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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.

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  • $\begingroup$ Please edit the question to show how you got to statement "... CdF2 to get s = 0.11720". Ksp is for molarity which is per liter. $\endgroup$ – MaxW Feb 17 '17 at 22:02
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    $\begingroup$ I got CdF2 s = 0.11720 using the below formulae: Ksp = [Cd^2+][F^1-]^2, therefore: 6.44 x 10^-3 = (s)(2s)^2 = 4s^3, solve for s, s = 0.11720, therefore: F M = 2s = 0.2344 M $\endgroup$ – Mike Feb 17 '17 at 22:31
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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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  • $\begingroup$ okay, so is the volume of liquid just a red herring? $\endgroup$ – Mike Feb 17 '17 at 22:29
  • $\begingroup$ @MikeW Please read my answer again. I did not say that CdF2 is 0.04290 M and that it has a Fluoride for 0.2344 M. $\endgroup$ – Pedro O'Verde Feb 17 '17 at 22:37
  • $\begingroup$ @Mike Yes, it is a red herring. If the problem were asking you to calculate which solution has the higher amount of fluoride then the volume would be necessary and you would calculate the number of moles from moles=V in L x M $\endgroup$ – Pedro O'Verde Feb 17 '17 at 22:39
  • $\begingroup$ Deleted 3 of my earlier comments. Your numbers are right. Your comment "MikeW Please read my answer again. I did not say that CdF2 is 0.04290 M and that it has a Fluoride for 0.2344 M." should have be to me... $\endgroup$ – MaxW Feb 17 '17 at 23:05
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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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