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Chemical hydrogeology 101

BO X
Solubility product and saturation index
3.6
The dynamic equilibrium between a mineral and its saturated solu- of X is 0 at equilibrium with positive values indicating supersatura-
tion when no further dissolution occurs is quantified by the thermo- tion and negative values undersaturation.
dynamic equilibrium constant. For example, the reaction: By calculating saturation indices, it is possible to determine from
hydrochemical data the equilibrium condition of groundwater with
CaCO j Ca 2+ + CO 2− eq. 1 respect to a given mineral. For example, a groundwater from the
3(calcite) (aq) 3(aq)
unconfined Chalk in Croydon, South London, gave the following
is quantified by the equilibrium constant, K, found from: results: temperature = 12°C, pH = 7.06, Ca 2+ concentration =
−
−1
−1
−1
121.6 mg L (10 −2.52 mol L ) and HCO = 217 mg L (10 −2.45 mol
3
−1
[Ca 2 + ][CO 2 − ] L ). By making the assumption that the concentrations are equal
K calcite = 3 eq. 2 to activities for this dilute groundwater sample, the first step in
[CaCO ] 2−
3 calculating a calcite saturation index is to find the CO 3 concen-
−
tration. The dissociation of HCO can be expressed as:
Since CaCO is a solid crystal of calcite, its activity is effectively con- 3
3
stant and by convention is assigned a value of one or unity. HCO j H + CO 2− eq. 5
+
−
The equilibrium constant for a reaction between a solid and its 3 3
saturated solution is known as the solubility product, K . Solubility
sp for which the approximate equilibrium constant at 10°C (Table 3.6) is:
products have been calculated for many minerals, usually using
pure water under standard conditions of 25°C and 1 atmosphere [][ 2−
+
HCO ]
.
pressure and are tabulated in many textbooks, for example data for K − = 3 = 10 − 10 49 eq. 6
HCO 3 HCO −
major components in groundwater are given by Appelo and Postma 3
(1994). The solubility product for calcite (eq. 2) is 10 −8.48 and equa- Rearranging equation 6 for the unknown CO 2− concentration and
tion 2 now becomes: + 3 −
substituting the measured values for H and HCO gives:
3
[Ca 2 + ][CO 2 − ] + − −
2
K = 3 = [Ca 2 ][CO 2 = ] 10 −8 .48 mol L −2 eq. 3 K − [HCO 3 ] 10 − 10 .49 ⋅ 10 − . 2 45
=
sp 3 2− HCO 3 − . 588 eq. 7
=
1 [CO ] = 10
+
3
[] 10 − . 706
H
−1
The solubility product can be used to calculate the solubility (mol L ) Now, using the result of equation 7, the calcite saturation index is
of a mineral in pure water. The case for calcite is straightforward since found from:
each mole of CaCO that dissolves produces one mole of Ca 2+ and
3
2−
2−
2+
one mole of CO . Thus, the calcite solubility = [Ca ] = [CO ] and [Ca 2 + ][CO 2 − ]
3
3
therefore calcite solubility = (10 −8.48 1/2 = 10 −4.24 = 5.75 × 10 −5 X calcite = log 10 K 3 eq. 8
)
−1
mol L . sp
The state of saturation of a mineral in aqueous solution can be If K for calcite = 10 −8.41 at 10°C, then:
expressed using a saturation index, where: sp
5
[10 − .252 ][10 − .88 ] = log 10 . 001 = . 0 01 eq. 9
= log
X
=
X IAP eq. 4 calcite 10 10 − .841 10
K
sp
Hence, the Chalk groundwater is marginally supersaturated and,
in which IAP is the ion activity product of the ions in solution given the assumptions used in the calculation of X , can be
calcite
obtained from analysis. A X value of 1 indicates that mineral satura- regarded as at equilibrium. A more accurate calculation using the
tion (equilibrium) has been reached. Values greater than 1 repres- chemical program wateq (Truesdell & Jones 1973) that accounts for
ent oversaturation or supersaturation and the mineral is likely to be the chemical activity and speciation of the sample as well as the
precipitated from solution. Values less than 1 indicate undersatura- actual sample temperature of 12.0°C gives a calcite saturation
tion and further mineral dissolution can occur. An alternative to index of −0.12, again indicating equilibrium with respect to calcite
equation 4 is to define X = log (IAP/K ) in which case the value for practical purposes (Mühlherr et al. 1998).
10 sp
they have a small grain size and therefore have a large adsorb. The adsorption may be weak, essentially a
surface area on which sorption reactions can occur. physical process caused by Van der Waals’ force, or
In addition, clays tend to be strong adsorbers since strong, if chemical bonding occurs. Divalent cations
they have an excess of negative charge at the surface are usually more strongly adsorbed than monovalent
due to crystal lattice defects on to which cations can ions as a result of their greater charge density, a