Fluoride concentration in LiF and CdF2 solutions

Trying to learn how to use $\ce{K\_{\mathrm{sp}}}$ values and not understanding. See question below:



> 
> Two saturated aqueous solutions are prepared at $\pu{25^\circ C}$
> 
> 
> **Solution1** is made by dissolving cadmium fluoride ($\ce{CdF2}$, $\ce{K\_{sp}} = \pu{6.44E−3}$) in $\pu{100.0 mL}$ of water until excess solid is present.  
> 
> **Solution 2** is prepared by dissolving lithium fluoride ($\ce{LiF}$, $\ce{K\_{sp}} = \pu{1.84E−3}$) in $\pu{200.0 mL}$ of water until excess solid is present.  
> 
> Which solution has a higher fluoride ion concentration?"
> 
> 
> 


I used the $\ce{K\_{sp}}$ given for $\ce{CdF2}$ to get $s = 0.11720$, so $\ce{[F]} = 2s = \pu{0.2344 M}$, is that right? I guess I'm not sure what to do after that to get find the concentration in $\pu{100 mL}$. Wouldn't it be $\pu{2.344 M}$? I am told this answer is incorrect.




The problem as stated is not asking you which solution has the higher amount of fluoride but which one has the higher **concentration** of fluoride, i.e. the higher molarity (as this is usually the concentration unit more often used by chemists).
You need to calculate the fluoride concentration for the $\ce{LiF}$ solution in the same way you did for the $\ce{CdF2}$ solution. My answer comes to 0.04290 M.
Thus, the $\ce{CdF2}$ solution has a fluoride concentration of 0.2344 M while the $\ce{LiF}$ solution has a fluoride concentration of 0.04290 M. Therefore, the answer to the problem is: the $\ce{CdF2}$ has a higher fluoride ion concentration.


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Ugh. enough nonsense from me...


* For $\ce{CdF2}$, $\mathrm{K\_{sp} = 6.44\ x\ 10^{−3}}$



> 
> Let $x$ = molarity of $\ce{CdF2}$
> 
> 
> $\ce{[Cd^{2+}]}$ = $x$   
> 
>  $\ce{[F^-]}$ = $2x$   
> 
> 
> 
> $\mathrm{K\_{sp} = 6.44\ x\ 10^{−3} = \ce{[Cd^{2+}][F^-]^2} = (x)(2x)^2 = 4x^3 }$  
> 
>  $x = 0.117$
> 
> 
> $\ce{[Cd^{2+}]}$ = $x = 0.117$  
> 
>  $\ce{[F^-]}$ = $2x = 0.234$  
> 
> 
> 
> 


* Lithium fluoride ($\ce{LiF}$), $\mathrm{K\_{sp} = 1.84\ x\ 10^{−3}}$



> 
> Let $x$ = molarity of $\ce{LiF}$
> 
> 
> $\ce{[Li^+]}$ = $x$   
> 
>  $\ce{[F^-]}$ = $x$   
> 
> 
> 
> $\mathrm{K\_{sp} = 1.84\ x\ 10^{−3} = \ce{[Li^+][F^-]} = (x)(x) = x^2 }$  
> 
>  $x = 4.29\times 10^{-2}$
> 
> 
> 


Thus the $\ce{CdF2}$ solution has a higher fluoride ion concentration.


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