# Solution Exercise 6

For the Ethiopian Rift Valley sample, the molal concentration product, that is, the product of the molality of Ca2+ times the molality of F squared is as follows (note: calcium has 40 atomic mass units and fluorite has 19):

$\displaystyle \left ( a_{Ca^{2+}}a_{F^{-}}^{2} \right )_{sample}=\left ( 14.1\frac{\mathrm{mg}}{\mathrm{L}}\frac{1\ \mathrm{L}}{1000\ \mathrm{mg}}\frac{1\ \mathrm{atom}}{40\ \mathrm{amu}} \right )$ × $\displaystyle \left ( 2.62\frac{\mathrm{mg}}{\mathrm{L}}\frac{1\ \mathrm{L}}{1000\ \mathrm{mg}}\frac{1\ \mathrm{atom}}{19\ \mathrm{amu}} \right )^{2}=6.7\times 10^{-12}$

Taking the logarithm results in log molal product = –1.17

SI = log molal product – log Ksp = –11.17 – (–10.6) = –0.57

This suggests undersaturated conditions. Using the PHREEQC code, SI = −0.66 which is slightly more undersaturated. The difference is caused by the use of activity coefficients in the code calculation.