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I am pondering over that what must be the reason that $\ce{AlF3}$ is insoluble or less soluble in anhydrous $\ce{HF}$? And pageants it is soluble to a good extent only in the presence of $\ce{KF}$? Is there any other reagent in presence of which $\ce{AlF3}$ is soluble in $\ce{HF}$?

I guess a probable answer for the cause of insolubility could be: "the nonavailability of $\ce{F-}$ ions because of the presence of intermolecular hydrogen bonding in $\ce{HF}$, which makes $\ce{AlF3}$ insoluble in $\ce{HF}$."

Can someone elaborate the cause and mention other such reagents like $\ce{KF}$ which can be used for the same purpose?

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    $\begingroup$ Could you clarify the conditions? I find that $\ce{AlF3}$ is okayish soluble in water, see Wikipedia, so I am curious under what conditions it is insoluble in HF, because if we talk about an aqueous solution, then I doubt that this is true. Otherwise the question is more or less why $\ce{AlF3}$ is so much more soluble in the presence of $\ce{KF}$. $\endgroup$ – Martin - マーチン Jan 7 '17 at 16:46
  • $\begingroup$ Do you mean pure hydrogen fluoride, or a concentrated aqueous solution? $\endgroup$ – hBy2Py Jan 8 '17 at 0:50
  • $\begingroup$ I mean pure HF. $\endgroup$ – Resorcinol Jan 8 '17 at 1:17
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$\ce{AlF3}$ is insoluble in the anhydrous $\ce{HF}$ because the $\ce{F-}$ ions are not available in intermolecular hydrogen bonded $\ce{HF}$, but it becomes soluble in the presence of $\ce{KF}$ due to the formation of soluble complex, $\ce{K3[AlF6]}$. $$\ce{AlF3 + 3KF -> K3[AlF6]}$$

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    $\begingroup$ Hmm, it looks like too strong simplification. $\endgroup$ – Mithoron Jan 8 '17 at 13:26
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    $\begingroup$ From the reaction equation it looks like you need a massive amount of KF. And while we are at it, what is a little amount? $\endgroup$ – Martin - マーチン Jan 12 '17 at 12:02
  • $\begingroup$ Probably there is more than one complex ion. But yes, complex ion formation is the key. It's like an aqueous system where zinc or aluminum hydroxide is almost insoluble in just water but dissolved easily when you add NaOH. $\endgroup$ – Oscar Lanzi May 29 '17 at 16:32
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Since it is pure $\ce{HF}$ we are talking about, there wouldn't be any hydrogen bonding in the first place. Therefore, there wouldn't be any dissociation to form $\ce{H+}$ and $\ce{F-}$ ions.

Whereas for the $\ce{KF}$ solution, it would easy dissociate to form $\ce{K+}$ and $\ce{F-}$ ions since it is dissolved in water. I believe the crux to this question is the water; since you need it to produce $\ce{F-}$ ions which can form the $\ce{[AlF6]^3-}$ complex. This is compounded by the fact that $\ce{AlF3}$ is ionic in character since fluorine is very electronegative, the $\ce{Al-F}$ is relatively weaker and will easily break in the presence of water to form the complex.

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