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Topic: Volatility of NaOH  (Read 11151 times)

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Offline Dan

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Re: Volatility of NaOH
« Reply #15 on: August 10, 2012, 09:50:26 AM »
Are they the temperatures at which the NH3 begins to boil?

Yes, it is the temp at which ammonia begins to biol off. Note that this is not the boiling point of pure ammonia. As has been said, as the ammonia boils off the bp of the solution will rise gradually until all the ammonia is gone and you are left with pure water boiling at 100°C.

what does the melting point refer to? Does this refer to the melting point of a block of ice that 25% NH3 by weight as in impurity?

Yes. It means that if you take an aqueous solution that is 25% ammonia by weight and gradually cool it down, it will freeze at -57.5°C
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Offline Omega Glory

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Re: Volatility of NaOH
« Reply #16 on: August 10, 2012, 09:53:22 AM »
If we consider a binary salt consisting of a cation and an anion (like ammonium hydroxide), in cases where the cation is relatively acidic (ammonium) and the anion is relatively basic (hydroxide), you will get an acid base equilibrium. Furthermore, if the conjugate base of the cation (ammonia) is less basic than the anion (hydroxide), the binary salt (ammonium hydroxide) will not be thermodynamically favoured.

Ammonium chloride on the other hand is a stable solid because chloride is a much weaker base than ammonia.

Sodium ions are not Bronsted acids, and therefore cannot participate in such an acid-base reaction.

This is fascinating. I had never pondered this before. I must say though, it seems almost comprehensible to me that dissociated liquid will always be thermodynamically favored, no matter the temperature. Surely there must be some temperature, perhaps even a temperature approaching absolute zero, where the lack of any substantial energy in the system causes these ions to solidify? Perhaps not into a crystal lattice structure, but at least into something amorphous?

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