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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.

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.

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    $\begingroup$ The author is Arthur H. Brownlow, not Brunlow, and it could be informative to copy in the page with all the data, there is maybe something missing in the equations. $\endgroup$
    – Andrew Kovács
    Commented Jan 16, 2022 at 20:37

2 Answers 2

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the ppm of fluorite in pure water (15.6ppm)

the ppm of fluorite in pure water is zero!

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

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.

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    $\begingroup$ The third-order polynomial has one real root, which has the correct order of magnitude to be a credible solution, and the OP's approach is correct. But just like him, I find it impossible to arrive at the same number as his book. $\endgroup$
    – Abel Friedman
    Commented Oct 21, 2014 at 16:13
  • $\begingroup$ @AbelFriedman I agree, either the book has the wrong answer or OP is misconveying the data in table 4-1 $\endgroup$
    – DavePhD
    Commented Oct 22, 2014 at 16:26
  • $\begingroup$ @DavePhD - Thank you for your answer. Regarding the first quote, what I was trying to indicate was the ppm of fluorite dissolved in pure water (with no common ions). Would it be valid to use the approach I suggest where I subtract ppm values? $\endgroup$
    – equant
    Commented Oct 22, 2014 at 16:33
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    $\begingroup$ @equant oh, I understand what you mean now, but no that would not be valid. If you take your proposed answer from that method, [Ca2+] and [F-] concentrations, they will not yield the solubility product. Your first method which gives the cubic equation is correct. $\endgroup$
    – DavePhD
    Commented Oct 22, 2014 at 16:55
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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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