Potential byproducts of reacting NaOH with HCl in water

[caters]

Given a 1M concentration of HCL and NaOH here are the pHs:

pH of HCL: 1

pH of NaOH: 13-14

and of course water has a pH of 7.

Here is a reaction that is most likely to occur:

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 OH⁻ taking a proton from another hydroxide anion thus giving you O2⁻ which could react with the sodium to give you Na₂O which is a sodium oxide.

I also think that theoretically Cl⁻ could steal electrons from the OH⁻ since Cl can exceed an octet of electrons. This would give you neutral OH which could then react with another chloride anion to give you 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 OCl also known as chlorine monoxide which could react with another chlorine to give you OCl2 which is dichlorine monoxide.

 

So why does the formation of sodium oxides, chlorine hydroxides, and chlorine oxides not happen to a small extent when you react NaOH with HCl in aqueous solution?

[hypervalent_iodine]

Because none of those mechanisms make any chemical sense and/or your intermediates are too high energy.

[Sensei]

Like for example theoretically sodium oxides could form from the OH⁻ taking a proton from another hydroxide anion thus giving you O2⁻ which could react with the sodium to give you Na₂O which is a sodium oxide.

See sodium oxide website:

http://en.wikipedia.org/wiki/Sodium_oxide

 

Na₂O + H₂O -> 2 NaOH

 

Even if Na₂O would be made in solution, it would quickly react like above.

 

It reacts with hydrochloric acid AFAIK:

Na₂O + 2HCl -> 2NaCl+H₂O

Here is a reaction that is most likely to occur:

 

HCl+NaOH+H₂O->H₂O+NaCl(aq)

 

Such reaction cannot occur, because simply it's not balanced.

Edited February 23, 2015 by Sensei

[caters]

Well the reason I have H2O+NaCl(aq) instead of 2 H2O+NaCl(aq) is because the water that reacts is equal to the water that is made

[hypervalent_iodine]

You shouldn't have had water on the reactant side of that equation. It isn't reacting, so it doesn't belong there.

[caters]

Oh but it does react. In particular it steals the proton from the HCL. In other words the water is acting like a base so water goes through an acid base reaction. Then the hydroxide ion autoionizes with the hydronium ion which is essentially like water by itself going through an acid base reaction. This frees the chloride and sodium ions which then dissolve in the water that has been produced. Thus the water that reacts is equal to the water that is made. This is why I had H2O + NaCl(aq) instead of 2 H2O + NaCl(aq).

Edited May 19, 2015 by caters

[hypervalent_iodine]

Oh but it does react. In particular it steals the proton from the HCL. In other words the water is acting like a base so water goes through an acid base reaction. Then the hydroxide ion autoionizes with the hydronium ion which is essentially like water by itself going through an acid base reaction. This frees the chloride and sodium ions which then dissolve in the water that has been produced. Thus the water that reacts is equal to the water that is made. This is why I had H2O + NaCl(aq) instead of 2 H2O + NaCl(aq).

The NaOH acts as the base. The water doesn't react here and your equation doesn't make chemical sense. If you have water on both sides of your equation, it isn't part of the overall reaction. Perhaps it is involved somewhere in the mechanism, but that is not the same thing.

[caters]

Just because the NaOH acts as a base does not mean that water isn't also acting as a base. So it is completely possible to have 2 acid base reactions happening simultaneously with the same acid. So both of these reactions could be happening:

 

HCl + H2O + NaOH -> H3O⁺ + Cl^- + NaOH -> 2 H2O + NaCl(aq)

 

HCl + NaOH + H2O -> 2 H2O + NaCl(aq)

 

The top one is a 2 step reaction with water acting as a base with the HCl and then the NaOH acting as a base with the H3O+. The bottom one is a 1 step reaction with the NaOH acting as a base with the HCl and the water dissolving the ions but not participating in the overall reaction.

 

 

And just like how you can have 1 acid reacting with 2 bases, you can have 2 acids reacting with 1 base. It would be something like this:

 

3 HCl + H2SO4 + NaOH -> 3 H2O + SCl2 + O2 + NaCl

 

Or

 

6 HCl + 4 H2SO4 + 2 NaOH -> 8 H2O + 2 S2Cl2 + 5 O2 + 2 NaCl

 

Or

 

HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2S + 2 O2 + NaCl

 

Or

 

HCl + H2SO4 + 3 NaOH -> 3 H2O + Na2SO4 + NaCl

 

Even Sodium Sulfite, sodium hyposulfite, sodium dithionate, sodium metabisulfate, sodium persulfate, thionyl chloride, sulfuryl chloride, pyrosulfuryl chloride, sulfur dioxide, sulfur trioxide, sodium superoxide, sodium oxide, sodium peroxide, chlorine monoxide, chlorine dioxide, chlorine trioxide, dichlorine monoxide, dichlorine trioxide, dichlorine tetroxide, dichlorine hexoxide, dichlorine heptoxide, or chlorine tetroxide

could form as 1 of the products.

 

All of these products other than the NaCl can then participate in acid base reactions with other acids or with more of the same acids.

Edited May 19, 2015 by caters

[hypervalent_iodine]

I am really not sure how else to explain this to you. You might like to read up a little on basic chemistry and how we write reactions. The way you have written it is incorrect. Whether the water forms a hydronium ion following dissociation of HCl or whatever is irrelevant to the overall reaction. You don't have water on both sides of the equation.

[Fuzzwood]

The very first reaction emphasizes that HCl in water indeed reacts to H₃O⁺ and Cl^-, since water at that moment is the strongest base. If, however, a stronger base than water is introduced in the equation, the reaction with water really doesn't matter anymore for the end result. In this case, a water molecule might be used, and another one is produced. Net gain or loss = 0 and thus H₂O should not be in the net reaction equation for the reaction between HCl and NaOH.

 

To put it even more elegantly, the only reaction that actually matters is either H⁺ + OH^- --> H₂O or H₃O⁺ + OH^- --> 2 H₂O, whichever suits your fancy.

 

The rest of your equations are non-sensible redox equations and have nothing to do with the initial problem. Except of course for the very last one with which I totally agree.

Edited May 19, 2015 by Fuzzwood