[gavindor]

When looking at this reaction H2O  + Na2O  --> 2NaOH     which I understand involves a bronsted acid and bronsted base..

It was mentioned to me that Na+ is a spectator ion

And that I should think in terms of net ionic reactions

If I start with a simpler example, a standard one

using this example AgNO3 + HCl --> HNO3 + AgCl(s)  and getting a net ionic equation from it.

Ag⁺(aq) + NO₃⁻(aq) + H⁺(aq) Cl⁻(aq) --> H⁺(aq) + NO₃⁻(aq) + AgCl(s)

The aqueous ions not in any phase change cancel. And we are left with this net ionic equation

Ag+(aq) + Cl-(aq) --> AgCl(s)

Ag and Cl are involved in a phase change, so aren't spectators.

The others(that I cancelled out), are just aqueous ions on both the left hand side and right hand side, with no phase change.. So are considered to be spectator ions and can be cancelled out.

The thing is though if I apply that to

Na₂O(s) + H₂O --> 2NaOH(aq)

The Na+ is involved in a phase change, so it doesn't seem like a spectator ion. It's going from solid form, to  solvated ions.

So I don't understand the statement of "I would not call NaOH neither Bronsted acid nor base. Na+ is just a spectator, it is OH- that is reacting."

I agree that the OH- is reacting.   I'm not sure whether one would say the bronsted base is NaOH or OH- or both.

{correction, the O2- reacting, not OH- reacting}

But if we say the NaOH isn't a bronsted base  only the OH- is, because Na+ is a spectator ion.. then I'd have that issue with saying that Na+ is a spectator ion.. especially if, as was suggested to me, that I  should "think in terms of net ionic reactions"  When I do that, then Na+ looks like not a spectator ion.

Thanks

[Hunter2]

Na+ stay at it is. Its the spectator.

The oxid ion O 2- is the only one what reacts.

O 2- + H2O => 2 OH-

[gavindor]

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)  ?
 

[Hunter2]

We dont have H+, we have H3O+

This react with OH-

The result is water.

H3O+(Acid1) + OH-(Base1) => H2O(Acid2) + H2O(Base2)

K+ and Cl- are spectator.

[gavindor]

We dont have H+, we have H3O+

This react with OH-

The result is water.

H3O+(Acid1) + OH-(Base1) => H2O(Acid2) + H2O(Base2)

K+ and Cl- are spectator.

Ah ok yes thanks that makes sense and that representation of autoionisation of water with H3O+ fits into bronsted lowry giving two conjugate pairs.  And K+ and Cl- as spectator ions..  So that explains a lot

When it comes to spectator ions in

Na₂O(s) + H₂O --> 2NaOH(aq)

which splitting up the aq is

Na₂O(s) + H₂O --> 2Na+(aq) + 2OH-(aq)

it seems i'm meant to  consider Na to be spectator ions because the Na+ on the right are aq, regardless of the fact that they were involved in a phase change and were solid on the left.

Then why is it that when we write

NaCl --> Na+(aq) + Cl-(aq)

we don't say that Na+ and Cl- are spectator ions?  (if we did there'd be no equation there!)

 Though the Na and Cl are both solvated ions on the right!

[Hunter2]

If solvated or not doesn't matter.

You mix up a chemical reaction with a normal dissociation.

NaCl => Na+ + Cl-  is dissociation. Of course here the ions the active guys. There not spectator.

Example

NaCl + AgNO3 => AgCl(s) + NaNO3

Here the main reaction is Ag+ + Cl- => AgCl (solid)

Spectator are Na+ and NO3-, because they stay in the solution without change.

[gavindor]

You mix up a chemical reaction with a normal dissociation.

fair point

I see, the concept of spectator ions applies to chemical reactions not dissociation reactions..

NaCl + AgNO3 => AgCl(s) + NaNO3

Here the main reaction is Ag+ + Cl- => AgCl (solid)

Spectator are Na+ and NO3-, because they stay in the solution without change.

If solvated or not doesn't matter.

It seems to me they have to stay solvated e.g. you say the spectator (ions)  Na+ and NO3- "stay in the solution without change" . So they are solvated ions.

I was considering

 Na2CO3 + HCl -> NaCl + H2O + CO2

that's

Na2CO3(s) + 2HCl(aq) -> 2NaCl(aq) + H2O(l) + CO2(g)

But the solid Na2CO3 will be  Na2CO3(aq) straight away when in water, and will share in the solution of the HCl(aq)

Na2CO3(aq) + 2HCl(aq) -> 2NaCl + H2O + CO2

Na+ ions and Cl- ions are spectator ions..

And they are solvated ions on LHS and RHS.

So it seems to me that spectator ions are always solvated ions on LHS and RHS.

Similarly,

Na₂O(s) + H₂O(l) --> 2NaOH(aq)

is

Na₂O(aq) + H₂O(l) --> 2NaOH(aq)

so

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

and then likewise you have Na+ ions as spectator ions. Being solvated ions on the left and right.

[Hunter2]

For me its doesn't matter still.

You can evaporate the water from the solvated NaOH and get solid NaOH.
Na+ from Na2O and NaOH is the same and not changed, even between they dissolved in water.

[gavindor]

In the previous discussion when speaking of  Na2O + H2O --> 2NaOH

I wrote " one NaOH is the conjugate acid, and one NaOH is the conjugate base?"   And you said correct..  I wasn't sure if you were saying correct to include that line.

So is it your view that it's OK to include the spectator ion Na there?  Or would you say no you really have to just say O2- is the base and leave the spectator ion out?

It seems to me they have to stay solvated e.g. you say the spectator (ions)  Na+ and NO3- "stay in the solution without change" . So they are solvated ions.

For me its doesn't matter still.

You can evaporate the water from the solvated NaOH and get solid NaOH.
Na+ from Na2O and NaOH is the same and not changed, even between they dissolved in water.

I see that there's not much trouble to go from (s) to (aq) and back

When you say the Na isn't changed, do you mean that if you look at Na2O as Na2O(s)  and you take NaOH and evaporate the water off then you have NaOH(s)  Then since you have Na ions in solid in Na2O.  And Na ions in solid in NaOH,  so then the Na isn't changed?

What would change look like?


In this case

NaCl(aq) + AgNO3(aq) -> AgCl(s) + NaNO3(aq)

The difference in phase does matter.  The AgCl  aren't spectator ions.. (the rest are).  And I figured that is because the Ag and Cl don't stay in the same phase of solvated ions throughout the reaction.

Or is it that it's not the difference in phase that matters.. Or not just the difference in phase.  It's the fact that AgCl(s) can't really be turned into AgCl(aq), because it's so insoluble? So it's the different phase plus inability to easily convert it?

Thanks

[Hunter2]

Sodium stays as Na+ in the whole process. It will not reduced. Its only spectator.
O2- changes to OH- these are not spectators.

AgCl will precipitate and goes out of the equilibrium. All others Na+, NO3- are spectators.

[gavindor]

Yes no doubt in H2O  + Na2O  --> 2NaOH,   Na+ is spectator and  O2- and OH- aren't.

And I think you wouldn't say that NaOH is a base or conjugate base.  Only OH- of NaOH is but you wouldn't also describe NaOH as a whole  as a base, right?

Thanks

[Hunter2]

I would it described as follows  in NaOH is the OH- the Base.

[gavindor]

I would it described as follows  in NaOH is the OH- the Base.

Thanks, I think  I got it..

NaOH is an arrhenius base..  not a bronsted base.. (to be a bronsted base it'd have to be a base in a conjugate pair, and it's not)

OH- is a/the bronsted base..

In this reaction  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)

O2- is a bronsted base.

Na2O  fits the less strict definition of arrhenius base https://www.khanacademy.org/science/chemistry/acids-and-bases-topic/acids-and-bases/a/arrhenius-acids-and-bases


And this one all works too HCl(aq) + KOH(aq) → H2O(l) + KCl(aq) ..  It should be written

H+(aq) + Cl-(aq) + K+(aq) + OH-(aq) --> H2O(l) + K+(aq) + Cl-(aq)

so that gives
H+(aq) + OH-(aq) --> H2O(l)

and we can write that  writing the H+ as H3O+  and adding an H2O to the right.

H3O+(aq) + OH-(aq) --> H2O(l) + H2O(l)

And we have conjugate pairs there.

This website https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Acids_and_Bases/Acid/Bronsted_Concept_of_Acids_and_Bases 

says that HCl is a bronsted acid, but really I suppose it's the H3O+ that is a bronsted acid.
(Also that site writes HCl(aq) but  from what I,  it should be written H+(aq) + Cl-(aq) )

Thanks

[Hunter2]

You got it.