The relation among fluoride and calcium concentrations in water

Fluoride (F^-) replaces hydroxyl (OH^-) in the mineral apatite ( Ca₅(OH,F)(PO₄)₃ ) which makes the mineral harder and brittler. Apatite is part of the bones and teeth of humans. Ingestion of too much F is the cause of fluorosis, an embrittling of the cartilage that leads to painful joints and easily breaking bones in older people. The maximal admissible concentration (MAC) for F in drinking water is 1.5 mg/L (depending on the consumption rate since the body accumulates the element). Drinking water may contribute 10 - 50% of the total nutrition of this element in humans, so it is important to be able to estimate where the concentrations in groundwater may be high.

If the mineral fluorite dissolves to equilibrium, CaF₂ ↔ Ca²⁺ + 2 F^- , the law of mass action yields:
        K = 10^-10.6 = [Ca²⁺] [F^-] ²
or
        [F^-] = (10^-10.6 / [Ca²⁺])^½.

This hyperbolic relation predicts that high fluoride waters will have low calcium concentrations and vice versa.
The Ca concentration in water is often regulated by calcite equilibrium. In high-alkalinity, high-pH waters the Ca concentration is low.
The pH and alkalinity may increase by dissolution of alumino-silicates (weathering), or by evaporation and escape of CO₂ gas.
These two processes, in combination with calcite and fluorite equilibria, probably explain the high F concentrations found in many African Rift Valley waters.

The graph on the right is calculated and plotted by the PHREEQC input file ca_f.phr. The file starts with calculating soil water (SOLUTION 1), with a CO2 pressure of 0.1 atm, and Ca and F concentrations in equilibrium with calcite and fluorite, respectively. This is the encircled composition in the graph, with (right-click on the chart calculated by PHREEQC, and select 'Show Point Values') 157 mg Ca/L and 2.3 mg F/L. Next, 7.5 mmol Na-feldspar dissolve in 15 steps (keyword REACTION), while calcite, fluorite, kaolinite and quartz react to equilibrium (keyword EQUILIBRIUM_PHASES). The graph shows the increase of F (red line) and pH (green line) as a result of the reactions.

The proton loss (pH increase), as a result of the dissolution of Na-feldspar:
        H⁺ + NaAlSi₃O₈ + 7 H₂O → Na⁺ + Al(OH)₃⁰ + 3 H₄SiO₄⁰
triggers the dissociation of HCO₃^-:
        HCO₃^- → CO₃^-2 + H⁺
causing calcite to precipitate:
        Ca⁺² + CO₃^-2 → CaCO₃
and fluorite to dissolve:
        CaF₂ → Ca⁺² + 2 F^-

Al and Si, released by the dissolution of feldspar, precipitate in kaolinite:
        2 Al(OH)₃⁰ + 2 H₄SiO₄⁰ → Al₂Si₂O₅(OH)₄ + 5 H₂O
and in quartz:
        H₄SiO₄⁰ → SiO₂ + 2 H₂O

Another process that can lower the Ca concentration in water is ion exchange of Ca²⁺ with 2 Na⁺.

Some Questions:

• What happens if you cross out Calcite in EQUILIBRIUM_PHASES in ca_f.phr? Can you explain? (Look at the speciation of the solution in the output; Ca-complexation increases with pH)
• The most important solute species of Al in the first step of the reaction is not Al(OH)₃⁰, but ...
• What happens if you cross out Kaolinite in EQUILIBRIUM_PHASES? Explain...

• Find the CO₂ pressure in the last step in the output file. Why is it lower than 0.1 atm?
• Can you model the effect of CO₂ degassing on the Ca and F concentrations, for example in a lake, thus, without feldspar dissolution?

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