Sciencemadness Discussion Board

Solid Sulfuric Acid

agorot - 17-3-2010 at 18:46

Today I did something quite cool :D

To make things short, I used Sodium Bisulfate and decomposed it at high temperatures. This made H2SO4, and assuming I got all of the sodium bisulfate to decompose it was 100% H2SO4 (this was done in a 24/40 ground glass distillation setup). As the coolant in the receiving flask, I had 91% Isopropyl alcohol and CO2 (s). Very cold, and it looked so cool!

When the sulfuric acid drops came over, they immediately froze! I had never seen anything like that before.

I would post the pictures I took, but for some reason my crappy camera malfunctioned and deleted them or something. I may be able to try this in a week or two again and then I will post pictures from a different camera.

woelen - 17-3-2010 at 23:40

I don't think you make 100% H2SO4. If you heated the solid in a glass distillation setup then I expect that you did not go far beyond 200 C. The commercial sodium bisulfate, used as pH-minus in swimming pools is NaHSO4.H2O. IIRC this looses its water at somewhere around 130 C. When it is heated much stronger, then it looses more water and Na2S2O7 is formed (sodium pyrosulfate). Besides that, SO3 is formed, but you need red hot situations before that is formed. There is a thread on sciencemadness about formation of SO3 where this reaction is covered.

So, I think that your drops are just dilute sulphuric acid or even mainly water.

chief - 17-3-2010 at 23:56

@woelen: In think he talks about the "persulfate" ... which is used for etching of circuitry ...

DJF90 - 18-3-2010 at 02:16

No chief, he most definately says
Quote:
I used Sodium Bisulfate and decomposed it at high temperatures
...

1281371269 - 18-3-2010 at 13:23

Though Wiki says that H2SO4 freezes at 10C, so there's no reason you couldn't do it, even with ice.

agorot - 19-3-2010 at 16:55

Quote: Originally posted by woelen  
I don't think you make 100% H2SO4. If you heated the solid in a glass distillation setup then I expect that you did not go far beyond 200 C. The commercial sodium bisulfate, used as pH-minus in swimming pools is NaHSO4.H2O. IIRC this looses its water at somewhere around 130 C. When it is heated much stronger, then it looses more water and Na2S2O7 is formed (sodium pyrosulfate). Besides that, SO3 is formed, but you need red hot situations before that is formed. There is a thread on sciencemadness about formation of SO3 where this reaction is covered.

So, I think that your drops are just dilute sulphuric acid or even mainly water.


I disagree. :)

Before I heated the sodium bisulfate to reaction temperature, I desiccated it for two weeks (occasionally stirring with a stirring rod that was also in the desiccator bag) and then the morning of the experiment I put it in the oven at about 160 degrees F. I then transfered the bisulfate directly to the boiling flask, which now that I think about it may have had a little bit of water vapor in it, but not much.

Where I could have gone wrong is that the Calcium Chloride dessicator was not powerful enough to remove the water from the bisulfate. I'm not sure on this. Is the bisulfate a more powerful desiccator than calcium chloride?

I used a bunsen burner of sorts (butane that burns much hotter than methane (natural gas), ethane, or propane (camping fuel)) that I know got up to at least 500 degrees C. Also, the bisulfate (or eventually pyrosulfate or sulfate) mixture was largely liquid at the time that I stopped the experiment. Bubbles could be seen coming up from the bottom of the flask at an even pace.

I did get the bisulfate as swimming pool ph-down, but the bottle said 99%

woelen - 20-3-2010 at 04:53

If the bottle sais 99%, then it means 99% NaHSO4.H2O. NaHSO4.H2O is a chemical entity on its own an you don't remove the water from this material by drying it with CaCl2, not even with P4O10! It is like CuSO4.5H2O, the water is part of the compound itself, it is not like it is humid or wet.

On heating, the H2O is lost, but only at 130 C or so. Then you get:

NaHSO4.H2O --> NaHSO4 + H2O

On further heating you get:

2NaHSO4 --> Na2S2O7 + H2O

and only on very strong heating you get:

Na2S2O7 --> Na2SO4 + SO3

If you have been heating to 500 C (which I doubt) then you might get some H2SO4 but it is dilute. But of course, you can test your acid. Take 1 ml of the acid and mix it with 1 ml of water, both at room temperature. If this mix becomes hot, then the acid has a high concentration.

agorot - 20-3-2010 at 09:23

You're right, I just did a titration using 4 M NaOH in a buret and Phenolphthalein in with the acid. I tested 10 mL of the acid and it took 38.5 mL NaOH to neutralize, meaning my acid was less than 8 M. Or did I also do that calculation wrong:

1*4*38.5 = 2*Ma*10 where Ma = molarity of the acid.

so Ma = 7.7 M

woelen - 20-3-2010 at 09:27

Yes, you did the calculation right and corretly took into account that H2SO4 is a dibasic acid. Based on your info, I would say that the acid is somewhere between 7M and 8M concentration. Still not bad, but not the concentrated acid which you described in your first post.

agorot - 20-3-2010 at 12:27

Alright. Next time I try this then, here's what I'll do.

1) Heat the NaHSO4.H2O to 150 C until no more H2O comes off, discard the H2O
2) Heat the NaHSO4 to 500 C until no more vapor comes off

According to Len1's experiments with NaHSO4, 500 should be sufficient to make the pyrosulfate and then the sulfate yielding (eventually) equal molar amounts of SO3 and H2O

2NaHSO4 --> Na2S2O7 + H2O
Na2S2O7 --> Na2SO4 + SO3
combined: 2NaHSO4 --> Na2SO4 + H2O + SO3

This should, in theory, give me 100% H2SO4, correct? Since concentrated acid is usually considered 98%, I should probably add the correct amount of water to dilute it.

1281371269 - 20-3-2010 at 12:52

Why not make the anhydrous salt, then do the experiment such that any gasses produced are bubbled through already conc. H2SO4 instead of water?

You would get: 18ml water / mol NaHSO4 = 1mol H2SO4 and any SO3 that didn't dissolve would go to making H2S2O7.

If well calculated, you could then work out the exact amount of water to add to make it all into 98% H2SO4.

itchyfruit - 20-3-2010 at 13:38

Interesting!!!
I have a reagent bottle labelled conc sulphuric acid in my lab which appears to have frozen, this is strange as i have several other bottles of 96% and 98% sulphuric acid in the same lab which have not frozen.

agorot - 20-3-2010 at 13:43

Quote: Originally posted by Mossydie  
Why not make the anhydrous salt, then do the experiment such that any gasses produced are bubbled through already conc. H2SO4 instead of water?

You would get: 18ml water / mol NaHSO4 = 1mol H2SO4 and any SO3 that didn't dissolve would go to making H2S2O7.

If well calculated, you could then work out the exact amount of water to add to make it all into 98% H2SO4.


Let me make sure I'm understanding you.

I get the anhydrous salt thing...that's what I was accomplishing by heating it to 150 C and getting rid of the water that comes off afterward.

But I don't get bubbling the gasses produced into concentrated acid instead of water. I wasn't bubbling the gasses/vapor through anything originally, I just collected the condensed vapor into a flask submerged in an ice bath.

And I do understand that decomposing one mol of NaHSO4 into sodium pyrosulfate will yield 18 mL of H2O (one mol).

And H2S2O7 is H2SO4 + SO3

and so you're saying that I figure out how much H2S2O7 is in there and add an equivalent amount of water (or more correctly add the acid to water)

1281371269 - 21-3-2010 at 13:52

Looking back on my post I being to doubt the chemistry behind it, but...this was the idea:

The reason industrially they don't do H2O + SO3 is because it's an extremely violent reaction. Instead, they firstly add the SO3 to H2SO4 making H2S2O7, then add water.

In your reaction, water will come off first. If this is condensed, you have the SO3 + H2O problem (though you say you cooled it with dry ice / acetone which I guess would make it fine). So instead, I thought it would be good to bubble the gasses that came off through already conc H2SO4. Water would come off first, diluting the acid a little, and then SO3 would come off, and if it all went nicely you'd end up with exactly one mole of sulphuric per mole of NaHSO4.

This, combined with the acid you put in to start with, would leave you with slightly over 98% sulphuric, to which you would need to add a tiny bit of water to bring it down to 98%.

That was the plan! :D
However, in reality it might not go like that - can someone more knowledgeable comment?

Maybe the best way would be to remove all the water first by heating until SO3 comes off (test with an indicator paper). Then start collecting the gasses given off, bubbling them through conc H2SO4. It would be easy to calculate exactly how much water to add to the created H2S2O7 to make your desired 98%

[Edited on 21-3-2010 by Mossydie]

chief - 5-4-2010 at 01:25

After all, the SO3-generation might be something analoguous to the bahaviour of pyrosulfite: Na2S2O5, which gives off SO2 ...
==> I have a whole 25 kg bag of the stuff ...

So why should not the purosulfate give off SO3 then ?

agorot - 5-4-2010 at 12:30

Quote: Originally posted by Mossydie  

Maybe the best way would be to remove all the water first by heating until SO3 comes off (test with an indicator paper). Then start collecting the gasses given off, bubbling them through conc H2SO4. It would be easy to calculate exactly how much water to add to the created H2S2O7 to make your desired 98%

[Edited on 21-3-2010 by Mossydie]


The only problem is, according to Len1 who also did this experiment, some of the water holds onto the bisulfate until the temperatures are raised to the levels where the SO3 starts coming off.