Solutions Manual for Groundwater Geochemistry and Isotopes, 1e By Ian
2 Clark
Table of Contents
Chapter 1: Water, Rocks, Solutes and Isotopes ............................ 3
Chapter 2: Thermodynamics of Aqueous Systems..................... 11
Chapter 3: Geochemical Reactions ............................................. 20
Chapter 4: Isotope Reactions ...................................................... 26
Chapter 5: Tracing the Water Cycle ........................................... 34
Chapter 6: CO2 and Weathering ................................................. 43
Chapter 7: Geochemical Evolution ............................................. 67
Chapter 8: Groundwater Dating .................................................. 75
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Water, Rocks,
1 Solutes and Isotopes
1. What are the four major cations and three major anions in most natural waters? How do these compare
with their crustal abundances. What are the three most abundant elements in the earth’s crust after
oxygen and why are they not abundant in most natural waters?
Abundance (ppm)
Ion/Element Groundwater1 Crustal
Ca2+ 93 41500
Mg2+ 32 23300
+
Na 40 23600
+
K 4 20900
HCO3– 380 380 (as C)
SO42– 54 575 (as S)
–
Cl 60 280
1
from Table 1.7, as example.
The three most abundant elements in the crust are:
Si 27.7% - low solubility of silicate minerals
Al 8.13% - low solubility of aluminosilicates – feldspars and clay minerals
Fe 5.0% - low solubility as oxy-hydroxides – limonite, goethite, ferrihydrite
2. What are the molal concentrations of Ca2+ and SO42– resulting from the dissolution of 2 mg of gypsum
in 1 kg of water?
CaSO4·2H2O Ca2+ + SO42– + 2H2O
2 mg CaSO4·2H2O
gfwgypsum = 40.1 + 96.1 + 2×18 = 172.2
2 mg / 172.2 = 0.0116 mmol CaSO4·H2O
= 1.16E–5 molal Ca2+
= 1.16E–5 molal SO42–
mCa2+ = 0.0116 mmol/kg or 1.16·10–5 mol/kg. This gives 0.47 ppm Ca2+.
mSO42– = 0.0116 = 0.0116 mmol/kg or 1.16·10 –5 mol/kg. This gives 1.12 ppm SO42–
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3. For each of the water samples in Table 1.7, determine the concentration for Cl– in the following units:
mg/L, ppm, M and m. By what percentage for each water type does Cl– in mg/L differ from the value
in ppm?
Recall that mg/L = ppm (density – TDS)
Rain River Groundwater Seawater Brine
chloride — Cl– ppm 0.31 2.4 58.6 19,350 185,000
Density (kg/L) 1.00 1.00 1.00 1.026 1.25
TDS measured (mg/L) 2.85 76.4 525 34,700 280,000
Cl - mg/L 0.31 2.4 58.7 19,182 179,450
Cl – m 8.73E-06 6.76E-05 0.00165 0.545 5.21
Cl – M 8.73E-06 6.76E-05 0.00165 0.540 5.05
Percent error 0.00% 0.00% 0.05% 0.87% 3.0%
4. What is the weight of the water in 1 L of the brine in Table 1.7?
TDS = 280,000 mg/L
= 0.28 kg/L
Density = 1.25 kg/L
Weight of water = 1.25 [kg/L] – 0.28 [kg/L solutes]
= 0.97 kg/L
5. Calculate the TDS in mg/L for the rain, river water, groundwater, seawater and brine in Table 1.7 and
compare with the measured TDS values. A correction is required for the loss of half of the HCO3–as
CO2. Note that for the high salinity waters, a correction using density is required to convert the TDS
calculated in ppm to mg/L for comparison with the measured TDS values.
The TDS is calculated by adding the concentrations for all major ions. Note that the high
concentrations of minor ions such as Sr2+ and Br– make them contributors to TDS. Approximately
half of the bicarbonate will be lost as CO2 during the measurement of TDS by evaporation
according to:
Ca2+ + 2HCO3– CaCO3 + CO2 + H2O
A correction for the loss of up to half of the bicarbonate as CO2 can be made by adding in only
50% the mass of HCO3–.
For seawater and the brine, a conversion of the TDS calculated from concentrations in ppm to
TDS in mg/L is made using density, and the measured TDS in kg/L (divided by 106):
mg/L = ppm ∙ (density [kg/L] – TDS [kg/L])
Rain River Groundwater Seawater Brine
TDS, measured (mg/L) 2.85 76.4 525 34,700 280,000
TDS, calculated (ppm) 2.94 104 668 35,100 291,600
TDS, calculated, CO2 corrected (ppm) 2.80 77.0 479 35,100 291,600
TDS, calculated, CO2 corrected (mg/L) 2.79 77.0 479 34,800 282,800
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Note the importance of CO2 correction for the river and groundwater samples where HCO 3– is the
major ion, and the importance of density for the seawater and brine to convert to mg/L for comparison
with measured TDS values.
6. Using the density relationships in Figure 1.4, plot a diagram with ppm on the x-axis vs mg/L on the y-axis
for Cl– concentration in solutions of NaCl, CaCl2 and MgCl2 up to solutions with 400 g/L TDS. Which
chloride salt has the greatest effect on density of the solution. Which has the greatest difference between
mg/L and ppm?
The TDS values are defined up to 400 g/L or 400,000 mg/L, at say 50 g/L intervals. For each salt, the
mg/L concentration of Cl– are determined by dividing by the gfw of the salt (23+35.5=58.6 for NaCl)
then multiplying by the number of Cl– in the salt’s structure (1 or 2) to give mmol/L Cl–, then
multiplying by the gfw for chloride (35.5) to give mg/L Cl– in the solution. The densities of the three
solutions for a given TDS are calculated from the exponential equations given in Figure 1.4. The
density and TDS for each solution are used to calculate ppm Cl – at the differing defined values for
TDS, and plotted on a scatter plot.
MgCl2 has the greatest effect on density, with a density of 1.35 kg/L at 300,000 mg/L TDS. However,
it is the NaCl salt that has the greatest difference between mg/L and ppm, with a 6% greater value for
Cl expressed as ppm than as mg/L. The MgCl2 solution has a higher density which exceeds the excess
weight from the TDS, and so the ppm value is 5% lower than the mg/L value at 300,000 mg/L TDS.
Note that the NaCl solution reaches saturation with halite at about 350 g/L TDS and so cannot attain
400 g/L TDS.
7. Colloids are amorphous clusters of ions in waters with elevated salinity. Could colloids contribute to the
measured solute concentrations if the water sample has been filtered with the standard 0.45 m pore-throat
filter paper?
Yes, the size range for colloids is from 0.001-1 μm. Therefore, some colloids will pass through the filter
and contribute to salinity.
8. Runoff water from a sulfur extraction plant has the following geochemical composition:
pH = 2.3; Ca2+ = 25 mg/L; SO42– = 300 mg/L
