The question was:

If you are given a $\pu{1 L}$ solution of $\ce{CaF_2}$ where there is $\pu{2.0E5 M}$ $\ce{Ca^{2+}}$. How many milliliters of $\pu{0.1 M}$ $\ce{NaF}$ would you need to precipitate $\ce{CaF_2}$? The solubility product of $\ce{CaF2}$ is $1.7\times10^{-10}$ .

The answer given is

$\pu{956 ml}$

What I attempted: I thought that if $x$ Litres of $\ce{NaF}$ is necessary to precipitate $\ce{CaF_2}$; then after the mix, the ionic product($K_{ip}$) of $\ce{CaF_2}$ will exceed the solubity product($K_{sp}$). $$K_\mathrm {ip} > K_\mathrm {sp}$$ $$\ce{[Ca^2+][F-]^2} > K_\mathrm{sp}$$ and from this equation I tried to solve the value of $x$.

Here's what I tried in details:-

step 1) Write down the dissociation reaction of $\ce{CaF_2}$

$$\ce{CaF_2 <=> Ca^{2+} + 2F-}$$

and from the solubility product formula $K_{sp} = \ce{[Ca^2+][F-]^2}$ I calculated the concentration of $\ce{[F-]}$ which I got was $2.91\times10^{-3}$.

step 2) Then I again calculated the concentration of $\ce{Ca^2+}$ and $\ce{F-}$ after adding $x$ Litres, using the formula $\ce{S_1\times V_1 = S_2\times V_2}$, because the volume now was not $1$ Litre, but $(1+x)$ Litres.

I got $$\ce{[Ca^2+] = \frac{2\times 10^{-5}}{(1+x)}}M$$ $$\ce{[F-] = \left( \frac{0.1x}{1+x} + \frac{2.9\times 10^{-3}}{1+x}\right)}M$$

step 3) Then as I stated before, the condition of precipitation would be as follows $$K_\mathrm {ip} > K_\mathrm {sp}$$ $$\ce{[Ca^2+][F-]^2} > K_\mathrm {sp}$$

So, from the equation above, I calculated the value of $x$ which I got was $1.8$ Litres (approximately) which was nowhere near the given answer in the book. How to solve this problem?

  • 2
    $\begingroup$ The question is weird. $\ce{CaF_2}$ is virtually insoluble. If the solution was a saturated solution made by dissolving $\ce{CaF_2}$ then the $\ce{[F^-] = 2[Ca^{2+}]}$. Also if the solution were saturated then adding any $\ce{F^-}$ would ppt $\ce{CaF2}$. Finally you'd never be able to add enough $\ce{F^-}$ to ppt all the $\ce{Ca^{2+}}$. $\endgroup$
    – MaxW
    Nov 18, 2016 at 18:44
  • $\begingroup$ But no where in the question they said it was a saturated solution, did they? $\endgroup$
    – Tiash
    Nov 19, 2016 at 4:04
  • $\begingroup$ As I said the problem statement is weird. The problem as you stated it has the phrase a "solution of $\ce{CaF2}$" which is odd. It sort of makes sense if you assume a saturated solution with a non-stoichometric quantities. But then it really isn't just a solution of $\ce{CaF2}$ is it?!? $\endgroup$
    – MaxW
    Nov 19, 2016 at 4:41
  • $\begingroup$ Also how can it be a solution of $\ce{CaF2}$ if there is no $\ce{F^-}$ in solution? So there is no way to determine the initial $[\ce{F^-}]$ unless it is $\sqrt{\frac{\ce{K_{sp}}}{[Ca^{2+}]}}$. $\endgroup$
    – MaxW
    Nov 19, 2016 at 4:52
  • $\begingroup$ Like you said, if the solution was saturated then adding any amount of $\ce{F-}$ will result in precipitating the $\ce{CaF_2}$, so shouldn't we assume that the solution wouldn't be saturated and calculate the initial concentration of $\ce{[F-] = 2[Ca^{2+}]}$ like @Zhe said in his answer? Though doing that also didn't help. $\endgroup$
    – Tiash
    Nov 19, 2016 at 5:10

2 Answers 2


First the question doesn't really make sense. $\ce{CaF_2}$ is virtually insoluble, so a "solution of $\ce{CaF_2}$" is a really odd phrase.

Skipping this nit-picking, the reactants are 1 L of a $2.0 \times 10^{-5}$ M $\ce{Ca^{2+}}$ and a $0.1$ M $\ce{F^-}$ solution and $\ce{CaF_2}$ is to be precipitated.

This is a sloppy use of significant figures since volume of 1L and the molarity of the $\ce{F^-}$ solution are only given to one significant figure.

Ignoring the significant figure problem too, let's work backwards from the given answer of 946 ml.

For the initial solution we can solve for the moles of $\ce{Ca^{2+}}$ present.

$\text{moles }\ce{Ca^{2+}} = 1 \rm{L} * 2.0 \times 10^{-5}\frac{moles}{L} = 2.0 \times 10^{-5}\text{ moles}$

Now if we add 946 ml of 0.1 M NaF, then the moles of $\ce{F^-}$ in solution is 0.0946 moles which vastly more than the calcium. So we can assume the precipitated $\ce{CaF2}$ has a negligible effect on the $[\ce{F^-}]$, and calculate the final concentration of $[\ce{F^-}]$ as:

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0946}{1.946} = 0.0486 \text{M}$

Now from the $\rm{K}_{\text{sp}}$ we have

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0486^2} = 7.20 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (7.20 \times 10^{-8})(1.946) = 1.40 \times 10^{-7}$

The initial moles of $\ce{Ca^{2+}}$ is $2.0 \times 10^{-5}$ which is +/- $0.05 \times 10^{-5}$. Thus so far as I am concerned the solution is a "magic number" and there is no way to get from the problem statement to that value. In other words why not $2.00 \times 10^{-7}$ or $3.00 \times 10^{-8}$ for the final $\ce{Ca^{2+}}$ concentration?!?

Thus you have to assume that you're going to precipitate some fraction of the $\ce{Ca^{2+}}$. So are you going to precipitate 99%, 99.9%, 99.95% or what? There just isn't anything in the problem statement on which you can pin a value.

Just out of curiosity the % $\ce{Ca^{2+}}$ recovered is:

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.40 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.3 \%$

Note that we can also assume that the original solution is saturated in respect to $\ce{CaF2}$. As shown in OP's post, this leads to a concentration of $2.91 \times 10^{-3} \rm{M}$ for $\ce{F^-}$ and given the 1L volume, $2.91 \times 10^{-3} \rm{moles}$ of $\ce{F^-}$. Adding that to the number of moles from the 946 ml of $\ce{F^-}$ solution gives a slightly different value for the total moles of $\ce{F^-}$:

total moles of $\ce{F^-}$ = $0.0946 + 0.00291 = 0.0975$

$\ce{[F^-]}_{\text{final}} = \dfrac{0.0975}{1.946} = 0.0501 \text{M}$

So the magic value for the textbook solution is to assume that: $\ce{[F^-]}_{\text{final}}= 0.0500$

$\ce{[Ca^{2+}]}_{\text{final}} = \dfrac{\rm{K}_{\text{sp}}} {[\ce{F^-}]^2}= \dfrac{1.7 \times 10^{-10}}{0.0501^2} = 6.77 \times 10^{-8}$

$\text{moles }\ce{Ca^{2+}} \text{ final}= (6.77 \times 10^{-8})(1.946) = 1.32 \times 10^{-7}$

% $\ce{Ca^{2+}}$ recovered = $(1-\dfrac{1.32 \times 10^{-7}}{2\times 10^{-5}}) \times 100\% = 99.34 \%$

So we get a slightly different value, but it still seems like a magic number of some sort.

Now if I were to write the problem I would have stated:

You are given $1.00$ Litre of solution which is saturated with $\ce{CaF_2}$ and has a $\ce{Ca^{2+}}$ concentration of $2.00 \times 10^{-5}$ M. How many mililitres of $0.100$ M $\ce{NaF}$ would you need to recover 99% of the $\ce{Ca^{2+}}$ as a precipitate of $\ce{CaF_2}$? The solubility product of $\ce{CaF_2}$ is $1.7\times10^{-10}$ .

Curious I played with this some more using the assumption of the $\ce{F^-}$ saturated starting solution. Obviously if you add vast amounts of $\ce{F^-}$ solution then any $\ce{CaF_2}$ which formed would dissolve agin. So there is some amount of $\ce{F^-}$ solution which precipitates the maximum amount of $\ce{CaF_2}$.

The max % $\ce{Ca^{2+}}$ recovery = 99.44% with 1913 ml of $\ce{F^-}$ solution. To +/- 0.01% any volume between about 1400 ml and 2500 ml would yield effectively the same result.


I think your $2.91 \cdot 10^{-3}$ value for fluoride concentration doesn't make sense. You are providing the concentration whereby the solution is immediately saturated. Instead, I think you just want double $2.0\cdot 10^{-5}$ for stoichiometric reasons, since both the fluoride and calcium presumably come from dissolution of calcium fluoride.


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