Neutralisation Reaction Confusion

What will happen when $$\pu{1 mol}$$ of $$\ce{H2SO4}$$ reacts with $$\pu{1 mol}$$ of $$\ce{NaOH}$$?

Will $$\pu{1 mol}$$ each of $$\ce{NaHSO4}$$ and water be formed?

Or will $$\pu{0.5 mol}$$ of $$\ce{H2SO4}$$ will remain yield $$\pu{0.5 mol}$$ of $$\ce{Na2SO4}$$ and $$\pu{1 mol}$$ of water?

I believe it will be the second way since H2SO4 is a strong acid and would again quickly react with the $$\ce{NaHSO4}$$ formed. But I am not quite sure how the mechanism will be and whether the Rate constant of the second reaction will be high enough to facilitate this.

What will happen and why?

• @andselisk Those two ways are my thinking only. It wasn't asked in any homework question. As I am confused which will happen, I asked the same – user226375 May 16 at 16:26
• well, compared to $\ce{H2SO4}$. $\ce{HSO4-}$ is only a weak acid so I'd assume that $\ce{NaHSO4}$ will be formed first, but i guess and say and equilibrium is probably stablished between this and $\ce{Na2SO4}$ – H.Linkhorn May 16 at 16:34

First, let's assume these reaction happen in aqueous solution and everything is soluble. The two reaction you are interested in are:

$$\ce{H2SO4(aq) + NaOH(aq) -> HSO4-(aq) + H2O(l) + Na+(aq)}\label{rxn:1}\tag{1}$$

$$\ce{H2SO4(aq) + 2NaOH(aq) -> SO4^2-(aq) + 2H2O(l) + 2Na+(aq)}\label{rxn:2}\tag{2}$$

Your saying that reaction \eqref{rxn:2} might happen even when $$\ce{NaOH}$$ is limiting, and you would end up with some sulfuric acid and some sulfate. In a thought experiment, let's say that is what happened. Now, we can discuss whether sulfuric acid and sulfate, which are present in equimolar amounts, will react with each other according to:

$$\ce{H2SO4(aq) + SO4^2-(aq) <=> 2HSO4-(aq)}\label{rxn:3}\tag{3}$$

Sulfuric acid is a strong acid, and sulfate is a weak (very weak) base. This is like mixing hydronium ($$\ce{H3O+}$$) and hydroxide ($$\ce{OH-}$$) ions, which (mostly) form water. So just like water, which occurs mostly undissociated, the equilibrium of reaction \eqref{rxn:3} lies on the side of the products when the reactants are present in equimolar amounts.

Of course, if you add more $$\ce{NaOH}$$, you would start making a lot of sulfate.

If you wanted to know all of the products and their concentrations, you would have to consider water auto-dissociation as well. The argument I made is not meant to be quantitative, but to show why we don't expect sulfuric acid to be left over. In aqueous solution, there is never much sulfuric acid (even in the absence of $$\ce{NaOH}$$) because, as a strong acid, it dissociates.

• Tip on formatting references to the equations/reactions: you can label them with \label{<label>} (I use eqn:<equation #> and rxn:<reaction #> as <label>), and then refer to the corresponding numbered item via \eqref{<label>} (see my edit). This way the reference is clickable and consistent (as long as labels are correct). – andselisk May 16 at 17:23
• @andselisk Yes, I just saw that - very cool. And congrats on becoming a moderator! – Karsten Theis May 16 at 17:26
• Thank you very much! That was one of the "pro tips" Martin taught me a year or so ago:) – andselisk May 16 at 17:33
• @andselisk: I agree with Karsten, it is very cool tip! So, I checked it myself, but accidentally save as an edit. I did nothing but it save you edit as as my edit anyway. Sorry. But, since you are now a moderator (Congratulations!), you may able to save it! Thanks for the great tip. – Mathew Mahindaratne May 16 at 17:55
• @MathewMahindaratne I'm not sure I understand the editing part (yours and mine are still two separate edits), so I'm probably going to leave it as is – the answer looks perfectly fine to me:) – andselisk May 16 at 18:00