Page 93 - Hydrogeology Principles and Practice

HYDC03 12/5/05 5:36 PM Page 76

76 Chapter Three

2+ 2+ −
3.3 Chemical composition of groundwater (Ca ), magnesium (Mg ), chloride (Cl ), bicarbon-
−
2−
ate (HCO ) and sulphate (SO ). These cations and
3 4
The chemical and biochemical interactions between anions normally comprise over 90% of the total dis-
groundwater and the geological materials of soils and solved solids content, regardless of whether the water
rocks provide a wide variety of dissolved inorganic and is dilute rainwater or has a salinity greater than sea-
organic constituents. Other important considerations water (typical analyses of rainwater and seawater are
include the varying composition of rainfall and atmo- given in Appendix 5). Minor ions include potassium
+ 2+ 2+
spheric dry deposition over groundwater recharge (K ),dissolved iron (Fe ), strontium (Sr ) and fluoride
−
areas, the modification of atmospheric inputs by (F ) while aqueous solutions commonly also contain
evapotranspiration, differential uptake by biological amounts of trace elements and metal species. The
processes in the soil zone and mixing with seawater in introduction of contaminants into groundwater from
coastal areas. As shown in Table 3.2, and in common human activities can result in some normally minor
with freshwaters in the terrestrial aquatic environ- ions reaching concentrations equivalent to major ions.
ment, the principal dissolved components of ground- An example is nitrate, as excessive application of nitro-
+
water are the six major ions sodium (Na ), calcium genous fertilizers can raise nitrate concentrations in

BO X
Concentration units used in hydrochemistry
3.1

Concentration is a measure of the relative amount of solute (the Chemical equivalence: the concept of chemical equivalence takes
dissolved inorganic or organic constituent) to the solvent (water). A into account ionic charge and is useful when investigating the pro-
list of atomic weights is supplied in Appendix 4. portions in which substances react. This aspect of chemistry is called
There are various types of concentration unit as follows: stoichiometry.
−1
Molarity (M): number of moles of solute dissolved in 1 L of solution Equivalents per litre (eq L ) = number of moles of solute multiplied
−1
(mol L ). For example, if we have 10 grammes of potassium nitrate by the valence of the solute in 1 L of solution. From this it follows
−1
−1
(molar mass of KNO = 101 grammes per mole) then this is that meq L = mg L × (valence/atomic weight).
3
−1
(10 g)/(101 g mol ) = 0.10 moles of KNO . If we place this in a As an example of the application of chemical equivalence, take
3
flask and add water until the total volume = 1 L we would then have the effect of ion exchange when a fresh groundwater in contact
a 0.1 molar solution. Molarity is usually denoted by a capital M, with a rock is able to exchange a chemically equivalent amount of
for example a 0.10 M solution. It is important to recognize that calcium (a divalent cation with atomic weight = 40 g) with sodium
molarity is moles of solute per litre of solution, not per litre of (a monovalent cation with atomic weight = 23 g) contained within
solvent, and that molarity changes slightly with temperature the aquifer. If the groundwater has an initial calcium concentration
−1
−1
because the volume of a solution changes with temperature. of 125 mg L and sodium concentration = 15 mg L , what will be
Molality (m): number of moles of solute dissolved in 1 kg of solvent the new groundwater sodium concentration if all the calcium were
−1
(mol kg ). Notice that, compared with molarity, molality uses mass exchanged with the clay material?
rather than volume and uses solvent instead of solution. Unlike
−1
molarity, molality is independent of temperature because mass Initial calcium concentration = 125 mg L × (2/40)
does not change with temperature. If we were to place 10 g of = 6.25 meq L −1
KNO (0.10 moles) in a flask and then add one kilogramme of water −1
3
we would have a 0.50 molal solution. Molality is usually denoted Initial sodium concentration = 12 mg L × (1/23)
−1
with a small m, for example a 0.10 m solution. = 0.52 meq L
Mass concentration: this unit of concentration is often used to New sodium concentration after ion exchange = 6.25 + 0.52
express the concentration of very dilute solutions in units of parts = 6.77 meq L −1
−1
per million (ppm) or, more commonly, mg L . Since the amount of
solute relative to the amount of solvent is typically very small, the = 6.77 × (23/1) = 155.71 mg L −1
density of the solution is approximately the same as the density of
the solvent. For this reason, parts per million may be expressed in Therefore, the extra sodium contributed to the groundwater by ion
the following two ways: exchange is 141 mg L .
−1
~
ppm = mg of solute/L of solution
ppm = mg of solute/kg solution