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Topic: Produce pure H3O+  (Read 1419 times)

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Offline Dck140130

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Produce pure H3O+
« on: April 08, 2022, 05:43:08 PM »
Hello,

Could I theoretically mix 1 M of HBr with 1 M of H20 and then boil off the Br (BP = 58.8 C) and be left with H3O+?
                                             Δ
HBr + H20   ::equil:: H3O+ + Br    :rarrow:   H30+

Any idea how long I would have to boil to ensure that all Br is boiled off?

Offline Aldebaran

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Re: Produce pure H3O+
« Reply #1 on: April 09, 2022, 07:41:45 AM »
The short answer is no . Points to note: Br in the solution is Br- ion . If Br could be released it would be Br2 but that would involve breaking the covalent bond in gaseous HBr which you would not achieve by boiling.

Offline Ebbe Loos

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Re: Produce pure H3O+
« Reply #2 on: April 10, 2022, 03:54:04 AM »
Hi

No, what you subcribed is not possible.

Another way to prove this, is by our knowlegde of solutions.

In a solution, this netto charge should be 0. This means if we add HBr to a solution, we get the next equation:
[Br-] + [OH-] = [H3O+]
If we remove the Br via gas, and keep the H3O+ the equation wouldnt be correct anymore, thus what you said doesnt happen.

But what happens if we boil a HBr solution? (If boiling implies 100°C)

The boilingpoint of water is 100°C, so the water (and thus solvent) will leave the reaction as gas. Leaving a oversatuared solution HBr.
This solution will push out HBr molecule in order to get rid of oversatuarating. Since the boiling point of HBr is something like -50°C, the HBr will also leave as gas.

Hope this helped!
« Last Edit: April 10, 2022, 05:23:59 AM by Ebbe Loos »

Offline Aldebaran

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Re: Produce pure H3O+
« Reply #3 on: April 10, 2022, 05:51:40 AM »
In fact hydrobromic acid forms a constant boiling point mixture at about 124 °C. So all the hydrogen bromide gas cannot be boiled off in a simple way.

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