# 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.

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 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. Commented Jan 16, 2022 at 20:37

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.

• 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. Commented Oct 21, 2014 at 16:13
• @AbelFriedman I agree, either the book has the wrong answer or OP is misconveying the data in table 4-1 Commented Oct 22, 2014 at 16:26
• @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? Commented Oct 22, 2014 at 16:33
• @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. Commented Oct 22, 2014 at 16:55

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.