Maybe I am wrong, but I thought acids were covalently bonded since hydrogen doesn't form ionic compounds. For example, wouldn't $\ce{HCl}$ look like:

structure of HCl

Since the electrons are shared, why does it break into $\ce{H+}$ and $\ce{Cl-}$ in solution?


2 Answers 2


enter image description here

Because it is energetically favourable ($\Delta{G}<0$) for hydrogen chloride to react with water to form hydronium( $\ce{ H3O+}$) and chloride ions.

Remember that $\ce{ H+}$ does not exist as $\ce{ H+}$ in water, but rather as $\ce{ H3O+}$.

  • $\begingroup$ Why is it energetically favorable for this reaction to occur? Is the enthalpy very low (negative), or is the entropy very high, or both? $\endgroup$
    – null
    Commented Oct 12, 2014 at 6:39
  • $\begingroup$ Both. for most cases, dissolving a compound to form an acid releases heat (negative enthalpy). Also, entropy is increased due to the addition of chloride (and also the breaking of H-Cl bond). $\endgroup$
    – t.c
    Commented Oct 12, 2014 at 6:43
  • $\begingroup$ Typically the curved arrows are used for electron pair transfers. $\endgroup$ Commented Apr 28, 2016 at 16:14

Your are right. But you have to remember that the bond between hydrogen and chlorine (for example) is covalent polar one. When you introduce this molecule into water which is a polar solvent with a dipole moment of 1.85 D, the polarization of the molecule $\ce{HCl}$ increases. We have then two solvated ions $\ce{H_3O^+}$ and and $\ce{Cl^-}$. And as water has also a high dielectric constant $\epsilon_r=80$, the electrostatic force between $\ce{H_3O^+}$ and $\ce{Cl^-}$ will decrease by a factor of 80. So, the the bond between $\ce{H_3O^+}$ and $\ce{Cl^-}$ will weaken and we have ionization of the acid.


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