Wouldn't this be more stable if they stayed as neutral molecules?

I've learnt that they will seek to reach the ratio of $[A]^a[B]^b/[C]^c$ that matches the $\ce{Ka}$ or $\ce{Kb}$ value.

But I do not understand why they wouldn't prefer to stay as uncharged molecules as I thought ions are unstable compared to neutral molecules?

  • 6
    $\begingroup$ Hint: Ions in a vacuum are very different from ions in solution. $\endgroup$ Feb 3, 2017 at 6:34
  • 5
    $\begingroup$ Hint2: Take a look at lattice energy of ionic compounds. Take a look at the definition of lattice energy. $\endgroup$ Feb 3, 2017 at 7:01
  • 3
    $\begingroup$ Hint3: Not everything ionizes when it dissolves. Salt (sodium chloride) does, but sugar doesn't. $\endgroup$
    – MaxW
    Feb 3, 2017 at 8:00
  • 1
    $\begingroup$ The answers already given here for a slightly different question should (dis)solve your problem: Why don't polar and non-polar compounds dissolve each other? $\endgroup$
    – Autumn
    Feb 3, 2017 at 14:51
  • $\begingroup$ MaxW, I disagree. The pka of, glucose for example, is 12. The pka of water is 14, or 15.7 depending on your reference. Either way.. Glucose will give up a proton - thus ionization occurs. $\endgroup$
    – Bob
    Feb 3, 2017 at 23:07

2 Answers 2


While a picture as in Bob’s answer is sometimes better than a thousand words, it may also help to add the thousand words[1] to explain what the picture is supposed to be showing or why it is relevant.

Water is a very polar solvent. The electronegativity difference between oxygen and hydrogen is rather high and the bent shape of the molecule additionally allows a higher dipole moment. Therefore, you can think of water molecules having a decidedly negative side — the oxygen atom — and a decidedly positive side — the space between the hydrogen atoms. In pure water, the molecules will arrange (at first approximation) in such a way that the positive sites are close to the negative ones and vice-versa.

At second approximation we need to introduce hydrogen bonds, which go one step further. Remember how each single hydrogen atom is strongly positively charged due to the electronegativity difference/the polar bond? Well, because hydrogen does not have any core electrons, this means it can act like a ‘slightly weakened proton’ and interact well with lone pairs on electronegative atoms such as oxygen. Thus, a better description of the intermolecular interactions in water molecules is that hydrogen bonds are formed from each hydrogen to one of another molecule’s oxygen lone pairs. These interactions are pretty strong for intermolecular interactions since they are caused by rather large charge differences (and need to stabilise these).

After having considered the water molecules by themselves, what happens if we submerge ions into them? In solid state, ionic compounds form extensive networks of cations/anions approximately alternating. Each cation will typically only have anions as closest neighbours and vice-versa; therefore each ion’s charge is stabilised by the charges of the opposite ions close by. If you wanted to dissolve this ionic compound into something like hexane, it would not work well. It is just much better for the ions to stick together in their solid-state form and be submerged in the hexane liquid phase — as your question implies.

In water, however, we have the positively charged and negatively charged areas not unlike those of the counterions. Taking a single ion out of the ionic crystal and surrounding it with the appropriate number and orientation of water molecules as shown in Bob’s answer, means that electrostatically almost nothing is gained or lost. True, the water molecules, not being charged particles, can stabilise a little bit less at first approximation but it is still comparable.

At second approximation, we can on the one hand reintroduce hydrogen bonds gonig from water molecules to anions such as chlorine. These may stabile the anions in solution, but note that having a hydrogen bond directed towards oxygen is typically better than one to a ‘typical halide’. The cation side is much more interesting, though: oxygen’s lone pairs can interact with the cationic centre to form a coordination complex. This lowers the energy of oxygen’s electrons at the expense of unoccupied cationic orbitals — a win-win situation for both partners. Thus, we can expect additional stabilisation due to this effect. (To be fully honest, one can and should also consider most ionic crystals as chained local coordination complexes. However, these are typically better stabilised if only one ligand sphere is involved, thereby preferring the solution.)

Thus we have:

  • A net average enthalpic no-effect concerning charge stabilisation (ionic crystal versus Bob’s picture
  • A slight net enthalpic gain due to the formation of better coordination complexes
  • A strong net entropic gain from the fact that a single, highly ordered crystal is broken up into gazillions of minute entities that can move freely in solution (I did not touch this point in this answer, but it is true for all solutions).

Considering all these, the conditions for ionic compounds to dissolve in aquaeous solutions are very favourable. And Bob’s your uncle.

[1]: Well all right. 645 words then. Not quite 1000 but close enough.


enter image description here

Intermolecular forces stabilize the ions of a chemical

  • $\begingroup$ Thanks for the answer. But why is this new stability more stable than the NaCl stability from the ionic bond? And so why does the ionisation even occur? $\endgroup$
    – K-Feldspar
    Feb 3, 2017 at 23:12
  • $\begingroup$ E.g in the question in @Autumn 's comment chemistry.stackexchange.com/questions/38260/… it states "when the solute dissolves into the solvent, they are able to be make dispersion forces with each other. The making of these forces releases very little energy. So simply put, very little energy is required to break the forces and very little energy released when making the forces. Hence overall everything balances out and the process occurs." I don't understand why it gives up a stable scenario to go to another stable scenario $\endgroup$
    – K-Feldspar
    Feb 3, 2017 at 23:14
  • $\begingroup$ It has to do with the way in which the molecules would need to be arranged, in solution. If NaCl (to keep the example consistent) were to maintain its crystalline structure it would need to maintain a form described as highly ordered. In being dissolved, this "order" is disrupted. Entropy always shifts from order to disorder. $\endgroup$
    – Bob
    Feb 3, 2017 at 23:20
  • $\begingroup$ Ah okay. That makes sense then. I didn't think about entropy, only stability. $\endgroup$
    – K-Feldspar
    Feb 3, 2017 at 23:22
  • $\begingroup$ It is not more stable though. Think of the polarity of water. And the magnetic forces it would exhibit on a salt-crystal. If the crystal were to maintain shape, it somehow would need to "create" energy in a direction that opposes the forces from the water. $\endgroup$
    – Bob
    Feb 3, 2017 at 23:23

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