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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: orgopete on November 21, 2016, 07:00:08 PM

Title: Why is HF a weak acid
Post by: orgopete on November 21, 2016, 07:00:08 PM
I want to resurrect this topic which has been discussed before:
http://www.chemicalforums.com/index.php?topic=19788.0
http://www.chemicalforums.com/index.php?topic=16293.0

I am aware of some of the past thoughts but I want to learn more. First of all, I do not wish to invoke electronegativity arguments as I believe this is a scale greatly biased by the redox reactions. Thus the electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. This is simply different than the electron withdrawing power. Thus HF is a weak acid and HI is strong. That simply restates the question, but it doesn't explain why.

It is easy to understand why HI is a strong acid. Its nuclear charge is large (+53) and thus it can pull its electrons inward. Compared to the other halogens, the repulsive field of an iodine nucleus grows faster than bond length. That is, if we set the repulsive force of HI to be equal to that if HF, then the bond length of HI would be 223 pm versus 161 pm actual, x^2=(53*92^2)/9. Granted, this is simplistic reasoning, but I think it is consistent with a general trend in bond lengths, e.g., CH>NH>OH>HF due to nuclear charge. For this reason, we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. This should be a general principle of physics that matches the acidity trends of simple acids, HI>HBr>HCl>>HF; HF>H2O>NH3>CH4.

You may see arguments that HF is a weak acid because of low entropy. While I think this is true, I don't agree with the interpretation of it. The entropy is lower because HF doesn't ionize, not that it doesn't ionize because its entropy is lower. I don't think of entropy as a normal force. Entropy can be small if a bonding force is present, otherwise molecules will behave as ideal substances, that is without interaction.

The issue I am trying to resolve is gas phase v solution phase reactions. I think they are simply different. McMurray states, “… bond dissociation energies refer to molecules in the gas phase and aren’t directly relevant to chemistry in solution.” The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase. I have used this argument for a solution phase Born-Haber cycle in a formation of NaCl(aq), -386 kJ/mol v -411 kJ/mol (s). In a solution phase reaction, no bond is formed between Na(+) and Cl(-). The energy is largely the energy of the redox reaction of Na->Na(+) and Cl->Cl(-). So, if you perform a reaction, it will be guided by Hess's Law.

I saw a post for the reaction of LiBr + MeI  ::equil:: LiI + MeBr, the net enthalpy of bonds formed versus broken would be +18 kJ/mol. However, I began to think the bonds of LiBr and LiI are actually not being broken and thus should not be part of the calculation. Now, the reaction enthalpy is a difference in the bond energy of methyl bromide and methyl iodide, -72 kJ/mol. However, as I understand this calculation, this is a measure of the homolytic bond energy. The homolytic bond energy is being used to calculate the thermodynamics of a heterolytic reaction.
 
Further, the commenter posted:
Quote
In another paper, they gave the following equilibrium constants for a (reversed Finkelstein) reaction LiBr + RI = LiI + RBr: MeI (K = 1.7), n-alkyl (K = 7), iPr (K = 30), t-Bu (K ca 100). They furthermore stated that for the reaction LiCl + RBr = LiBr + RCl, the equilibrium should even be further on the RCl side (K larger). (See: DOI: 10.1039/JR9550003173)

Now, I realized I could calculate a Gibbs free energy from the equilibrium. The methyl bromide reaction has a rather modest enthalpy of -1.32 kJ/mol and isopropyl -8.4 kJ/mol by my calculations. As an organic chemist, these values seem about right. I am finding these calculations daunting. The enthalpy values should include all of the factors in a reaction, not just the bond energy. I don't know how to adjust for an entropy change, though I would have thought it to be small and toward equilibrium in this substitution reaction.

Now I've seen the heat of neutralization of HF is -68 kJ/mol while HCl, HBr, and HI are -57 kJ/mol. The simplicity of HF makes this an ideal molecule to understand these energy differences. I can understand HCl, HBr, and HI as they simply form the same H3O(+). The enthalpy of neutralization is the neutralization of hydronium ion. This is analogous to my ignoring the bonds of LiI and LiBr as they are broken upon dissolving in acetone.

One analysis that seems to be in line as to how this problem may be solved has been contributed by Jim Clark (http://chem.libretexts.org/Core/Inorganic_Chemistry/Descriptive_Chemistry/Elements_Organized_by_Block/2_p-Block_Elements/Group_17%3A_The_Halogens/1Group_17%3A_General_Reactions/The_Acidity_of_the_Hydrogen_Halides). This example illustrates the problem I have with understanding the data. It illustrates a Born-Haber cycle with enthalpies of vaporization, bond energy, ionization, electron affinity, and hydration. I find it difficult to verify any one of these data points and if I can, they may differ significantly from this source.

I don't know if I am trying to understand that which cannot be readily understood. I am dubious of gas phase acidity as it can often differ from solution phase acidity. For me, a clear warning to using values from one phase to predict reactions in another is the endothermic gas phase ionization of sodium. If sodium preferred to hold its electron, I would expect to find sodium in its ground state, yet metallic sodium reacts exothermally in solution.

I hope someone can clarify this problem.
Title: Re: Why is HF a weak acid
Post by: AWK on November 21, 2016, 07:11:00 PM
http://www.ch.imperial.ac.uk/rzepa/blog/?p=16320
Title: Re: Why is HF a weak acid
Post by: orgopete on November 21, 2016, 07:25:09 PM
http://www.ch.imperial.ac.uk/rzepa/blog/?p=16320
I had seen that, but I didn't find it helpful. I guess I should have specified I was interested in being able to calculate the energy differences. Otherwise, I find the acidity of HF to be as I would have expected, greater than H2O and less than HCl.
Title: Re: Why is HF a weak acid
Post by: AWK on November 21, 2016, 07:44:59 PM
Liquid HF (anhydrous) is a superacid
Title: Re: Why is HF a weak acid
Post by: Irlanur on November 23, 2016, 06:28:53 AM
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The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase.

Wut?

I find the post not very well structured and honestly have a hard time what your actual question is...
Title: Re: Why is HF a weak acid
Post by: Enthalpy on November 23, 2016, 03:17:18 PM
A part of the answer is how water eases the formation of the halide ion, because without the solvent, there would be extremely little ionization.

And then, one simple reason is the diameter of the halide ion. The bigger iodide concentrates less the charge of the excess electron than the smaller fluoride does, and the less concentrated charge needs less electric energy.

Modelling water as a continuous insulator with permittivity already tells it. q (puts signs as you like) concentrated at the surface of a radius r sphere means an energy varying like 1/(r×ε) where both the big r and water's big ε help the ionization.

Or if you prefer to imagine the discrete water molecules, there are more of them near the bigger ion, so they mitigate more easily the effect of the excess electron.
Title: Re: Why is HF a weak acid
Post by: Enthalpy on November 23, 2016, 03:38:25 PM
[...] The electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. [...]
If you check the bond strength of the halogens with other atoms like C and H, it's not very different from C-C or C-H bond strength. C-H and C-Cl are almost equally strong bonds despite Cl being more electronegative, and the electronegative Br makes a weaker bond than C-H. What does make the light halogens reactive is rather that the X-X bond is weak, and this acts both on the reaction enthalpy and on the activation energy.

[...] we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. [...]
I don't see a difference (except in reaction direction and energy sign) between the addition of an electron to a halogen atom (the electronegativity) and the withdrawal of an electron from a halide ion.
Title: Re: Why is HF a weak acid
Post by: orgopete on November 23, 2016, 11:36:59 PM
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The conservation of energy principle means the same energy difference would exist whether a reaction were gas or solution phase.

Wut?

I find the post not very well structured and honestly have a hard time what your actual question is...

I agree. I should have kept it simple at why is the heat of neutralization of HF is -68 kJ/mol while HCl, HBr, and HI are -57 kJ/mol, but ...
Title: Re: Why is HF a weak acid
Post by: orgopete on November 23, 2016, 11:50:25 PM
A part of the answer is how water eases the formation of the halide ion, because without the solvent, there would be extremely little ionization.[/unquote]
This is little ionization in water

Quote
And then, one simple reason is the diameter of the halide ion. The bigger iodide concentrates less the charge of the excess electron than the smaller fluoride does, and the less concentrated charge needs less electric energy.
A lot of people like to make that argument. I am thinking electrons are negative, in the case of iodide, there are 54 of them. How do you figure the charge is less than fluoride?

Quote
Modelling water as a continuous insulator with permittivity already tells it. q (puts signs as you like) concentrated at the surface of a radius r sphere means an energy varying like 1/(r×ε) where both the big r and water's big ε help the ionization.

Or if you prefer to imagine the discrete water molecules, there are more of them near the bigger ion, so they mitigate more easily the effect of the excess electron.
If I understand this correctly, the big "r" means the ion is being treated as a Gaussian shell. I think that is only true at a distance further from an anion. If electrons have a negative charge and we invoke the inverse square law, then at close distances, electrons will behave as local charge. I reason that is why you can see the tetrahedral structure of fluorine at close distances.

Am I construing this correctly, you do not think the repulsive effect of an iodine's nucleus is a factor?
Title: Re: Why is HF a weak acid
Post by: orgopete on November 24, 2016, 12:49:34 AM
[...] The electronegativity of fluorine is large because fluorine is very reactive and iodine is small because it is much less so. [...]
If you check the bond strength of the halogens with other atoms like C and H, it's not very different from C-C or C-H bond strength. C-H and C-Cl are almost equally strong bonds despite Cl being more electronegative, and the electronegative Br makes a weaker bond than C-H. What does make the light halogens reactive is rather that the X-X bond is weak, and this acts both on the reaction enthalpy and on the activation energy.
Enthalpy of formation:
HF (-272), HCl (-92.3), HBr (-36), HI (+26)

Bond energy
HF (570), HCl (432), HBr (366), HI (298)
CF (452), CCl (352), CBr (293), CI (236)

I know these are not exactly the same thing, but really, don't you think the enthalpy of formation for HF is high because fluorine is more reactive than the other halogens? The bond energy is a kind of reverse of the formation. If we are using Pauling electronegativity, he assumed the bonds were stronger (which I think may be partly true), but I also think fluorine is a lot more reactive. I believe it is this assumption that leads to the awkward result of HF being more ionic than HI, but does not ionize. HI is a stronger acid and iodine is more electron withdrawing.

Quote
[...] we might then expect that the electrons of fluoride to be more accessible, more basic, than those of iodide. [...]
I don't see a difference (except in reaction direction and energy sign) between the addition of an electron to a halogen atom (the electronegativity) and the withdrawal of an electron from a halide ion.
This is two different things. I'm arguing that in the case of the electrons of fluoride, if we likened them to the electrons of a carbon (anion), nitrogen, and oxygen, there are two factors in their basicity. One, the smaller the nuclear charge, the further the electrons are to the nucleus. As an inverse square force, the nuclear field diminishes exponentially. The force of a pair of electrons would create an inverse square force. Comparing fluoride to the other electron pairs, it would be closest to the nucleus, the larger nuclear charge and the least basic. If compared to iodide, it would be the most basic.

I argue that this should be a general observation. For carbon, the shorter the CH bond, the more acidic. I use this to argue why ammonia is a stronger base than fluoride. Even though ammonia has one more proton, they are much further from the basic electrons.

The other point about adding an electron to a halogen versus knocking one off. I agree they may be quite similar. That could be reassuring, but the gas phase loss of an electron from sodium is +496 kJ/mol. So we should assume sodium will release 496 kJ/mol to reform sodium metal. What happens if you add sodium to water? It is a different reaction. The bonds to sodium ions are long and weak. A sodium cation is surrounded by 10 electrons. They are negative. It is only at a greater distance than the greater nuclear charge is felt. Up close, the electron are repelling. Sodium does not hold its electrons. I am imagining a different trend here.
Title: Re: Why is HF a weak acid
Post by: orgopete on November 24, 2016, 07:57:07 AM
I'm not sure I have explained my bond length argument well. I reason that in a Bronsted acid, the attractive force should be an inverse square force to the nearest electron pair. The closer a proton to an electron pair, the greater the force. If you compare two acids, the stronger acid should correspond with the weaker force and via the inverse square law, a greater distance between the proton and its electron pair. If we compare HF and H2O, fluorine with a larger nuclear charge exerts a greater pull on its electron pairs. The electron pairs are normally repelling to one another so they resist being pulled in. This is a continuing effect for all atoms. While fluorine pulls its electrons in more than oxygen, the proton follows and a shorter bond results. However, the gap between the electron pair and a proton increases due to the repulsive effect of the fluorine nucleus. The larger gap results in a heterolytically weaker bond, and stronger acid.

I expect a similar trend should exist for the other haloacids, except the position of the electrons is masked by an increase in the number of electrons in its valence shell. I would expect iodide to exert a greater pull on its electrons than fluoride. It would also have a greater repulsive field to a proton. The net effect is the gap between an electron pair and a proton would be the greatest for iodide.

I argue it isn't the bond length that matters, but the proton-electron pair distance. Thus, HF has the shortest bond and is the strongest acid of C, N, O, and F. HI is the longest bond of F, Cl, Br, and I, but it has the longest bond because we have added an additional layer of valence electrons to which a proton is attracted. It isn't the bond length that matters, it is the gap.

In the case of carbon, the bond of an acetylene is the shortest. I interpret this to indicate that with a triple bond, the electrons are being pulled away from the HC bond and in toward the nucleus. This shift decreases the bond length and increases the proton-nuclear repulsion. The result is a larger gap between the proton and electron pair, weaker force, and greater acidity. With a tertiary hydrogen, the electron donation of carbon reverses this effect. The electrons directed to the proton are pushed out further, the bond is longer, but the proton-electron pair gap is smaller. The result is a greater force, a heterolytically stronger bond, and weaker acid.
Title: Re: Why is HF a weak acid
Post by: mjc123 on November 24, 2016, 08:19:43 AM
How about this from Wikipedia https://en.wikipedia.org/wiki/Hydrogen_fluoride:
"The weak acidity in dilute solution is sometimes attributed to the high H—F bond strength, which combines with the high dissolution enthalpy of HF to outweigh the more negative enthalpy of hydration of the fluoride ion.[19] However, Giguère and Turrell[20][21] have shown by infrared spectroscopy that the predominant solute species is the hydrogen-bonded ion-pair [H3O+·F], which suggests that the ionization can be described as a pair of successive equilibria:
H2O + HF  :requil: [H3O+·F]
[H3O+·F]  :lequil: H3O+ + F
The first equilibrium lies well to the right (K ≫ 1) and the second to the left (K ≪ 1), meaning that HF is extensively dissociated, but that the tight ion pairs reduce the thermodynamic activity coefficient of H3O+, so that the solution is effectively less acidic.[22]"
Title: Re: Why is HF a weak acid
Post by: orgopete on November 24, 2016, 02:35:22 PM
How about this from Wikipedia https://en.wikipedia.org/wiki/Hydrogen_fluoride:
"The weak acidity in dilute solution is sometimes attributed to the high H—F bond strength, which combines with the high dissolution enthalpy of HF to outweigh the more negative enthalpy of hydration of the fluoride ion.[19] However, Giguère and Turrell[20][21] have shown by infrared spectroscopy that the predominant solute species is the hydrogen-bonded ion-pair [H3O+·F], which suggests that the ionization can be described as a pair of successive equilibria:
H2O + HF  :requil: [H3O+·F]
[H3O+·F]  :lequil: H3O+ + F
The first equilibrium lies well to the right (K ≫ 1) and the second to the left (K ≪ 1), meaning that HF is extensively dissociated, but that the tight ion pairs reduce the thermodynamic activity coefficient of H3O+, so that the solution is effectively less acidic.[22]"

Here is my take. A lot of what is being offered is an attempt to counter the implications of Pauling's electronegativity theory. Even Pauling seemed surprised at its failure when he states, “…the bond type has no direct connection with ease of electrolytic dissociation in aqueous solution.” If we were discussing H2O, I doubt anyone would be surprised at its low acidity. HF should have followed the trends of its row. If electronegativity theory had not been advanced, I doubt its acidity would have been such an issue. Although I tried to discount electronegativity theory, but I cannot prevent others from bringing it up.

I might think the Wikipedia article might be written as:
 HF + H2O   :requil: H2O·HF   :lequil: H3O+ + F-

That would seem reasonable. HF is a good proton donor for hydrogen bonding. I'd guess even better than water. Even though it is hydrogen bonded, it still doesn't ionize. The difference is I have failed to write a negative charge on the fluorine atom. My perspective is if HO-, H2O, and H3O+ are compared, the formal charges reflect the net number of hydrogens added (irrespective of hydrogen bonding to other water molecules). The charge of the oxygen remains unchanged at +8. Adding a negative charge to the hydrogen bonded structure doesn't change the charge of the fluorine. The difference between the Wikipedia article and what I have written is the Wikipedia article expects HF to ionize hence it is written with an negative charge, yet it doesn't ionize.
Title: Re: Why is HF a weak acid
Post by: Irlanur on November 25, 2016, 01:35:01 PM
Quote
I reason that in a Bronsted acid, the attractive force should be an inverse square force to the nearest electron pair. The closer a proton to an electron pair, the greater the force.

This is simply an extremly unhelpful language in the context of chemical bonds. The forces on the atoms at the equilibrium distance are simply 0. And the force constant in a harmonic approximation doesn't say too much about the dissociation energy. It becomes even more useless if a proton is hoping back and forth between different bases, which is quite often the case with acidic protons in "normal" pH solutions.
Title: Re: Why is HF a weak acid
Post by: Enthalpy on November 26, 2016, 04:07:31 AM
I am thinking electrons are negative, in the case of iodide, there are 54 of them. How do you figure the charge is less than fluoride?
You wrote that. I didn't. I wrote "the excess electron". The net charge of F-, Cl-, Br- and I- is one excess electron.

About any angular structure in the electron shell: at least F- has spherical symmetry. Adding the probability densities for three 2p orbitals removes all the angular dependencies. I suppose, but didn't check, that this holds for halides with bigger orbitals too.
Title: Re: Why is HF a weak acid
Post by: Enthalpy on November 26, 2016, 04:27:59 AM
Bond strength compared with electronegativity and reactivity:

439kJ/mol CH3−H
432kJ/mol H-Cl
350kJ/mol CH3−Cl
218kJ/mol H-H
121kJ/mol Cl-Cl

On this example you see that bond strength with carbon doesn't result from electronegativity nor from the difference of electronegativity between the atoms, and that chlorination happens because the Cl-Cl bond is weak.

Also, the stronger H-H bond is what distinguishes hydrogen from the halogens.
Title: Re: Why is HF a weak acid
Post by: orgopete on November 28, 2016, 09:50:10 PM
I am thinking electrons are negative, in the case of iodide, there are 54 of them. How do you figure the charge is less than fluoride?
You wrote that. I didn't. I wrote "the excess electron". The net charge of F-, Cl-, Br- and I- is one excess electron.

... one simple reason is the diameter of the halide ion. The bigger iodide concentrates less the charge of the excess electron than the smaller fluoride does, and the less concentrated charge needs less electric energy.
I was referring to the notion that the iodide concentrates less charge. I thought it was actually the opposite. There are 54 electrons per atom compared to 10.

I think perhaps my thought that the nuclear charge is important is not being given enough credit. If protons are being repelled by the nuclear charge, then it seems plausible that one should compare the charges and distance for HF and HI. HF, 9 protons, has a bond length of 92 pm while HI, 53 protons, has a bond length of 161 pm. By my calculations, the proton of HI has a 1.9 times greater repulsive force from the nucleus. I find this a compelling argument. I use a similar argument to explain why ammonia should be more basic than fluoride. It isn't the net charge per se, but the collection of forces of independent particles. That is, I use the charge and distance of a nitrogen nucleus plus the charge and distance of the three more remote protons to calculate the repulsive force a proton would feel at the distance of an ammonium ion.

I further find this plausible that a greater charge from a larger nucleus can compress the inner electrons to a greater extent than lighter atoms. If the electrons are being pulled in further, a proton must approach closer to the nucleus to be attracted to the electrons. I am prone to believe this nuclear charge may accommodate nuclear attacks on iodine in net reductions by iodide and iodo compounds forming halogen bonds as electron acceptors.
Quote
About any angular structure in the electron shell: at least F- has spherical symmetry. Adding the probability densities for three 2p orbitals removes all the angular dependencies. I suppose, but didn't check, that this holds for halides with bigger orbitals too.

Poly HF has a zig-zag structure. This makes me think it would have a similar tetrahedral geometry as borohydride, methane, ammonia, and water. How do you know it is spherical?
Title: Re: Why is HF a weak acid
Post by: orgopete on November 28, 2016, 10:02:40 PM
Quote
I reason that in a Bronsted acid, the attractive force should be an inverse square force to the nearest electron pair. The closer a proton to an electron pair, the greater the force.

This is simply an extremly unhelpful language in the context of chemical bonds. The forces on the atoms at the equilibrium distance are simply 0. And the force constant in a harmonic approximation doesn't say too much about the dissociation energy. It becomes even more useless if a proton is hoping back and forth between different bases, which is quite often the case with acidic protons in "normal" pH solutions.

In the model I was imagining, I was proposing something like that shown below. In that model, a pair of sp3 electrons can both attract a proton and to be attracted to a nucleus. If the proton were an acid, then I'm further imaging the attraction of the proton should be an inverse square force to the electron pair. If I compare the second row elements, LiH, BeH2, and BH4(-), the electrons remain with the proton and are hydride donors. Carbon is in between. As the nuclear charge increases, the electrons are pulled closer and I reason the force between the sp3 electrons and a nucleus is greater than to a proton and so they are acids. The greater the nuclear charge, the stronger the acid. I reason it is because the proton-electron pair distance is larger and its force is smaller.
Title: Re: Why is HF a weak acid
Post by: orgopete on November 28, 2016, 10:15:58 PM
Bond strength compared with electronegativity and reactivity:

439kJ/mol CH3−H
432kJ/mol H-Cl
350kJ/mol CH3−Cl
218kJ/mol H-H
121kJ/mol Cl-Cl

On this example you see that bond strength with carbon doesn't result from electronegativity nor from the difference of electronegativity between the atoms, and that chlorination happens because the Cl-Cl bond is weak.

Also, the stronger H-H bond is what distinguishes hydrogen from the halogens.

I'm not following the argument (nor the bond strength data posted). Are you agreeing with me or disagreeing? I am of the opinion that gas phase and solution phase data are simply different. I would conclude that even though the gas phase data is similar, HCl has a heterolytically weaker bond than chloromethane. If that is what you are arguing, then I agree. If you are saying they are similar, then I disagree.
Title: Re: Why is HF a weak acid
Post by: Irlanur on November 29, 2016, 04:50:21 AM
Quote
I reason it is because the proton-electron pair distance is larger and its force is smaller.

doesn't change anything about the fact that the force is 0 at Equilibrium. If it the force wouldn't be zero, then the proton would move closer.

The only possibility I see to include forces in any way is by discussing the force constants, which come from a taylor approximation around the equilibrium position. But again this has rather little to do with acidity.
Title: Re: Why is HF a weak acid
Post by: orgopete on November 29, 2016, 09:56:56 AM
Quote
I reason it is because the proton-electron pair distance is larger and its force is smaller.

doesn't change anything about the fact that the force is 0 at Equilibrium. If it the force wouldn't be zero, then the proton would move closer.
I think I get this argument. The forces on any subatomic particle will be found in the valley of a Morse curve, zero. I agree that will describe the equilibrium condition, I'm not sure I agree with the implications. It suggests an equal energy level (forces) for two different acids. I think even though the protons may be found in a valley, I don't think the valleys have the same energy levels (forces) nor the same shapes, but I think I'll let you explain this to us.

If the forces acting on a proton in HF and HI are 0 at equilibrium, why is HI a stronger acid?
Title: Re: Why is HF a weak acid
Post by: Enthalpy on November 30, 2016, 02:34:15 PM
I was referring to the notion that the iodide concentrates less charge. I thought it was actually the opposite. There are 54 electrons per atom compared to 10.

I stop here. Good luck.
Title: Re: Why is HF a weak acid
Post by: Irlanur on December 01, 2016, 03:35:10 AM
Quote
I think I get this argument. The forces on any subatomic particle will be found in the valley of a Morse curve, zero. I agree that will describe the equilibrium condition, I'm not sure I agree with the implications. It suggests an equal energy level (forces) for two different acids. I think even though the protons may be found in a valley, I don't think the valleys have the same energy levels (forces) nor the same shapes, but I think I'll let you explain this to us.

If the forces acting on a proton in HF and HI are 0 at equilibrium, why is HI a stronger acid?


Quote
energy levels (forces)

Why do you put the (forces) in brackets? The force is the negative gradient of the potential energy. You can deduce nothing about the energy from the fact that the force is 0 at equilibrium.
The acidity is determined by the Gibbs free energy difference between the protonated and the deprotonated form. you cannot make any (rigorous) conclusions whatsoever about this energy difference from the knowledge about the forces. 
Title: Re: Why is HF a weak acid
Post by: orgopete on December 01, 2016, 10:00:58 AM
Quote
Why do you put the (forces) in brackets?

It is because I perceive a difference between chemistry and physics over force v energy. Chemists refer to and calculate energy or energy differences but without referring to the force that creates it. I argue you can only have an energy difference as the result of a force. I think this is the source of why you may find, "you cannot calculate the energy difference with classical physics, you must use quantum theory". I perceive this to mean, we know there is an energy difference, but we haven't been able to describe the forces that actually cause the energy difference. For example, an aromatic ring has a lower energy level than predicted for a cyclohexatriene. What is the force leading to the lower energy level? (This is a very old idea. Gilbert Lewis suggested the discredited Parson's magnetron model. I think anyone able to elucidate an equation for the force between electrons at subatomic distances would receive a Nobel prize.)

Quote
The force is the negative gradient of the potential energy. You can deduce nothing about the energy from the fact that the force is 0 at equilibrium.
The acidity is determined by the Gibbs free energy difference between the protonated and the deprotonated form. you cannot make any (rigorous) conclusions whatsoever about this energy difference from the knowledge about the forces.

I think we may be in agreement here. At equilibrium, the difference in force acting on two atoms would be zero and this does not predict a result of a collision with another atom. My analogy here is that in the gas phase, molecules can experience no difference in the forces acting on its bonds. If a collision were to occur, the bonds may break as a result. The energy required to break a bond in the gas phase can be quite different from the energy to break a bond in solution.

It should be obvious that I cannot explain my thinking on this subject. I am interested in the implications of this difference. I'll wait to see whether you even agree with this point of view or how I might understand yours.
Title: Re: Why is HF a weak acid
Post by: Irlanur on December 01, 2016, 10:44:04 AM
Quote
It is because I perceive a difference between chemistry and physics over force v energy. Chemists refer to and calculate energy or energy differences but without referring to the force that creates it. I argue you can only have an energy difference as the result of a force. I think this is the source of why you may find, "you cannot calculate the energy difference with classical physics, you must use quantum theory". I perceive this to mean, we know there is an energy difference, but we haven't been able to describe the forces that actually cause the energy difference.

Quote
I argue you can only have an energy difference as the result of a force

I don't think it is very useful to think this way. You say the the force is the cause for an energy difference. You can also say it the other way around: an energy difference leads to a force. It's just a functional relationship, which one can express in differential or integral form. The question here is not which one is "truer", but which formalism is easier to use. You can of course neglect the last roughly 100 years of research in theoretical chemistry, but let's not do this for now. What we can do (with various accuracy) quite well at the moment is calculate the energy of a chemical system where we set the positions of the nuclei as fixed (Born-Oppenheimer). We can then parametrically scan what energy we get for different nuclear positions. This gives us what we call an energy hyper-surface. The fixed nuclei are also needed for what we usually called a "structure". Usually we don't need to calculate the force to predict anything. In case we do (e.g. for IR-spectra), we can simply differentiate the energy.

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My analogy here is that in the gas phase, molecules can experience no difference in the forces acting on its bonds.
I don't understand the sentence. A force cannot really act on a bond, only on the elementary particles. or do you mean the "bonding electrons"?

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The energy required to break a bond in the gas phase can be quite different from the energy to break a bond in solution.

Yes of course, but this is simply because these are two different reactions. Let me clarify this:

Let's say we have an acid in the gas phase at very low pressure, we can then approximate the system as an ensemble of non-interacting subsystems. One subsystem is just a single acid molecule A-H. Now there is an energy difference between the states A-H and A(-) + H(+). The transition between the two is continuous. the second state in principle only exists in our minds, and what we usually calculate is the energy of a state where the proton and the base are infinitely far away from each other, so that E(A(-)+H(+))=E(A(-))+E(H(+)).

The approximation of isolated centers is pretty good in the gas phase at low pressures. It is absolutely terrible in solution. What we then usually write as A-+H+ or sometimes as A-+H3O+ is extremely far away from reality. The energy of the system and the energy differences can vary substantially if there is a solvent. One can probably say that this is one of the main reasons why Compuational Chemistry is not what we would like it to be.
Title: Re: Why is HF a weak acid
Post by: orgopete on December 08, 2016, 09:57:32 AM

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I argue you can only have an energy difference as the result of a force

I don't think it is very useful to think this way. You say the the force is the cause for an energy difference. You can also say it the other way around: an energy difference leads to a force. It's just a functional relationship, which one can express in differential or integral form. The question here is not which one is "truer", but which formalism is easier to use. You can of course neglect the last roughly 100 years of research in theoretical chemistry, but let's not do this for now. What we can do (with various accuracy) quite well at the moment is calculate the energy of a chemical system where we set the positions of the nuclei as fixed (Born-Oppenheimer). We can then parametrically scan what energy we get for different nuclear positions. This gives us what we call an energy hyper-surface. The fixed nuclei are also needed for what we usually called a "structure". Usually we don't need to calculate the force to predict anything. In case we do (e.g. for IR-spectra), we can simply differentiate the energy.

I rather liked my model of an acid. I thought it was easy to understand the forces holding a proton to the electrons of an anion and I think it is actually "easier to use". I argue it is a simple inverse square force, the further a proton is from an electron pair, the weaker the force. I think it explains why ammonia is a stronger base than fluoride. I think it generally holds, but I am interested in knowing where it fails also.

If I didn't know what the force actually was or how to calculate it, then I too would say it is unimportant. I'd consider Newton's connection of gravity and mass together to be more important than anyone who may have built a scale to weigh things. Just because you can weigh something or calculate it doesn't explain what the force was that caused it.

Anytime you determine there is an energy well, don't you just wonder what caused it? Can't you make a guess? Shouldn't everyone at least make a guess? We may not succeed as Newton had, but even simple logic tells me there must be a force for an energy well to exist. This is what Gilbert Lewis thought over 100 years ago. I think there is a lot of merit to this kind of thinking. I cannot imagine anyone telling Newton they have a scale for measuring weight (mass), so they don't need a formula to calculate it.
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The energy required to break a bond in the gas phase can be quite different from the energy to break a bond in solution.

Yes of course, but this is simply because these are two different reactions. Let me clarify this:

Let's say we have an acid in the gas phase at very low pressure, we can then approximate the system as an ensemble of non-interacting subsystems. One subsystem is just a single acid molecule A-H. Now there is an energy difference between the states A-H and A(-) + H(+). The transition between the two is continuous. the second state in principle only exists in our minds, and what we usually calculate is the energy of a state where the proton and the base are infinitely far away from each other, so that E(A(-)+H(+))=E(A(-))+E(H(+)).

The approximation of isolated centers is pretty good in the gas phase at low pressures. It is absolutely terrible in solution. What we then usually write as A-+H+ or sometimes as A-+H3O+ is extremely far away from reality. The energy of the system and the energy differences can vary substantially if there is a solvent. One can probably say that this is one of the main reasons why Compuational Chemistry is not what we would like it to be.
Two things. Yes, the difference between solution and gas phase energy is what I am interested in. Pauling's electronegativity theory is based upon bond energy as a sum of covalent and ionic portions. I agree there is a difference between gas and solution phase. By a conservation of energy principle, one should be able to make a calculation. I want to know how to do it. HF seems like an ideal example. It is small, I am pretty familiar with the properties of the rows and columns. It is at an end of an extreme which can be helpful to understanding or discovering any principles. So why is the enthalpy of neutralization of HF -68 kJ/mol while H3O+ is -57 kJ/mol? Who determined -68 kJ/mol? If HF is a weaker acid, than I don't see why more energy should be released. However, I find these calculations complicated.

The second thing is that it seems paradoxical that I should rely upon computational methods and that I should also realize that computational chemistry is not what we would like it to be.
Title: Re: Why is HF a weak acid
Post by: orgopete on December 08, 2016, 10:47:43 AM
I think I understand the Born-Haber cycle for NaCl. You can also find a value for a solution phase Born-Haber cycle and it is slightly lower. I reason that in a solution phase Born-Haber cycle, no bond is formed between NaCl so the net energy is mostly the energy of the redox reaction. By a conservation of energy principle, there shouldn't be any difference in the net energy of a reaction even if a different mechanism. I think I'm still running up against the  “… bond dissociation energies refer to molecules in the gas phase and aren’t directly relevant to chemistry in solution.” McMurry, 8th edition of organic chemistry. McMurry still has some examples. The ones that are gas phase an analogous to the measurement of the heats of formation, I expect are reasonable. Those that are entirely ionic is what I'm expecting McMurry is referring to. It doesn't seem to me that the values shouldn't be reliable from a conservation of energy point of view, it is simply a matter of how.
Title: Re: Why is HF a weak acid
Post by: Irlanur on December 09, 2016, 09:11:14 AM
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I rather liked my model of an acid.
Well. I don't find this very fruitful and will also leave the thread at this point.
Title: Re: Why is HF a weak acid
Post by: Scitalk on December 10, 2016, 04:51:23 PM
Great question. Here's my shot at it. Fluorine is very electronegative which makes it the ideal halogen for having a very acidic hydrohalide( HX). However, fluorine has a little problem when it comes to acidity...and I do mean "little"..its size doesnt allow for stabilization of the added electron from hydrogen. After dissociation in a aqueous solution, the H+ and -F are on their own. -F is too unstable because its not successfully delocalizing that electron over its relatively small nucleus. In order for reactions to make sense, they end game is stability. Just remember stability governs reactions even LIFE.  :)
Title: Re: Why is HF a weak acid
Post by: orgopete on December 11, 2016, 02:49:28 AM
Great question. Here's my shot at it. Fluorine is very electronegative which makes it the ideal halogen for having a very acidic hydrohalide( HX). However, fluorine has a little problem when it comes to acidity...and I do mean "little"..its size doesnt allow for stabilization of the added electron from hydrogen. After dissociation in a aqueous solution, the H+ and -F are on their own. -F is too unstable because its not successfully delocalizing that electron over its relatively small nucleus. In order for reactions to make sense, they end game is stability. Just remember stability governs reactions even LIFE.  :)
Although this is a little different than the kind of data I'm seeking, I'll try to explain. 10 electrons is the most stable state of fluorine. If comparing HF and F(-), each has 10 electrons in each so I don't imagine any instability. Boron, carbon, nitrogen, oxygen, fluorine, neon, sodium, and magnesium are all in their most stable states with 10 electrons. Except for boron, these atoms are all less stable if they have less than 10 electrons. Adding an electron to fluorine increases its stability. Adding an electron is a redox reaction and is a different reaction with different energies than the forces holding a proton to an electron pair (my question). I further argue that for methane, ammonia, and water, the electrons are arranged in pairs in a tetrahedral arrangement. By analogy, HF and F(-) should have the same structure. If this is so, then all electron pairs may be treated equally and delocalization shouldn't change the structure of the atom or the basicity of fluoride (or any atom). Basicity is a property of an electron pair. I argue basicity should be an inverse square force, the closer, the stronger. If you compare the bond lengths, CH>NH>OH>HF. The nuclear charge of carbon is less so its electron pair extends further from its nucleus than the others, hence its electron pair should be more basic to a proton. By analogy, HF would be the least basic of this row.

If this model holds (and I think it does), then it should hold for all atoms. For the series BH4(-), CH4, NH3, H2O, and HF, they all contain 10 electrons so the difference in forces should be found in nuclear-electron pair, nucleus-proton, and electron pair-proton distances. We do not know the locus of electrons, but I argue they should still follow the laws of physics, namely inverse square forces. Anyone can do the calculations as I have and discover HF should be the weakest acid. It should be weaker than HI because of the greater charge of iodine pulls its valence electrons proportionately closer to its nucleus increasing its nucleus-proton repulsion and reducing the electron pair-proton attraction or acidity.

I know how to calculate the energy differences in reactions. What I do not know how to do is to adjust for the various changes that must be made to accommodate for the difference between gas phase and solution phase energies. For example, HF is a strong bond in the gas phase but weaker in solution. HCl, HBr, and HI are similar. The loss of an electron by sodium is endothermic in the gas phase but exothermic in solution. We should still be able to calculate the energy differences in reactions. Acidity is a solution based force and is different from gas phase forces. How do we adjust for the values to be used?