# How to calculate mass of calcium fluoride that will dissolve in sodium fluoride solution?

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?

• 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.
– Manu
May 24 '20 at 17:36
• @Manu: Why don't you put as an answer since OR has showed his/her work (although this is very much like a homework). . May 24 '20 at 18:50
• 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. May 24 '20 at 19:22

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

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

• 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. May 24 '20 at 20:24
• 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. $\pu{5\!\bar{0}\bar{0} cm^3}.$ May 24 '20 at 20:24

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 : $$\ce{m(CaF_2) = 7.3·10^{-9} mol· 78.1g/mol = 5.7·10^{-7}g }$$