Geology Reference
In-Depth Information
are twice as many as calcium) captures one extra elec-
tron to become a singly charged negative ion or
anion
.
The dissolution of solid fluorite can therefore be writ-
ten as a chemical reaction leading to an equilibrium:
Given the small size of this number, it is often more
convenient to write
K
CaF
2
in logarithmic form. Since
log (4.26 × 10
−11
) = −10.37:
log
K
CaF
=−
10 4
.
(4.15a)
10
2
+
2
−
CaF a F
2
+
2
(4.12)
10
10 4
−
.
whichisequivalenttowriting
K
=
(4.15b)
solid
solution
CaF
2
(Because that is a
heterogeneous equilibrium
like
those considered in Chapter 2, one needs to specify in
which phase each reactant or product resides when
writing the reaction.) As dissolution proceeds, the con-
centration of Ca
2+
and F
−
ions in solution builds up, and
increasingly the ions will react with each other to pro-
duce solid CaF
2
again (the reverse reaction). Saturation
of the solution is an equilibrium in which the gross rate
of dissolution of solid CaF
2
is equalled by the rate of
precipitation from solution, so that no further
net
change is observed. One can formulate an equilibrium
constant for reaction 4.12 (cf. Equation 4.11):
or, by analogy with
pH
notation (Appendix B),
p
K
=
.
10 4
(4.15c)
CaF
2
Solubility products are available in published tables
(see references at the end of this chapter). Like all equi-
librium constants, they vary with temperature (Box 4.2).
If, in a solution of CaF
2
, the observed ion activity
product
⋅
(
)
2
a
a
2
+
−
Ca
F
has a value (e.g. 10
−12
) that is numerically smaller than
the solubility product
K
CaF
2
for the appropriate tempera-
ture, the solution is not saturated with CaF
2
. Any solid
CaF
2
introduced will be unstable, and will tend to dis-
solve. But if circumstances produce an ion activity prod-
uct greater than 10
−10.4
(say 10
−8
) at 25 °C, the solution has
become
supersaturated
with CaF
2
, and it will precip-
itate solid CaF
2
until the activity product has fallen to the
equilibrium value indicated by the solubility product.
Table 4.1 shows the solubility products for a few
important minerals. More data relevant to geochemis-
try are given by Krauskopf and Bird (1995).
⋅
(
)
2
a
a
−
2
+
F
K
=
Ca
(4.13)
CaF
X
2
CaF
2
Because the solid phase is pure CaF
2
, the
mole fraction
X
CaF
2
=
.
100
. Thus:
⋅
(
)
2
Ka
=
a
(4.14)
CaF
2
+
−
Ca
F
2
where
a
Ca
2
+
and
a
F
−
are the activities of Ca
2+
and F
−
ions
in a solution saturated with CaF
2
.
This kind of equilibrium constant provides an alterna-
tive way of expressing the solubility of a slightly soluble
salt (in this case CaF
2
) in water. It is called the
solubility
product
of CaF
2
. It conveys the same information as the
solubility, but in a different - and more versatile - form.
The value of
K
CaF
2
can be calculated from the solu-
bility data given above. Reaction 4.12 tells us that one
mole of solid CaF
2
dissolves (in sufficient water) to
yield one mole of Ca
2+
ions and two moles of F
−
ions.
Thus if 0.00022 moles of CaF
2
can saturate 1 kg of water
at 25 °C, the activities of Ca
2+
and F
−
in the saturated
solution will be 0.00022 and 0.00044 respectively.
Therefore:
Interaction between ionic solutes:
the common-ion effect
We have been concerned up to now with solutions con-
taining only a single salt (CaF
2
). Natural waters, how-
ever, are mixed solutions containing many salts, and in
such circumstances the question of solubility ceases to
be a simple matter, because the solubility of CaF
2
, for
example, is now affected by contributions of Ca
2+
and
F
−
from other salts present which contain these ions
(such as CaSO
4
and NaF respectively).
Consider the salt barium sulfate, BaSO
4
, which forms
the mineral barite (another vein mineral). On dissolv-
ing in water, BaSO
4
ionizes as follows:
⋅
(
)
2
Ka
=
a
CaF
2
+
−
Ca
F
2
2
= ×
=×
0 00022
.
( .
0 00044
)
+
2
2
−
BaSO
Ba
+
SO
(4.16)
4
4
−
11
42610
.
crystal
solution
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