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I have read that the actual definition of a strong acid is one which will fully dissociate in water, so please do not turn the problem on its head when answering. It is also my understanding that every acid reaction is in fact reversible (just as with any other reaction) and yet the thing that is different about strong acid forward reactions is that they are shifted so far forward (the equilibrium constant is very high) to the extent that hardly any of the reactants remain after reaction. The acid reactions follow the laws of chemical reaction kinetics and the reason for this is that the Gibbs free energy decrease of the acid reaction is fairly high and so according to a Maxwell-Boltzmann Curve of energy distributions, only a very small fraction of the molecules will have enough energy to actually revert back to their undissociated form.

My real question is what is it about the electronic (or otherwise) structure of these compounds, for example $\ce{HCl}$, as opposed to weak acids, e.g. $\ce{CH3COOH}$, that makes them more easily dissociated in water. It might also be helpful to learn the same about strong and weak bases. I think it might have something to do with the electronegativity of the ions involved.

If you can help answer the above, please could you also explain the weak/strong conjugate rule (that if a weak/strong acid/base reacts it will produce a strong/weak conjugate base/acid respectively).

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The easy answer:

It comes down to the strength of the bonds. Or that covalent bond strength is directly related to the bond dissociation energy and thus dissociation.

A little Physical Chemistry: (The following is how Pauling explained this.)

In physical chemistry, they have something called bond dissociation ($D_o$).

From Atkins (9th Ed 10.4): "bond dissociation energy, $D_o$, the energy required to separate the atoms to infinity or by the well depth $D_e$." - What is meant here by well is a potential well – for now just think of that as the 2D box talked about in physics.

Now lets shift gears. You may be familiar with bond enthalpy (sometimes just called bond energies). A common formula used in general chemistry is $\Delta H_\mathrm{rxn} = n(\text{bonds broken}) - n(\text{bonds formed})$

It turns out that $\text{bonds broken}$ or $\text{bonds formed}$ is actually referring to the bond energy or bond enthalpy.

It actually works out that there is a direct relationship to bond dissociation energy and bond enthalpy; which makes sense. It is elegantly related by:

$\Delta H^o_\mathrm{bond} = D_o + \frac32 RT $

I guess the natural question to ask now is why do some bonds have a greater enthalpy of formation?

This was not well understood for a long time. In reality there is a lot of factors that can come into play. The way that Pauling initially came up with dissociation energy, $D_o$, in The Nature of the Chemical Bond followed this protocol:

  1. Spectroscopy, (I did this with FTIR for example, but there are several choices) to get a molecular potential energy curve.
  2. Extract the well depth, $D_e$
  3. Find the dissociation Energy using $D_o = D_e - \frac12 h\nu$

In modern thermo:

Because of the relationship with enthalpy, we can measure enthalpy with calorimetry and back calculate.

But the best description uses statistical thermodynamics … Which I will not get into.

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  • $\begingroup$ It's not what OP asked - he asked why some of those bonds are weaker... Also your answer looks rather impolite. $\endgroup$
    – Mithoron
    Commented Apr 4, 2015 at 22:17
  • $\begingroup$ Was not my intention to be impolite; my apologies. Will edit this up. $\endgroup$
    – PhysicalChemist
    Commented Apr 4, 2015 at 22:32
  • $\begingroup$ @PhysicalChemist Hi, thanks for your reply! Firstly, just wanted to ask what you were referring to by the '2D box in physics'? Also, am I right in saying that when you refer to '3/2RT' this is equivalent to the kinetic energy of ideal gaseous molecules - this would therefore mean that the equation you have stated means the bond enthalpy is equal to the energy needed to separate the atoms in a stationary state accounting only for electrostatic forces + the energy needed to overcome their ideal kinetic energy? Just need some back-up on my deductions from what you wrote in that sense? $\endgroup$
    – Resquiens
    Commented Apr 5, 2015 at 20:09
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The rule is not "weak/strong conjugate rule" . The strength of an acid is expressed by its acidity constant, Ka. The higher the value of the constant, the stronger the acid. Likewise, the strength of a base is expressed by its basicity constant, Kb. For a conjugate acid-base pair, in water, it is Ka . Kb = Kw = 10*-14 (25 ºC). So, the rule is "the stronger the acid, the weaker the conjugate base and vice-versa", which is not exactly the same thing. Just try with "weak" acetic acid (Ka=10*-5); its conjugate base, the acetate ion has a basicity constant Kb= 10*-9. Which is the weak and which is the strong?

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