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Given a 1 M concentration of $\ce{HCl}$ and $\ce{NaOH}$ here are the pHs:

pH of $\ce{HCl}$: 1

pH of $\ce{NaOH}$: 13–14

and of course water has a pH of 7.

Here is a reaction that is most likely to occur:

$\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$

However I think that other reactions are theoretically possible.

Like for example theoretically sodium oxides could form from the $\ce{OH-}$ taking a proton from another hydroxide anion thus giving you $\ce{O^2-}$ which could react with the sodium to give you $\ce{Na2O}$ which is a sodium oxide.

I also think that theoretically $\ce{Cl-}$ could steal electrons from the $\ce{OH-}$ since $\ce{Cl}$ can exceed an octet of electrons. This would give you neutral $\ce{OH}$ which could then react with another chloride anion to give you $\ce{OHCl-}$(not sure if formula is written right or if it would be negatively charged) or in other words chlorine hydroxide. The proton could then be taken by another hydroxide anion giving you $\ce{OCl}$ also known as chlorine monoxide which could react with another chlorine to give you $\ce{OCl2}$ which is dichlorine monoxide.

Do these theoretical reactions that lead to oxyanions($\ce{O^2-}$ to be specific), chlorine oxides, chlorine hydroxides, and sodium oxides actually happen to some extent when you have $\ce{HCl}$ and $\ce{NaOH}$ in aqueous solution

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Given a 1M concentration of $\ce{HCL}$ and $\ce{NaOH}$ here are the pHs:

pH of $\ce{HCL}$: 1

pH of $\ce{NaOH}$: 13-14

First, a 1 molar concentration of hydrochloric acid solution, $\ce{HCl}$, has a pH of 0.0 if we take ${-log[H_3O^+]=pH}$ to be true. (And it is mostly true for dilute solutions).

and of course water has a pH of 7.

Pure water at 25 degrees Celsius has a pH of 7.0. Boil the water and the water will no longer have a pH of 7.0. Cool the water and pH will similarly change. The only condition for neutrality is that the concentration of hydronium ion equals the concentration of hydroxide ion. It just so happens that water at normal lab conditions has a hydronium ion and hydroxide ion concentration of ${1.0*10^{-7}}$ M, which corresponds to a pH of 7.0. Your statement only holds at a specific temperature. At other temperatures, this reaction - the auto-ionization of water - which controls the pH of pure water - will go to varying extents:

$\ce{2H_2O <=> H_3O^+ + HO^-}$

Here is a reaction that is most likely to occur:

$\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$

Since we are talking about solutions of HCl and NaOH, your equation isn't reflective of what's actually going in on the system. HCl and NaOH both dissolve in solution (unless the solution is saturated with either). So you don't have HCl or NaOH in solution, but rather hydronium ion and chloride anions in the case of HCl and sodium ions and hydroxide ions in the case of NaOH solution. A more accurate equation would be:

$\ce{H_3O^+ + HO^- ->2H_2O}$

Regarding:

$\ce{NaCl(aq)}$

Sodium chloride has a very high solubility in water and dissociates unless the solution is saturated. So it would be more accurate to write simply sodium ions and chloride ions, instead of "aqueous" sodium chloride (unless you are referring to the ion-pair effect). Ion pairs often occur in solution; the name describes exactly what is implied - two oppositely charged ions are attracted to each other. For sodium chloride, ion pairing is likely minimal, but in other cases, such as between a carbocation and a halide anion, the ion pair effect has noticeable effects on chemical reactions (i.e. partial racemization rather than full racemization).

Like for example theoretically sodium oxides could form from the $\ce{OH-}$ taking a proton from another hydroxide anion thus giving you $\ce{O^2-}$ which could react with the sodium to give you $\ce{Na2O}$ which is a sodium oxide.

So hydroxide ion reacts as an acid and abstracts a proton from another hydroxide ion.

This is at least theoretically possible, but thermodynamically unfavorable. The oxide anion is very unstable in water solution to the point of being able to react with the solvent (water) itself to a large extent.

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  • $\begingroup$ but the oxide anion could also react with the positive sodium ions and form sodium oxides and the chlorine could react with neutral OH to form chlorine hydroxide which can react with another hydroxide anion and chlorine oxides can form. $\endgroup$ – Caters Aug 3 '14 at 1:25
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    $\begingroup$ @caters, the reaction you suggest of $\ce{2NaOH -> Na2O + H2O}$ will not occur spontaneously under any reasonable conditions. Using Hess's law and standard thermodynamic data on enthalpies of formation, the reverse reaction is highly exothermic: $\Delta H^{\circ}_r \approx -151.9 \mathrm{\frac{kJ}{mol}}$. This is not even factoring in enthalpies of solvation, while the entropy change will be small either way. Intuitively, this should make sense. It should not be possible to spontaneously react two highly basic molecules to make an even stronger conjugate base and a stronger conjugate acid. $\endgroup$ – Greg E. Aug 3 '14 at 3:21
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    $\begingroup$ Chlorine hydroxide is likely not a product. Especially not through the mechanism you propose (through a "neutral OH" or OH radical). The hydroxyl radical does not form spontaneously except under extremely high temperatures. Regarding the acid/base stuff - I would consult an introductory chemistry text. $\endgroup$ – Dissenter Aug 3 '14 at 4:42
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    $\begingroup$ @caters, I'm referring to the fact that in your proposed reaction $\ce{2NaOH -> Na2O + H2O}$, you have hydroxide acting as an amphoteric molecule, one of the hydroxides a base, the second one an acid. The resulting oxide is a stronger base than hydroxide, and water is a stronger acid than hydroxide. I wasn't even referring to your hypothetical chlorine equilibria. $\endgroup$ – Greg E. Aug 3 '14 at 5:32
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    $\begingroup$ @caters, you can actually get an approximation of the (idealized) thermodynamic favorability of these reactions by calculating $\Delta G^{\circ}_r$ from standard molar entropies and enthalpies of formation with Hess's law. NIST is a good source for the thermodynamic data. $K_{eq} = e^{-\frac{\Delta G^{\circ}_r}{RT}}$. A quick calculation on your proposed reaction gives $K_{eq} \approx 1.28 \cdot 10^{-26}$, and that would be a generous estimate as it fails to account for solvation and aqueous equilibria. $\endgroup$ – Greg E. Aug 3 '14 at 18:10

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