Why HF (hydrogen fluoride) is not soluble in water?

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Discussion Overview

The discussion revolves around the solubility and dissociation behavior of hydrogen fluoride (HF) in water, exploring the reasons why HF does not fully ionize despite its high hydration enthalpy. Participants examine various factors including hydrogen bonding, bond strength, and the nature of the H-F bond.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Conceptual clarification

Main Points Raised

  • Some participants note that the hydration enthalpy of the fluoride ion is high, suggesting it could promote ionization of HF in water.
  • Others argue that HF does not ionize well in water due to the covalent nature of the H-F bond and the strong hydrogen bonding capabilities of HF itself.
  • It is mentioned that HF can form strong hydrogen bonds, which may contribute to its solubility but also to its weak acid behavior.
  • Some participants highlight that HF's dissociation varies with concentration, indicating that it is a weak acid compared to other hydrogen halides.
  • A later reply discusses the thermodynamic aspects of dissociation, emphasizing the role of entropy and the stability of the fluoride ion in solution.
  • There is a suggestion that the question may be misphrased, as HF is soluble in water but does not dissociate completely.
  • One participant explains that the strong hydrogen bonding in HF creates a highly ordered environment that inhibits dissociation, contrasting it with other hydrogen halides.

Areas of Agreement / Disagreement

Participants express multiple competing views regarding the solubility and dissociation of HF in water. There is no consensus on the primary reasons for its behavior, with various factors being proposed and debated.

Contextual Notes

Some limitations in the discussion include the dependence on definitions of solubility and dissociation, as well as the complexity of thermodynamic factors influencing the behavior of HF in aqueous solutions.

Ahmed Abdullah
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The Attempt at a Solution



The hydration enthalpy of Floride ion is very high.
We know that HF react violently with NaOH to form floride salt and water. The enthalphy of this reaction is greater than Neutralization enthalphy (H+ + OH-=H2O) of other acid base. That proves that the hydration enthalpy is sufficient enough to supply required energy for H-F bond breaking, beside producing extra heat.
So I don't understand why HF don't ionise in water, where hydration enthalphy is large enough to promote ionization.
 
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WHAT? Where did this come from?
Ahmed Abdullah said:

Homework Equations



none

The Attempt at a Solution



The hydration enthalpy of Floride ion is very high.
We know that HF react violently with NaOH to form floride salt and water. The enthalphy of this reaction is greater than Neutralization enthalphy (H+ + OH-=H2O) of other acid base. That proves that the hydration enthalpy is sufficient enough to supply required energy for H-F bond breaking, beside producing extra heat.
So I don't understand why HF don't ionise in water, where hydration enthalphy is large enough to promote ionization.
 
hydrogen bonding

Your enthalpy approach was very good. i would have done the same, but for a molecule to dissolve in water, it must be able to hydrogen bond with itself. that is hydrogen bond with an identical molecule like itself. this very impossible for a molecule like hydrogen flouride. you see flourine is a highly electronegative atom. It wouldn't like to have another hydrogen bonded to it. This is why it wouldn't pass for a strong acid, which can completely ionize in water. Also the electron density in the flourine atom on the HF molecule is very high. these electrons are strongly attracted to the nucleus of the atom reducing it's ability to solubilize in water.
 
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Consider the bonding approach. F is the halogen that shall have orbitals of most similar size to H in all halogens. Bond overlap would be strongest. Hence it doesn't readily give H+ , thus is a weak acid .
However disassociation varies a lot with concentration of HF in solution
 
Ahmed Abdullah said:

Homework Equations



none

The Attempt at a Solution



The hydration enthalpy of Floride ion is very high.
We know that HF react violently with NaOH to form floride salt and water. The enthalphy of this reaction is greater than Neutralization enthalphy (H+ + OH-=H2O) of other acid base. That proves that the hydration enthalpy is sufficient enough to supply required energy for H-F bond breaking, beside producing extra heat.
So I don't understand why HF don't ionise in water, where hydration enthalphy is large enough to promote ionization.

Why do you think hydrogen fluoride is not soluble in water. It is commonly sold as a 70% and 50% solution in... water!

Perhaps the question was "Why is hydrogen fluoride not completely dissociated in water?" If that is the question, I think you should argue that the nature of the H-F bond is very covalent (as opposed to ionic).
 
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Jide said:
Your enthalpy approach was very good. i would have done the same, but for a molecule to dissolve in water, it must be able to hydrogen bond with itself. that is hydrogen bond with an identical molecule like itself. this very impossible for a molecule like hydrogen flouride. you see flourine is a highly electronegative atom. It wouldn't like to have another hydrogen bonded to it. This is why it wouldn't pass for a strong acid, which can completely ionize in water. Also the electron density in the flourine atom on the HF molecule is very high. these electrons are strongly attracted to the nucleus of the atom reducing it's ability to solubilize in water.

Actually, hydrogen fluoride forms very strong hydrogen bonds. In gas phase, HF aggregates to H2F2, H4F4, etc, resulting in a higher than expected boiling point for HF.
 
Why is the nature of H-F bond covalent?
 
hydogen bonding

I guess i didn't read his question carefully. yes, up to 70% of HF will dissociate in water. the other 29.9% is because it is a weak acid compared its other buddy hydrohalogens (HCl, HI) and so on. this also makes HF an exception in the class of strong acids
 
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HF is soluble (miscible) but it remains nearly undissociated in the solution.
The dissociation constant(pKa=3.15) of hydrofloric acid is too low to be dissociated up to 70%.
 
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  • #10
Hi,

HF forms strong hydrogen bonds, so it is quite soluble in water. If the question was as you stated it, then it is wrong. HF is however a much weaker acid than other hydrogen halides. This because the F- ion is so small, the negative charge is concentrated on a small area and its entropy of solvation in water is therefore strongly negative. This makes the dissociation of HF thermodynamically unfavourable compared to its bigger cousins of halogen using family. Also, because of the greater bond strength of HF, its dissociation is also kinetically inhibited but that is not important in discussing its aqueous phase chemistry.

Molu
 
  • #11
Jide? Stop (and listen). People's exam grades are at stake here.

The molecule HF *dissolves* freely in water (even more than 70%!), but it is a weak acid (does not *dissociate* well in water) for two reasons: 1) because hydrogen bonding is FON, and 2) because delG = delH - TdelS.

...

N, O and F atoms have 1, 2 and 3 lone pairs whose electrons are concentrated enough to be attractive to H atoms in other molecules. So N, O and F hydrogen bond. No other lone pairs on the periodic table can do that.

Not Cl, not Br, not I (think HCl, HBr, HI). Hydrogen bonding is FON.

Yes, the HF bond is very strong (covalency has to do with proper orbital overlap, and the sizes of H and F are very appropriate for each other). But when you turn one mole of HX into H+(aq) and X-(aq), it actually won't matter whether X is F, Cl, Br or I: they will each have about the same heat of formation (delH). And it will be slightly negative (favored). Darn. So HF bond-breaking is not the reason that it doesn't dissociate in water.

DeltaG is equal to a couple of things. Firstly, delG = RTlnK, where K let's you know how well something dissociates. But that's the frosting on the cake, so save it for later. Secondly, delG also = delH - TdelS. DelG is free energy (how well a reaction will go forward), and it is helped by heat (enthalpy) and hindered by order (entropy) at a given temperature.

When you put HCl, HBr or HI in water, the lone pairs on the O of a nearby water kindly take the H+ away from the Cl- or Br- or I- because... hydrogen bonding is FON. Or, at least it's more FON than Cl-. Now the Cl- kinda hangs out with the H3O+ to keep things neutralized. And some other waters weakly hang out with both the Cl- and the H3O+ so that there's no jealousy.

But here's the punch. When you put HF in water, it starts taking hostages. Remember those 3 lone pairs on F that can hydrogen bond? Well, when a water wanders on over to take away the H+ from HF, the F- starts shoving other waters in its back pocket. That is, it hydrogen bonds with H's of other waters behind its back. This creates a real frenzy of strong order around the F-. Then the lifeguard blows the whistle. Everyone in the pool stands still. The head count (delG) reveals that there is too much order for that reaction (dissociation) to go forward, and water is commanded to give back the H+ to the F- that has 3 waters sticking out of its back pocket.

That is, when HCl dives in the pool, delH (from a few paragraphs above) is medium negative, delS is not too big negative, and so delG is overall negative, and the dissociation goes forward. But when HF dives in, everything gets real tense and there's lots of order. So delH is negative, but delS is really negative, and so delG is overall positive, and the dissociation does not go forward well.

delG = RTlnK. Solve for K with a negative delG (HCl) and you have a large dissociation constant. Solve for K with a positive delG (HF) and you have a small dissociation constant.

So, HF is a weak acid because hydrogen bonding is FON, and because delG = delH - TdelS.

(http://www.chemguide.co.uk/inorganic/group7/acidityhx.html)
 

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