Solution Exercise 5
If SIcalcite = 0.36 = log IAP − log Ksp and log IAP = SI + log Ksp
then, [latex]\displaystyle 0.36-8.48=\mathrm{log}\left ( a_{Ca^{2+}}\ a_{CO_{3}^{2-}} \right )=-8.12[/latex]
and [latex]\displaystyle \mathrm{log}\left ( a_{Ca^{2+}} \right )=-8.12-\mathrm{log}\left ( a_{CO_{3}^{2-}} \right )=-8.12+4.61=-3.51[/latex]
hence, [latex]\displaystyle \mathrm{log}\left ( a_{Ca^{2+}} \right )\left ( a_{F^{-}} \right )^{2}=-3.51+2(-3.57)=-10.65[/latex]
and SI = −0.05 thus fluorite is essentially at solubility equilibrium