[kapital]

I have two questions regarding this topic:
If we add small amount of NaOH in water, the reactions that takes place is NaOH --> Na+ + OH-. The amount of OH- increases. But if we look the concentration of H₃O+, it should be lower, after we add NaOH in water. But there is no H₃O+ in reaction. What happen with the H₃O+(amount that has gone after we add base)?

And second questions(not connected to previous): Why can addition of small base cannot ionize even very weak acids(like for example aliphatic alcohols)? Because if  we have for example some weak acid in water, and there is 0,1% of it in ionized form, when we add base(even if it is weak), we add OH- ions in solution which should react with H_4
30+ ions to form water, which would than force a little more of acid to ionize(so there would be 0,1 %ionised), and whit enough addition, the whole amount of weak acid should be ionised?

[Borek]

I have two questions regarding this topic:
If we add small amount of NaOH in water, the reactions that takes place is NaOH --> Na+ + OH-. The amount of OH- increases. But if we look the concentration of H₃O+, it should be lower, after we add NaOH in water. But there is no H₃O+ in reaction. What happen with the H₃O+(amount that has gone after we add base)?

You can't ignore another reaction - water autoionization:

H₂O H⁺ + OH^-

which will shift to the left, consuming H⁺.

Why can addition of small base cannot ionize even very weak acids(like for example aliphatic alcohols)?

It does, just the extent is so low it is negligible.

[kapital]

About the second answer, that is exatcly my question. When you add more base(for example you do the experiment like titration), with every base addon it should be more acid ionezd, which should at the end result in completly ionised acid), not only neglible.

[Borek]

If the acid is substantially weaker than the water, you have a leveling effect, and the change in ionization is negligible.

When the acid is stronger than water, it becomes more and more ionized.

This is not some kind of either-either situation, there is a whole continuum of acids, and the extent to which they ionize depends on their strength.

Propose some examples and we will see what can be said about them.

[kapital]

For example you have substituted phenol solution in water(0,1 M, pkA = 15). Very small amount is in ionised form, other is an-ionised. You add one milliliter of ammonium (0,1 M). With that you add some amount of OH- ions, so the reaction OH- + H₃0+     2H₂O would happen, which would then force a little more ionisation of phenol. And so by adding more ammonia all acid should be ionised.

[Borek]

For example you have substituted phenol solution in water(0,1 M, pkA = 15). Very small amount is in ionised form, other is an-ionised.

Assuming pKa of 15, 0.1 M concentrations, ionized form is around 10^-9 M.

You add one milliliter of ammonium (0,1 M).

Doesn't make much sense not knowing what was the original volume Let's say we mix them 1:1

With that you add some amount of OH- ions, so the reaction OH- + H₃0+     2H₂O would happen, which would then force a little more ionisation of phenol.

Yes, when mixed 1:1 with ammonia ionized form concentration goes up, to about 9×10^-4 M - or 0.9%.

And so by adding more ammonia all acid should be ionised.

No, it doesn't work this way.

[tex]K_a = \frac {[H^+][A^-]}{[HA]}[/tex]

rearranging

[tex]\frac {[HA]}{[A^-]} = 10^{pK_a-pH}[/tex]

The higher the pH, the higher the ionization. 50% ionization means [HA]=[A^-], that's when pH=15. Good luck getting pH that high with ammonia (actually - with anything).

[kapital]

Yes, that is what equation says.
But if we look at the each reaction species, it doesn't seems consistent to me.
Let say we do similar experiment than before. We have 0,1 M phenol(volemes here aren't rely important, lets say 50 ml) and this time we add gas ammonium in solution. So, in terms of reaction, with each ammonia addon, also some OH- is generated. And since that, it should react with H3O+ to form water. That would be driving force for reaction. But why does this not happen? What is limiting factor? Possibly solubility of ammonium? Or I made some wrong assumption before?

[Borek]

Just because it goes against your intuition doesn't mean it is wrong.

It is an equilibrium, reaction doesn't have to go to completion, quite the opposite.

[kapital]

I know that. But nevertheless I think I still did get complete explanation. To sum up: My assumptions are:
1. If we remove  part of oxonium ions from solution(of weak acid), it more acid will be ionised.
2. Adding more and more ammonia will "remove" H3O+"
3. With enough ammonia complete phenol would be ionise because of H₃O+ + +OH- 2H₂O

Witch leads to wrong conclusion. And I still do not know at what point I make mistake. Like I said before, the only think that I see could be limited solubility of ammonia.

[Borek]

limited solubility of ammonia.

That's putting things on the head, but if you want to follow that route - maximum concentrations of everything are limited. There are limits to how much substance you can put in a given volume, which is why we don't care about what happens when the concentration goes up to infinity.