I'm just going to try to clear up what's happening here, hopefully you'll see that the sodium ions just stay as sodium ions all the way through
First we put sodium ethanoate into water, this is an ionic compound, soluble in water, so it exists in solution as a sodium ion and an ethanoate ion.
CH
3COONa(s) ----> CH
3COO
-(aq) + Na
+(aq)
In this situation the sodium ion is quite stable, it is wrapped up in a layer of solvent water and it doesn't react with water.
Ethanoate however, can gain a proton from water to become ethanoic acid, so the ethanoate enters equilibirum with ethanoic acid (it does this because it is a weak acid so readily exchanges protons with water)
CH
3COO
-(aq) + H
2O(l) <----> CH
3COOH(aq) + OH
-(aq)
When we first added the ethanoate, there was no ethanoic acid present, in the process of reaching equilibrium, the concentration of reactants falls and the concentration of product rises.
This mean the concentration OH
-(aq) has risen and the product is alkaline!
I think you might be getting confused with the fact that the reaction taking place can also be written as
CH
3COONa(aq) + H
2O(l) <-----> CH
3COOH(aq) + NaOH(aq)
This is just the same as the previous equation I wrote but now the sodium ions are included! The confusion is arising because you see NaOH is formed and think that the sodium has reacted to form sodium hydroxide.
But remember, Sodium hydroxide is an ionic compound, very soluble in water, so it exists as ions (just like sodium ethanoate)!
If you think about what happens to the sodium through the reaction, it starts out as Na
+(aq) stable and wrapped up in water and it finishes up as Na
+(aq) still stable and wrapped up in water!
In other words it plays no part in the reaction! it is what you would call a spectator ion.
So the NaOH(aq) arises because the ethanoate reacts with water leaving OH
-(aq) meanwhile the sodium ion sits in solution, not doing anything!
Hope that helps