# Dissociation of HCl in aqueous solution

My question is regarding $\ce{HCl}$. In my book it states that:

$$\ce{HCl (aq) -> H+ (aq) + Cl- (aq)}$$

I understand that when added to water the $\ce{H}$ leaves its electron to the $\ce{Cl}$ atom forming the $\ce{Cl-}$ and the $\ce{H}$ attaches to water to form the hydronium ion ($\ce{H3O+}$). Alltogether:

$$\ce{HCl + H2O -> H3O+ + Cl-}$$

Why does this dissociation actually occur scientifically? What causes the $\ce{H}$ to leave its electron behind to the $\ce{Cl}$ and form the hydronium ion? Why does water ($\ce{H2O}$) take the additional $\ce{H}$ ion? What is the general concept here in dissociation? Electronegativity?

Water is a polar species in which the oxygen atom in $\ce{H2O}$ has a partial negative charge and the two hydrogen atoms have partial positive charges. As you suggested, this is because oxygen is highly electronegative (i.e. it has a high affinity for electrons) and so the negative electrons tend to have a higher density around the oxygen atom than the hydrogen atoms, creating this separation of charges due to water's molecular structure. Take a look at the molecular structure of water, which can be shown below. It has an asymmetry bent geometry and therefore the charge distribution in water is also not symmetric.

The hydrogen in $\ce{HCl}$ does not actually give its electron to the chlorine atom. Rather, the more electronegative chlorine ion is attracted to the partially positive hydrogen atoms in water, and the hydrogen ion in $\ce{HCl}$ is attracted to the partially negative oxygen atom in water. This separates the $\ce{H+}$ from the $\ce{Cl-}$ in $\ce{HCl}$ so that the acid dissociates into its respective ions. You therefore get the equation $\ce{HCl -> H+ + Cl-}$ when in water. If you include water on the reactant side, it can be equivalently written as $\ce{HCl + H2O -> H3O+ + Cl-}$. Here, $\ce{H3O+}$ is simply indicating that the $\ce{H+}$ ion is in an aqueous solution.

• Thanks so much for your answer. Yes, it was late I corrected the question format. What I still dont get is how the Hydrogen in HCl becomes an ion and vice versa for the Cl atom. To me NaCl which is ionic is an easy concept since the sodium already transferred an electron to the Cl (when they break apart they become Na+ and Cl-), that I get. But in the HCl how exactly does the hydrogen become and ion and how does the Cl become an ion? The fact that the Cl is attracted to the partial delta chrage hydrogen still doesnt explain how it became an ion (Cl-) to begin with considering it is covalent. – Atticus283blink Jun 28 '16 at 19:46
• You are very welcome! As for your follow-up, HCl is a strong acid. This means that in solution, it donates a proton (the hydrogen ion) to water to form the hydronium ion. The electron that you think of as originally being on the hydrogen in HCl is left behind on the Cl atom, making the hydrogen atom become the H+ ion and the chlorine atom become the Cl- ion. Look at the image on this Wikipedia article about acid dissocation for a helpful graphic using acetic acid as an example. – Argon Jun 28 '16 at 19:58
• I have deleted your first para because the errors you mentioned in your answer has been rectified by Jan. – Nilay Ghosh Jun 30 '16 at 10:41
• @AndrewR. Thanks for your additional feedback, now that make sense. Just a follow up, how exactly is water asymmetrical ? This may seem ignorant due to my lack of knowledge but if one were to draw a line through a water molecule it appears perfectly symmetrical. Is it asymmetrical due to the distribution of electrons and if so, how? – Atticus283blink Jun 30 '16 at 19:37
• @AndrewR. Any ideas? – Atticus283blink Jul 4 '16 at 4:09

The water molecule has a 3D structure based on a $$\ce{AX2E2}$$, which transforms into $$\ce{AX3E1}$$ for the oxonium ion (refer to Gillespie's VSEPR theory). Water has a two symmetry planes, one containing both hydrogen atoms and the tilted plane where those two atoms face each other, with in the intersection the axis permitting a Debye dipole moment of $$P = \pu{6.2e-30 C m}$$ or circa $$\pu{1.85 D}.$$ The compounds with partially ionic bounds, such as $$\ce{H-Cl}$$, for which the $$\ce{Cl}$$ atom is much more electronegative than the hydrogen atom, can be considered as a three step process:

• Ionisation step, formation of a $$\ce{H+}$$ (proton) and a $$\ce{Cl-}$$.

• Dissociation step, separation of the proton from the $$\ce{Cl-}$$ anions, note that chlorine ions keep the electronic charge because they are halogens.

• Finally, a solvation step, with the introduction of Coulomb interaction between the ions and the water molecules.

I enclose a simple model to explain this with hands...