This might be really simple question but I have no idea how to proceed to solve such kind of question.

The solubility product of $\ce{SrF2}$ in water is $\pu{8E-10}$. Calculate its solubility in 0.1M $\ce{NaF}$ aqueous solution.

I do have solution to this question, but I don't understand it.

$$\ce{NaF -> Na+ + F-}$$
Since $\ce{NaF}$ is strong electrolyte, the concentration of $\ce{Na+}$and $\ce{F-}$ is 0.1M.
$$\ce{SrF2 -> Sr^{2+} + 2 F-}$$
Let the concentration of $\ce{SrF2}$ be S. Then, the concentration of of $\ce{Sr^{++}and F^-}$ is S and 2S respectively.
Now, $$K_\mathrm{sp} = \ce{[Sr^{2+}][F-]^2} \\ \pu{8E-10} = S (2S+0.1)^2 \\ \pu{8E-10} = S \times (0.1)^2 \\ S = \pu{8E-8 mole/L}$$

I don't understand why 0.1 is added to 2S and squared and later 2S in removed and how doing this exactly gives solubility of $\ce{SrF2}$ in $\ce{NaF}$. Can anyone help me in making me understand what is being exactly done in the solution and why does it gives the solubility?

  • $\begingroup$ OK, then do you understand the line before that? I mean, $K_\mathrm{sp} = \ce{[Sr^{++}][F^-]^2}$? $\endgroup$ Commented Aug 12, 2023 at 7:00
  • $\begingroup$ I read that it is ionic product of $\ce{Sr}$ and $\ce{F}$. $\endgroup$
    – izack
    Commented Aug 12, 2023 at 9:24

1 Answer 1


In the expresion $\ce{S(2S + 0.1)^2}$ the first $\ce{S}$ is for the concentration of $\ce{Sr^{2+}}$ which is dissolved. There are two ions of $\ce{F^-}$ for each atom of $\ce{Sr^{2+}}$. So $\ce{2S}$ is the concentration of $\ce{F^-}$ from the dissolved $\ce{SrF_2}$. However there is also 0.1 moles per liter of $\ce{F^-}$ from the $\ce{NaF}$. Thus the total $\ce{F^-}$ is $\ce{(2S + 0.1)}$.

Now you have to consider significant figures. Very little $\ce{SrF_2}$ is going to dissolve, so all the $\ce{F^-}$ essentially comes from the $\ce{NaF}$. Thus the simplification from $\ce{(2S + 0.1)^2}$ to $\ce{(0.1)^2}$.

You have to realize that in chemistry math there are limited significant figures and making reasonable assumptions greatly simplifies the math. Without the assumption that all the $\ce{F^-}$ essentially comes from the $\ce{NaF}$ you'd be trying to solve a messy cubic equation. Not a problem for a computer, but very very messy to do by hand. So in any kind of book problem it would be very very rare to need anything beyond a quadratic equation.

  • $\begingroup$ Thank you very much for the clear explanation. I would like to ask one more thing. Is $K_\mathrm{sp}$ for any electrolyte is constant for a given temperature in any solvent or is it different for different solvent? Like in this case, does the $K_\mathrm{sp}$ of $\ce{SrF2}$ changes in $\ce{NaF}$? $\endgroup$
    – izack
    Commented Aug 12, 2023 at 9:45
  • $\begingroup$ In general, the $\ce{K_{sp}}$ will change with temperature. // The $\ce{K_{sp}}$ for aqueous solutions is only good for aqueous solutions. For example if the solvent was changed to methanol, then there would be an entirely different $\ce{K_{sp}}$. // This is getting more advanced but different fluoride salts would effect the equilibrium. That is because using concentrations of the species is a simplification. The equilibrium should actually be calculated using the chemical activities of the species. $\endgroup$
    – MaxW
    Commented Aug 12, 2023 at 10:08
  • 1
    $\begingroup$ Solubility and Ksp are different in different solvents. Ksp does change with changes in ion content in a solvent [or the activities of the ions change] Most salts become more soluble in higher ionic strength solutions than Ksp predicts. $\endgroup$
    – jimchmst
    Commented Aug 12, 2023 at 10:10

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