Solution Exercise 4

Dissolution reaction for fluorapatite:

Ca5(PO4)3F = 5Ca2+ + 3PO43− + F

the ion-activity product is:

[latex]\displaystyle a_{Ca^{2+}}^{5}a_{PO_{4}^{3-}}^{3}a_{F^{-}}=\left ( 3.895 \times 10^{-5} \right )^{5}\left ( 3.065\times 10^{-10} \right )^{3}\left ( 2.009\times 10^{-4} \right )=10^{-54.28}[/latex]

log Ksp = −55.1

[latex]\displaystyle SI=\mathrm{log}\frac{10^{-54.28}}{10^{-55.1}}=0.82[/latex]

The solution is oversaturated with respect to fluorapatite.

Return to Exercise 4

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