Caclulating the Solubility of Fluorite in River Water

**Source:** Chapter 4 of Brownlow'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.
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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.





> 
> the ppm of fluorite in pure water (15.6ppm)
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the ppm of fluorite in pure water is zero!



> 
> $K\_\mathrm{sp}=[\ce{Ca\_{original}} + \ce{Ca\_{added}}][2\ce{Ca\_{added}}]^2 = 4[\ce{Ca\_{original}} + \ce{Ca\_{added}}][\ce{Ca\_{added}}]^2$
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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](http://en.wikipedia.org/wiki/Cubic_function), which can be solved exactly or by approximate methods. 


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I would start from DavePhD's equation :


$\ce{𝐾\_{sp} = [Ca\_{original} + Ca\_{added}][2Ca\_{added}]^2 = 4[Ca\_{original} + Ca\_{added}][Ca\_{added}]^2}$ = $\pu{10^{-10.5}}$


But at a difference from him, I would admit that : $\ce{[Ca\_{original} + Ca\_{added}] = 0.375×10^{−3}}$ M.


As a consequence, [$\ce{Ca\_{added}}$] = $\frac{10^{-10.5}}{4~ · ~0.375~·~10^{-3}}$= $\pu{2.43 10^{-8}}$M


This is the solubility of $\ce{CaF2}$ in the river water containing some calcium ions. It is much smaller than in pure water.


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