# Caclulating the Solubility of Fluorite in River Water

Source: Chapter 4 of Brunlow's 1979 Geochemistry text.

Calculate the solubility of fluorite in river water with the average composition given in table 4-1. Assume that no fluoride complexes form.

Table 4-1 lists ppm, mEq/L and moles/L values for constituents, anions and cations found in the water sample. There is no value for $\ce{F}^-$ but there is for $\ce{Ca}^{2+}$. The molarity of $\ce{Ca^{2+}}$ is given to be $\pu{0.375E-3M}$

My attempt::

$$\ce{CaF2 -> Ca^{2+} + 2Fl^{-}}$$

$K_\mathrm {sp}$ of $\ce{CaF2}$= $10^{-10.5}$ (from a table in the book)

$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{F_{\text{added}}}]^2$

$[\ce{F_{\text{added}}}] = [2\ce{Ca_{\text{added}}}]$

$K_\mathrm{sp} = [\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][2\ce{Ca_{\text{added}}}]^2 = 4[\ce{Ca_{\text{original}}} + \ce{Ca_{\text{added}}}][\ce{Ca_{\text{added}}}]^2$

Am I on the right track? I end up with a polynomial to solve for. The answer in the back of the book is $10^{-4.64}$ which I can't arrive at.

The other method I've tried is to calculate the ppm of fluorite dissolved in pure water ($\pu{15.6ppm}$) and then subtract the ppm value for the river water $\ce{Ca^{2+}}$ given in the table ($\pu{15.0ppm}$) and then convert the result ($\pu{0.6ppm}$) to $\pu{7.69e-6 moles/L}$ which doesn't equal the answer in the back of the book ($10^{-4.64}$), so I believe that I'm missing something conceptually.

$K_\mathrm{sp}=[\ce{Ca_{original}} + \ce{Ca_{added}}][2\ce{Ca_{added}}]^2 = 4[\ce{Ca_{original}} + \ce{Ca_{added}}][\ce{Ca_{added}}]^2$
Yes, this is correct and can be solved, you say you know $[\ce{Ca_{original}}]$ is $\pu{0.375E-3M}$ and $K_\mathrm{sp}$ is $10^{-10.5}$ so there is one equation and one unknown, just a matter of solving a cubic equation, which can be solved exactly or by approximate methods.