Solubility of SrF2 in aqueous solution of NaF

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?

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

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.
• 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 $\ce{NaF}$? Commented Aug 12, 2023 at 9:45
• 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.