The most straightforward way to approach this would be to consider several equilibria constituting (1) dissociation of the oxide and (2) reactions between the oxide ion, hydroxide ion, and water. At equilibrium free oxide concentrations would be very low, perhaps not even measurable.

A more advanced picture would consider heterogeneous equilibria at the metal oxide surface. In aqueous suspension, most oxides exhibit various degrees of protonation at the surface. Dissolution of the oxide is a complex process involving different crystalline faces, defect cites, and so forth.

This is only dissociation  of sodiumchloride, no chemical reaction. Here we don't use the word spectator ions.

For exqmple

AgNO3 + NaCl => AgCl(s) + NaNO3.

Sodium and nitrate are spectator, because they stay in solution. Only Silver and chloride form silverchloride as precipitate.

NaCl(aq) + AgNO3(aq) ⟶ AgCl(s) + NaNO3(aq)

NaCl is soluble ionic as is AgNO3 so should be broken up as


Na+(aq) + Cl-(aq) + Ag+(aq) NO3-(aq) ⟶ AgCl(s) + Na+(aq) + NO3-(aq)

So fine I can see Na+(aq) and NO3-(aq) are solvated ions on both sides..

no phase change.

I think that's the normal definition of spectator ions that i'm used to.

Applied where the spectator ions are compound that ionise, so "in" soluble ionic compounds so i.e. as solvated ions, (or perhaps floating around without a counter-ion, still a solvated ion).   Or an ionising polar covalent compound like HCl.

You don't count the silver ion and chloride ion in the precipitate , as spectator ions. because they're not solvated ions.

So I agree with that

The solid Na2O will not dissolve as Na2O(aq) it reacts directly to form Na+( aq) and OH- (aq) ions.

I agree, we don't have Na2O(aq)   or 2Na+(aq) + O2-(aq).  It's not like HCl(which is covalent and ionises) or NaCl(a soluble ionic compound, so ionises).  It doesn't "ionise"/dissociate. And is not like NH3(aq)(covalent, but doesn't ionise).   It's Na2O(s), a crystal (or amorphous solid).
 
So in agreement so far.

Na+ is spectator because it is not changed.

The main reaction is
O2- + H+ OH- ( H2O)  => 2 OH-

You get from sodiumoxide sodiumhydroxide.

I think perhaps most would disagree with the idea that Na+ is a spectator.

I agree that Na+ hasn't changed oxidation state.  And, that a reaction occurred. But it has changed physical state, which means it took part in the reaction.

Also , if you see here

How to Write the Net Ionic Equation for Na2O + H2O = NaOH
Wayne Breslyn
https://www.youtube.com/watch?v=MLNbUC9fhr8

He speaks of the "complete ionic equation"  (which includes spectator ions)

And he speaks of the "net ionic equation" (which excludes spectator ions)

And he says there are no spectator ions, so for that equation,  Na2O(s) + H2O(l) --> 2Na+(aq) OH-(aq)


I agree that Na2O does split/dissociate/ionise, as part of the reaction. But I think to be a spectator ion, it has to split before the reaction. i.e. it has to be in split form as a reactant. (not just in the products).

Thanks

Yes, but I meant sodium  stays as a Na+, doesnt matter as solid or in solution. So it's a spectator. The oxide anion is in water not stable and changes to hydroxide by adding one hydrogen from the water.

Thanks.. Isn't one of the ideas with spectator ions, that you can cancel them out because they're solvated ions on both sides?

Or is your definition of spectator ion that it doesn't have to be solvated ions on both sides. Just on one side, particularly the right side?

So NaCl(s) + H2O(l) --> Na+(aq) + Cl-(aq)

would you say that Na+ and Cl- are both spectator ions on the right there?

Na+ is spectator because it is not changed.

The main reaction is
O2- + H+ OH- ( H2O)  => 2 OH-

You get from sodiumoxide sodiumhydroxide.

The solid Na2O will not dissolve as Na2O(aq) it reacts directly to form Na+( aq) and OH- (aq) ions.

The Na+ did go from solid phase (before reaction)  as it's part of Na2O(s).

To aqueous phase (after reaction), where it's Na+(aq). 

Isn't that a change?

Let's suppose you have MgO or CaO or Na2O

from what I understand, they are all reactive and form hydroxides.. MgO would react at 120C..  CaO would react at eg  25C  . 
 

but let's suppose the temperature was low enough that they don't react.

So presumably they're then insoluble.  (in the insoluble range) e.g. sparingly soluble or practically insoluble.

And you just get crystals?

To the extent that they are soluble , would you get solvated O^2- anions?

And if there's O^2- anions there there'd be, in the case of Na2O,   Na+ cations too?  (albeit at tiny levels we don't factor in).

In an earlier post I wrote

" H2O(l)  + Na2O(s)  --> Na+(aq) + Na+(aq) + OH-(aq) + OH-(aq)

Writing Na2O(s) as aq (and spiltting it up since it's ionic), Na+(aq) + Na+(aq) + O^2-(aq)     (And they would split before reacting so that's fine)

Then certainly there's no phase change, it's clear the Na+ ions  are spectator ions

H2O(l) + O^2-(aq) --> OH-(aq) + OH-(aq)
"

I meant

taking

H2O(l)  + Na2O(s)  --> Na+(aq) + Na+(aq) + OH-(aq) + OH-(aq)

and writing

H2O(l)  + 2Na+(aq) + O2-(aq)  --> Na+(aq) + Na+(aq) + OH-(aq) + OH-(aq)



But The dissociation of Na2O is part of a reaction

Without a reaction, it wouldn't be dissociating.  (or at least not to an extent that a chemist would be interested in).

So, is it wrong to have said that Na+ is a spectator ion there?

And was it wrong to have split up Na2O(aq) into 2Na+(aq) + O2-(aq) ?

'cos e.g. there are no solvated O2- anions. (other than perhaps in the middle of a reaction so then it's not a reactant).

Like the state symbol for insoluble ionic compounds (not yet reacted), (or any insoluble compound) is not (aq). And it's (s) here.  So no solvated ions in the reactants. So no spectator ions?

But then also

Would you then say there's no bronsted lowry reaction? 

'cos if I remove the Na+ then   Na2O + H2O --> 2Na + 2OH-      becomes O2 + H2O --> OH- + OH-

But if I can't cancel out the two Na+  as spectator ions, then I can't get O2 + H2O --> OH- + OH-   So then I can't identify a bronsted lowry reaction


Thanks