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The solubility product of calcium fluoride, $\ce{CaF2},$ is $\pu{1.46E-10 mol^3 dm^{-9}.}$

What mass of calcium fluoride will dissolve in $\pu{500 cm^3}$ of $\pu{0.10 mol dm^{-3}}$ sodium fluoride solution? (Molar mass of $\ce{CaF2}$ is $\pu{78.1 g mol^{-1}}.)$

I tried using the following method:

$$ \begin{align} [\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\ x(2x)^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\ x &= \pu{3.3E-4 mol dm^{-3}} \end{align} $$

Solubility $s$ of $\ce{CaF2}$ can be found as such:

$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{3.3E-4 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{2.6E-2 g dm^{-3}}.$$

Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be

$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{2.6E-2 g dm^{-3}}\times\pu{500 cm^3} =\pu{1.3E-2 g}.$$

However, my answer differs from the given answer, that is $\pu{5.7E-7 g}.$

How should I approach this question instead?

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    $\begingroup$ While writing the equation for $K_{SP}$ in place of [F-] concentration you have to take 2s+(0.1) instead of only 2s due to common ion effect. So equation becomes $s(2s+0.1)^2=K_{SP}$. As s is very less than 0.1 you can neglect s and equation becomes $s(0.1)^2=K_{sp}$. Then you can calculate s which is the concentration of dissolved $CaF_2$ in solution and then use the volume given to calculate it's mole and then its mass. $\endgroup$
    – Manu
    May 24 '20 at 17:36
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    $\begingroup$ @Manu: Why don't you put as an answer since OR has showed his/her work (although this is very much like a homework). . $\endgroup$
    – Mathew Mahindaratne
    May 24 '20 at 18:50
  • $\begingroup$ I personally like Manu's argument. By common ion effect the total fluoride ion concentration will be 0.1+2s where s is the solubility of the sparingly soluble calcium fluoride. And normally since common ion effect reduces solubility of such salts drastically, 0.1 will be much greater than s and a useful approximation is that 0.1+2s = 0.1. Then the remaining unknown term will be the s standing for calcium ion concentration and a useful answer can be got. $\endgroup$
    – Amayubo Ben
    May 24 '20 at 19:22
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In the $\pu{500 cm^3}$ solution there are is already $\ce{F^-}$ dissolved coming from the $\ce{NaF}$ $\pu{0.10 mol dm^{-3}}$ that was already there. This means that the solubility of $\ce{CaF2}$ is reduced due to the common ion effect.

NB $\ce{NaF}$ can be considered completely dissociated because it has a very high $K_\mathrm{sp};$ therefore the initial concentration of $\ce{F^-}$ is $\pu{0.10 mol dm^{-3}}.$

$$ \begin{align} [\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\ x(2x+\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\ \end{align} $$

The issue here is that to find $x$ you need to solve a cubic equation…

However, $K_\mathrm{sp}$ of $\ce{CaF2}$ is very small and $x$, that is its solubility, must be much less than $\pu{0.10 mol dm^{-3}}!$ This allows you to make the approximation that $2x + \pu{0.10 mol dm^{-3}} ≈ \pu{0.10 mol dm^{-3}}.$

$$ \begin{align} [\ce{Ca^{2+}}][\ce{F^-}]^2 &= K_\mathrm{sp}(\ce{CaF2})\\ x(\color{red}{\pu{0.10 mol dm^{-3}}})^2 &= \pu{1.46E-10 mol^3 dm^{-9}} \\ x &= \pu{1.46E-8 mol dm^{-3}} \end{align} $$

Solubility $s$ of $\ce{CaF2}$:

$$s(\ce{CaF2}) = xM(\ce{CaF2}) = \pu{1.46E-8 mol dm^{-3}}\times\pu{78.1 g mol-1} = \pu{1.14E-6 g dm^{-3}}.$$

Finally, the mass of $\ce{CaF2}$ dissolved in $\pu{500 cm^3}$ would be

$$m (\ce{CaF2}) = s(\ce{CaF2})\times V = \pu{1.14E-6 g dm^{-3}}\times\pu{500 cm^3} =\pu{5.7E-7 g}.$$

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  • $\begingroup$ You did a stellar job at explaining the answer, just a couple of things (that I already corrected). First, when you copy-paste markup from the question, make sure to stick with the same notations, and, whenever possible, units: don't use two different schemes or randomly omit all formatting at all. $\endgroup$
    – andselisk ♦
    May 24 '20 at 20:24
  • $\begingroup$ Second, you are not allowed to replace $\pu{500 cm^3}$ with $\pu{0.500 dm^3}.$ You don't know for sure whether zeros in $\pu{500 cm^3}$ are significant figures or not, so at best you can only guarantee $\pu{0.5 dm^3}.$ Note that normally there is a bar on top of a zero if it is significant, e.g. $\pu{5\!\bar{0}\bar{0} cm^3}.$ $\endgroup$
    – andselisk ♦
    May 24 '20 at 20:24
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Gc3941d has mixed the solubilities of $\ce{NaF}$ and $\ce{CaF_2}$. Let's start the calculation from the beginning.

The concentration of fluoride ion is : [$\ce{F^-}$] = $0.1$ M. The concentration of Calcium may be calculated from the solubility product $\ce{K_{sp}}$ and [$\ce{F^-}$] according to : :$$\ce{[Ca^{2+}] = K_{sp} /[F^-]^2} = 1.46·10^{-10}/(0.1)^2 = 1.46·10^{-8} M$$ It means that in $0.5$ liter, $ 0.5·1.46·10^{-8}$ mol $\ce{Ca}$ = $7.3·10^{-9}$ mole $\ce{Ca}$ is dissolved. And of course, the same number of mole of $\ce{CaF_2}$ is dissolved .
As the molar mass of $\ce{CaF_2}$ is $\ce{78.1 g|mol}$, the mass of $\ce{CaF_2}$ dissolved in this solution is : $$\ce{m(CaF_2) = 7.3·10^{-9} mol· 78.1g/mol = 5.7·10^{-7}g }$$

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