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:
Since the electrons are shared, why does it break into $\ce{H+}$ and $\ce{Cl-}$ in solution?
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2 Answers
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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 $\ce{ H3O+}$.
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$\begingroup$ Why is it energetically favorable for this reaction to occur? Is the enthalpy very low (negative), or is the entropy very high, or both? $\endgroup$– nullOct 12, 2014 at 6:39
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$\begingroup$ 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.cOct 12, 2014 at 6:43
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$\begingroup$ Typically the curved arrows are used for electron pair transfers. $\endgroup$ Apr 28, 2016 at 16:14
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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.
answered Oct 12, 2014 at 6:56