People! I have a question! (1 Viewer)

[Dreamerish*~]

People! I Have A Question! (About Chemistry)

Which of the following equations correctly represents the products when NaOH reacts with H₂SO₄?

a) H₂SO_4(aq) + NaOH_(aq) → NaHSO_4(aq) + H₂O_(l)
b) H₂SO_4(aq) + 2NaOH_(aq) → Na₂SO_4(aq) + 2H₂O_(l)

If you chose b), could you please tell me how sodium hydrogen sulphate (NaHSO₄) is formed?

Thank you very much.

 

[wanton-wonton]

According to google, it's b), although I don't know why it can't be a).

im just wondering it can be both a and b can't it.. because sulfuric acid is diprotic so it can donate both the H's or just one? or am i off track? sorry if i am!

I think you're right.

http://www.coursework.info/i/68808.html

 

[fantasy27]

im just wondering it can be both a and b can't it.. because sulfuric acid is diprotic so it can donate both the H's or just one? or am i off track? sorry if i am!

 

[Jumbo Cactuar]

Yes sulfuric acid is diprotic, though in solution bisulfate ions aren't present in any great quantity (unless you want to add a load of HCl as well).

This is a limiting reagents question. In excess NaOH it is certainly a).

Though consider;

2H2SO4(aq) + 2NaOH(aq) → 2Na+ 2SO4 2- + 2H+ + 2H2O(l) *ignore HSO4 2- for now

Then we evaporate the solution such that all the water has been removed. *just accept it ~ please!

The question is... would you have either;

Na2SO4(s) + H2SO4(l) or 2NaHSO4(s) ?

Well what is more thermodynamically favourable?

∆_fH(Na2SO4(s) + H2SO4(l)) = -1387 + -814 kJ/mol = -2201 kJ/mol
∆_fH(2NaHSO4(s)) = 2 * -1126 kJ/mol = -2252 kJ/mol

So under limited NaOH, NaHSO4 is a product...

 

[Dreamerish*~]

Jumbo, so when around 30 mL of NaOH is used in a titration, and the end point is reached with around 5 mL of H₂SO₄, would the product be NaHSO₄?

This was a question in my trial, and we had to calculate the concentration of H₂SO₄. I wasn't sure which equation to use, so I used the one that had sodium hydrogen sulphate as a product - would this be marked wrong?

 

[Casmira]

is there more of HSO4- present than SO4^2- present which is why it forms NaHSO4 ? wouldnt SOME Na2SO4 form?

 

[Jumbo Cactuar]

It is important to understand that the second proton is more tightly bound, though to put it into perspective;

A neutralisation reaction involving sulfuric acid reaches an end point at ~pH 7. The solution contains 2000000 more SO4 2- ions than HSO4 - ions.

This may seem to condradict my previous post, but remember that an aqueous solution and solid structures are completely different ball games.

It may be conveinient to say we form NaHSO4(aq), though, as I understand it, this is really Na+(aq) + H+(aq) + SO4 2-(aq) (with some HSO4-).