reaction. Since the activity of the solid M A is 1 by definition (see
x y
previous section), the equilibrium constant for this reaction is:
a
m
K [ M ] x [ A ] y (2.44)
s
where K = the equilibrium constant for dissolution –precipitation reactions, also referred to as
s
the solubility product . The solubility product increases with the solubility of a mineral. It is
important to stress that solubility and the solubility product refer to a reversible equilibrium
condition. However, dissolution reactions are often characterised by slow reaction rates,
so that ‘non-equilibrium’ conditions prevail. The extent and direction of departure from
saturation can be evaluated using the saturation index (SI), which is defined as:

(2.45)

where Q = the reaction quotient (see Equation 2.33) [-], and K is the solubility product
s
[-]. If SI is less than zero (Q/K < 1), the system is undersaturated and may move towards
s
saturation by dissolution . Conversely, if SI is greater than zero (Q/K > 1), the system is
s
supersaturated and should be moving towards saturation by precipitation.
Example 2.9 Saturation index
2+
-1
-1
A leachate from coal ash contains 0.03 mg l barium (Ba ) and 90 mg l sulphate
2-
(SO ). The ionic strength is 0.06. Calculate the saturation index SI of barite (BaSO )
4 4
given the log solubility product logK for barite = –9.97 at 25 °C.
s
Solution
-1
2-
2+
First, calculate the concentrations of Ba and SO in mol l :
4
2+
2+
The atomic weight of Ba = 137.33 (see Appendix I), so the Ba concentration is
-1
-1
-7
-1
0.03 mg l /1000 mg g /137.33 g mol = 4.37 · 10 mol l -1
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