[peterpan1372]

I thought that since H2 are prton donators they donate their protons to NaOH, giving them two molecules... right?

[Borek]

If by H2 you mean two acidic protons in the diprotic acid, you are to some extent right, but you misuse the nomenclature in a way that makes it almost impossible to follow your posts. Nothing is "giving two molecules", I have no idea what you even mean by that.

[peterpan1372]

H2 donates its protons.. thats what I meant...

[Borek]

H2 (or rather H₂) is a molecule of hydrogen and it doesn't donate protons.

Proton is not a molecule.

As I said - you are misusing words and symbols making it very difficult to follow what you mean.

[peterpan1372]

could you tell me then why NaOH has two molecules please?

[Borek]

could you tell me then why NaOH has two molecules please?

NaOH doesn't have two molecules, I have no idea what you are asking about.

[peterpan1372]

well, why is n1=2 and n2=1 (2NaOH and one H2)...
the answer to the question is 120ml ( as the solutions states)

[Borek]

It is all in the stoichiometry of the neutralization of the diprotic acid.

What diprotic acids do you know?

How does an acid react with a base?

You have already wrote reaction equation of a reaction between HCl and NaOH. Can you write similar reaction equation for a diprotic acid, any of those you know?

And stop writing H2 when you mean diprotic acid, I told you several times it is wrong and confusing. Common way (and one that will be almost always correctly understood) of denoting diprotic acids is H₂A.

[peterpan1372]

2NaOH + H2SO4 = Na2SO4 + 2H2O...
The way I understood it is that there will be two NaOH because of the fact that you also have to equate the reaction.. right?

[Borek]

Yes. Or, to put it slightly different - NaOH contains a single OH^- anion, capable of neutralizing a single acidic proton. Diprotic acid has two such protons, so you need two OH^-  anions (or two NaOH molecules) to neutralize one molecule of the acid. <- pay attention to the nomenclature and the way it is used.