Table of Contents - Chapter 2: Thermodynamics of Aqueous Systems.....

2 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 Solutions Manual for Groundwater Geochemistry and Isotopes, 1e By Ian Clark 1 / 4

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Water, Rocks, Solutes and Isotopes

• 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 Groundwater 1 Crustal Ca 2+

93 41500

Mg 2+

32 23300

Na +

40 23600

K +

4 20900

HCO3 – 380 380 (as C) SO4 2– 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

• What are the molal concentrations of Ca

2+ and SO4 2– resulting from the dissolution of 2 mg of gypsum in 1 kg of water?

CaSO4·2H2O  Ca 2+

• SO4

2–

+ 2H2O

• mg CaSO4·2H2O

gfwgypsum = 40.1 + 96.1 + 2×18 = 172.2

• mg / 172.2 = 0.0116 mmol CaSO4·H2O

= 1.16E–5 molal Ca 2+

= 1.16E–5 molal SO4 2–

mCa 2+ = 0.0116 mmol/kg or 1.16·10 –5 mol/kg. This gives 0.47 ppm Ca 2+ .mSO4 2– = 0.0116 = 0.0116 mmol/kg or 1.16·10 –5 mol/kg. This gives 1.12 ppm SO4 2–

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• 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%

• 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

• 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 Sr 2+ 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:

Ca 2+

+ 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 10 6

):

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 HCO3 – 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.

• 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.

• 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.

• Runoff water from a sulfur extraction plant has the following geochemical composition:

pH = 2.3; Ca 2+ = 25 mg/L; SO 4 2– = 300 mg/L

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