Reaction of HF in water [closed]

Question

I was wondering, if we put $\ce{HF}$ into a glass of water the acid will be dissolved according to this reaction:

$$\ce{HF->F- + H+}$$

Now there is a proton travelling across the water.

All $\ce{H2O}$ molecules have equal chance of getting the proton so it travels through the liquid.

Now for the reaction

$$\ce{F- + H2O ->HF + OH-}$$

the bond strength between O and H is lower than the bond strength of F and H (fluorine is has more electronegative and has a smaller atomic radius than O).

According to Bronsted-Lowry theory for acids and bases the second reaction happens more often than the first one. But how is this happening? Why does the second reaction happen at all? Oxygen is more negatively charged than fluorine so the proton would have a very small chance to reach fluorine and then the electrons would be rearranged to form the HF bond.

Answer 1

The reason why $\ce{HF}$ doesn’t tend to disassociate in aqueous solution can be rationalized as follows:

In aqueous solution, all ions (cation or anion) are immediately solvated by a solvent “shell” that stabilizes the ion by drawing some charge density away from it. This stabilizes larger ions such as $\ce{Cl-}$ or $\ce{K+}$ more than smaller ions like $\ce{Li+}$ and $\ce{F-}$; as a matter of fact the solvation energy for $\ce{HF}$ is less than the energy needed for the $\ce{H-F}$ bond to break (bond enthalpy, if you will).

As such, $\ce{HF}$ doesn’t tend to disassociate in aqueous solution. It should be highlighted that pure, liquid $\ce{HF}$ is a much stronger acid than its aqueous form as the acid disassociates into $\ce{FHF-}$ and $\ce{H2F+}$ instead of $\ce{F-}$ and $\ce{H3O+}$.

EDIT 7 MAY

It appears that what you are asking is “why the H-F bond is weaker than the H-O bond”. If that’s what you’re asking, here’s what I have to say:

We will consider the resonance integral, $\cs{Hij}$, between atomic orbitals i and j. This resonance integral can be interpreted to be the energy of an electron associated with these 2 orbitals.

The one-electron relation (Wolfsberg-Helmholz Equation) tells us that $\cs{Hij = 0.5K(Hii +Hjj) Sij}$, and it so happens that the F orbitals are more contracted and can overlap less well. In addition, since the degree of stabilisation of the bonding MO is inversely proportional to the energy difference between the AOs, we see that there is less stabilisation from the original F 2p Orbital in a MO involving HF compared to a MO involving HO, and the HF bond is weaker.