If $\ce{NaCl}$ is such a strongly bonded ionic compound, why it does it disassociate so readily in $\ce{H2O}$? One would think that because of the very strong ionic bond it would not disassociate it all. Further, why do other compounds like $\ce{AgCl}$ or $\ce{HF}$ not disassociate as readily even though they also have strong bonds?

Although others have marked this as duplicate, I prefer the wording of this question as it explicitly refers to NaCl has having a strong bond (which includes the search term I used to find this website). I have been writing a manuscript in which I have been attempting to answer this question (among others). The real answer becomes a bit complicated, but I will try to give a short answer.

It is my opinion that thermodynamic data is not always properly understood. The first thing we should understand is that the same thermodynamic change in energy should result whether the reaction were to take place in the gas phase (Born-Haber cycle) or in solution. This is a principle of conservation of energy, Hess's Law. In the Born-Haber Cycle, the sum of the energies for vaporization, ionization, and lattice energies are determined. If a hypothetical solution phase Born-Haber calculation were to be performed, the same amount of energy would exchange, except for the solvation energy, which is small. That is what the poster was asking.

If you were to think about the conditions of a solution phase Born-Haber cycle, then you would take the elements and convert them directly to solvent separated ions. What this should tell you is that you have just calculated the solution phase energy of a redox reaction of sodium and chlorine to give sodium and chloride ions. This bypasses the implication of sodium chloride having a strong affinity for its ions. Yes, it would take a lot of energy to separate sodium and chloride ions in the gas phase. That is why the lattice energy is so large, but if we did the reaction in solution, we will not have a gas phase lattice energy component in our calculations. Of course this seems plausible as the solution phases of HCl and NaOH are ionic. This is consistent with our expected properties of HCl to act as an acid (chloride having a low affinity for cations) and NaOH to act as a base (sodium having a low affinity for electrons). The sum is we should not expect NaCl to have a strong affinity (not none) for the ions of the opposite charge.

By the way, I think you will also find a good correlation of heterolytic solution phase bond strength with bond length (actually, ion gap). This is different from the homolytic bond energies you may find reported in many tables, including the poster's question, "Why does NaCl dissolve in H2O despite its strong ionic bond?"


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    $\begingroup$ Dr OChem's edit of this question is an answer and should have been posted as such, not as an edit to the question. :-( $\endgroup$
    – MaxW
    Sep 10, 2016 at 19:45

1 Answer 1


The electrostatic forces between the positive cations and anions are strong. Coulomb's Law expresses the potential energy of two charges separated by a distance.

$$E_p = \frac{Q_1Q_2}{4\pi \epsilon_0 r} $$

As you may already know, water molecules stick to each other through dipole-dipole forces due to the permanent dipole that water faces due to the high electronegativity of the oxygen atom.

As ionic solids are added to water, water molecules proceed to surround each ion on the surface of the solid, forming a sphere of hydration. In the process, ions are separated from each other.

The $\delta^-$ charge on the oxygen atoms of water are attracted to cations and inversely, repels the $\delta^+$ hydrogen atoms. Thus, for cations, the oxygens of water point inward, and for anions, the hydrogens face inward respectively. The most important thing is that the ion-dipole interactions and separation of ions with little change in energy.

We can relate the potential energy of the ions to the two partial charges of a polar molecule like water:

$$ E_p \propto - \frac{|z|\mu}{r^2}$$

Z is the charge number of the ion and $\mu$ is the dipole moment of the polar molecule. Potential energy is lowered by the interaction between the solvent molecules and the ion. The $r^2$ term indicates that the interaction between ions and dipoles depends more on distance than the charges between two ions.

Thus, for hydration to occur, ion-dipole interactions must occur at the surface of the ion, and thus, ion-dipole interactions are strong for small, highly charged ions such as $\ce{Mg^{2+}}$, $\ce{Li^{2+}}$ etc.

$\ce{AgCl}$ is very slightly soluble in water and will not dissociate into it's ions. $\ce{HF}$ is a weak acid thus it does not deprotonate easily.

  • $\begingroup$ Ok, this is what I understood so far based on your answer and the link in the comment to my question. When NaCl is added to water, the electronegative oxygen is attracted to the Na+ and the electropositive hydrogen is attracted to the Cl-. However, this is still not strong enough to break the ionic bond so external energy is needed and this is thus an endothermic reaction. However, the reaction is still spontaneous because there is an increase in entropy. $\endgroup$
    – 1110101001
    Mar 29, 2014 at 3:13
  • $\begingroup$ @user2612743 Yes, looking at spontaneity is one way of looking at the reaction. Another way is too look at potential energy as matter will always try to achieve the lowest possible energy state. $\endgroup$
    – user3735
    Mar 29, 2014 at 3:17
  • $\begingroup$ How does being disassociated have a lower energy state than remaining as a solid? Because doesn't becoming disassociated requires an addition of energy, making it have a higher energy state? $\endgroup$
    – 1110101001
    Mar 29, 2014 at 3:21
  • $\begingroup$ @user2612743 Yes, in general that is correct. However, the partial charges of the water molecules replace the charges of the ionic neighbors and thus there is a minimal change of energy. In looking at the potential neergy between the full charge of an ion and the two partial charges of a polar molecule, we can relate it through the proportion above. As the solvent molecules and the ions interact, the potential energy is indeed lowered. $\endgroup$
    – user3735
    Mar 29, 2014 at 3:25
  • $\begingroup$ @Jun-Goo Kwak, CaCl2 gives off heat when it dissolves. I would say that there is a significant change in energy. Also, When KI dissolves, the water cools down at least 10 degrees. $\endgroup$
    – LDC3
    Mar 29, 2014 at 3:50

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