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Topic: Electrons in an acid-base reaction?  (Read 2275 times)

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Offline QuiteThePredicament

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Electrons in an acid-base reaction?
« on: January 24, 2019, 02:04:02 PM »
In this reaction, HF + NaOH :rarrow: NaF + H2O , when the F separates from the H, does it give back the electrons it had accepted to the H and then accept the one from the Na when forming NaF? Or does it leave with the electron it took, as F- ion, then forming NaF with the Na+ ion instead?         
In the latter situation, F- and Na+ have complete octets already, so there would be no electron exchange between them right? Then how would there be an ionic bond?     

Offline chenbeier

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Re: Electrons in an acid-base reaction?
« Reply #1 on: January 24, 2019, 03:40:48 PM »
The second one takes place  + and - touch together and get the ionic bond. Na+ and H+ will be exchanged. The same like in normal cooking salt NaCl.

Offline QuiteThePredicament

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Re: Electrons in an acid-base reaction?
« Reply #2 on: January 24, 2019, 04:23:00 PM »
The second one takes place  + and - touch together and get the ionic bond. Na+ and H+ will be exchanged. The same like in normal cooking salt NaCl.
     
I see. Few more simple questions, when H+ and OH- bond to yield water, H+ has no electrons to share in the covalent bond, so does the OH- has to supply both the electrons? Does that make it a dative bond?   
And the reason HF being a Lewis acid is that, H+ receives electrons from the OH- anion while F- doesn't exchange any electrons therefore making the molecule an electron acceptor?

Much appreciated.

Offline Enthalpy

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Re: Electrons in an acid-base reaction?
« Reply #3 on: January 25, 2019, 11:50:23 AM »
Do you mean the reaction between the aqueous solutions?

Then you have been mislead by the overly simplified writing of HF and NaOH. Their not too concentrated aqueous solutions are already dissociated as solvated H+, F-, Na+ and OH-. The only reaction is between H+ and OH- to produce H2O.

I suppose that without water, the reaction between gaseous HF and solid NaOH happens, but then HF is not ionic.

Offline QuiteThePredicament

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Re: Electrons in an acid-base reaction?
« Reply #4 on: January 25, 2019, 12:24:29 PM »
Do you mean the reaction between the aqueous solutions?

Then you have been mislead by the overly simplified writing of HF and NaOH. Their not too concentrated aqueous solutions are already dissociated as solvated H+, F-, Na+ and OH-. The only reaction is between H+ and OH- to produce H2O.

I suppose that without water, the reaction between gaseous HF and solid NaOH happens, but then HF is not ionic.
     
   
Yes, I meant that. When the aforementioned reaction between OH-(aq) and H+(aq) happens, does the OH- share two of its electrons with H+ to create the covalent bond? If it does, would that make it a coordinate covalent bond?     
Also, in the reaction between gaseous HF and solid NaOH to form NaF and H2O, could we say the F- or the Na+ doesn't undergo through any electron exchange after they split from the H+ and F- respectively? Would it be the same as the reaction that occurs between their aqueous solutions, but the spectator ions Na+ and F- also come together to form the NaF salt? But then, Na+ has no valence electrons to give and F- doesn't want to take any.     
   


Offline chenbeier

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Re: Electrons in an acid-base reaction?
« Reply #5 on: January 25, 2019, 02:02:57 PM »
If you have the ions, the fish is already eaten. The electron exchange took place aready.

You have Na+ and F- which swimming in the aquaeous  solution, + and - touch together and crystals are formed.

Offline QuiteThePredicament

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Re: Electrons in an acid-base reaction?
« Reply #6 on: January 25, 2019, 02:23:21 PM »
If you have the ions, the fish is already eaten. The electron exchange took place aready.

You have Na+ and F- which swimming in the aquaeous  solution, + and - touch together and crystals are formed.
     
What about the OH- giving two of the electrons when forming the covalent bond with H+? Is it true?

Offline chenbeier

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Re: Electrons in an acid-base reaction?
« Reply #7 on: January 25, 2019, 02:40:03 PM »
It  is the same The electrons from both Hydrogen are catched by the Oxygen.

The exchange can take place in this way

As 2 H2 + O2 => 2 H2O

and

H2 + F2 => 2 HF

or 2 Na + 2 H2O  => 2 NaOH + H2

and

NaOH + HF => NaF + H2O

The reult is in both case the same. The electron exhange takes place, if the elements react each other.

Offline QuiteThePredicament

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Re: Electrons in an acid-base reaction?
« Reply #8 on: January 25, 2019, 03:51:25 PM »
It  is the same The electrons from both Hydrogen are catched by the Oxygen.

The exchange can take place in this way

As 2 H2 + O2 => 2 H2O

and

H2 + F2 => 2 HF

or 2 Na + 2 H2O  => 2 NaOH + H2

and

NaOH + HF => NaF + H2O

The reult is in both case the same. The electron exhange takes place, if the elements react each other.
     
   
In the equation 2 Na + 2 H2:rarrow: 2 NaOH + H2, Na gives its single valence electron to OH and forms an ionic bond. In  H2 + F2  :rarrow: 2 HF, the F2 splits and shares one electron with H to form the HF molecule. In 2 H2 + O2  :rarrow: 2 H2O Hydrogen atoms split from each other, each having one valence electron, then bind to the oxygen atom, with oxygen providing one electron and the hydrogen providing the other.     
     
But in NaOH(s) + HF(g) :rarrow: NaF(s) + H2O(l) reaction;   
I) HF splitting into H+ and F- and NaOH into Na+ and OH- before the formation of water and salt, then the H+ that split from the HF has no electrons to share with OH- so the oxygen has to provide both of the electrons to form the covalent bond. Is this true or not, this is precisely what I'm asking.     
II) You said there is an electron exchange, but when the NaF(s) forms, Na+ and F- ions are already stable. How do they exchange electrons? Or do they form an ionic bond without the electron exchange?

Offline mjc123

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Re: Electrons in an acid-base reaction?
« Reply #9 on: January 28, 2019, 07:09:47 AM »
i. Yes
ii. Ionic bonding arises from the electrostatic interactions between ions. "Electron exchange" as such has nothing to do with it. Whether the ions were already there (as when MX(s) precipitates from MX(aq) solution) or whether they are formed by electron exchange between neutral atoms (as when MX is formed by the reaction of M and X2) makes no difference to the bonding in MX. Likewise the covalent O-H bond in H2O is exactly the same whether it was formed from H+ and OH- ions, or from H· and OH· radicals. In the former case, of course, both electrons must come from OH-, and in the latter, one from H· and one from OH·, but the bond is the same.

Offline Enthalpy

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Re: Electrons in an acid-base reaction?
« Reply #10 on: January 28, 2019, 08:37:56 AM »
RT water dissolves 40g/L of NaF. If the solutions are a bit dilute, no crystals form. Na+ and F- stay in solution and do nothing at all.

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In a hypothetical reaction between gaseous HF and solid NaOH, I have no idea of the processes, but would not suppose the formation of ions. Ions form in solutions and in solids. Don't expect important amounts in gases at reasonable temperatures.

Ions take much energy to form, because of the attraction. They exist in solutions and solids because there, they are immediately surrounded by ions of the opposite sign, or at least by polarized molecules. A lone H+ or F- is ultra-rare in a gas. Around 6000K, such species get more abundant.

So writings like Na+ are extremely simplified and misleading. Keep in mind that other atoms with at least partial charges surround it, to avoid misinterpretations.

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