Na+ stay at it is. Its the spectator.
The oxid ion O 2- is the only one what reacts.
O 2- + H2O => 2 OH-
I agree that with Na2O it's the O2- of Na2O that reacts when the O2- receives an H+, making the O2 an OH-.
I might have thought one could say O2- is the base, and that Na2O is the base?
I agree O2- is the base. Would you say Na2O is not the base?
I agree the Na doesn't react with anything but it does break off from the Na2O(s) So is it that since it doesn't form anything other than solvated ions, it's a spectator ion? (So it doesn't have to be solvated ions on both sides to be a spectator ion)
And by that logic of identifying and removing spectator ions, would you say that in
HCl(aq) + KOH(aq) → H2O(l) + KCl(aq)
no doubt the (aq) should really be written split up
H+(aq) + Cl-(aq) + K+(aq) +OH-(aq) --> H2O(l) + K+(aq) + Cl-(aq)
the reaction is
H+(aq) + OH-(aq) --> H2O(l)
The Cl- and K+ are both spectator ions.
So the Bronsted Lowry Acid is H+?
My issue there is that if we say that, then we don't get two conjugate acid base pairs. We have
OH- and H2O as one pair
But for our other pair, I get H+ and blank! (that blank would be Cl- but I made Cl- a spectator ion! Since it's just a solvated ion on the RHS, and was even a solvated ion on the left too, though I see it's being a solvated ion on the RHS that is what counts to make it a spectator ion).
So what would you make of what the conjugate pairs are in
H+(aq) + Cl-(aq) + K+(aq) +OH-(aq) --> H2O(l) + K+(aq) + Cl-(aq) ?