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.