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Topic: Volatility of NaOH  (Read 11157 times)

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Offline Omega Glory

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Volatility of NaOH
« on: August 09, 2012, 01:56:18 PM »
Hi all,

Can someone please explain to me why pure NaOH is so involatile whereas other ionic compounds like ammonium hydroxide evaporate readily at temperatures below 100F?

Thanks!

Offline fledarmus

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Re: Volatility of NaOH
« Reply #1 on: August 09, 2012, 02:00:29 PM »
What are the products formed when you heat up ammonium hydroxide? What types of bonds are broken?

Offline AWK

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Re: Volatility of NaOH
« Reply #2 on: August 09, 2012, 02:10:12 PM »
Ammonium hydroxide exists in solution only in minute amount. the rest is water solution of unreacted (but solvated) ammonia.
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Offline Omega Glory

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Re: Volatility of NaOH
« Reply #3 on: August 09, 2012, 02:12:47 PM »
NH4+ and OH-, I would imagine -- this would involve the breaking of ionic bonds

Offline fledarmus

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Re: Volatility of NaOH
« Reply #4 on: August 09, 2012, 02:28:45 PM »
Actually, no - most of the ammonium hydroxide is simply ammonia (NH3) dissolved in water (H2O). Only a small amount of it exists as ammonium (NH4+) hydroxide (OH-). It is a liquid at room temperature. As you heat it up, you drive off the ammonia, and due to the operation of LeChatelier's Principle, any ammonium hydroxide that is actually present turns back into ammonia and water. The bond that you are breaking is the bond between NH3 and H+, which is a coordinate bond rather than an ionic bond - both of the electrons in the bond are coming from the lone pair of electrons on the nitrogen.

Offline AWK

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Re: Volatility of NaOH
« Reply #5 on: August 09, 2012, 03:04:06 PM »
Quote
The bond that you are breaking is the bond between NH3 and H+, which is a coordinate bond rather than an ionic bond
Not exactly.
You are breaking rather a dipole interaction between ammonia and water. After removing some ammonia from solution an equilibrium in solution
NH4+ + OH- = NH3 + H2O
is moving to the right loosing some ammonia.
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Offline fledarmus

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Re: Volatility of NaOH
« Reply #6 on: August 09, 2012, 03:11:44 PM »
After removing some ammonia from solution an equilibrium in solution
NH4+ + OH- = NH3 + H2O
is moving to the right loosing some ammonia.

Yes, that reaction was what I was referring to as including the breaking of a coordinate bond

Offline Omega Glory

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Re: Volatility of NaOH
« Reply #7 on: August 09, 2012, 03:19:47 PM »
So say I begin with a lump of pure, solid ammonium hydroxide, NH4+OH-. As I heat it past its melting point, it undergoes a phase transition and becomes a liquid. At the same time that happens, it is also ceasing to be NH4+OH-. It is becoming a solution of NH3 in H2O. As the solution heats up, the NH3 boils away, which shifts the equilibrium to create more ammonia, until there is no more NH4+ or OH- left in solution, and all you are left with is pure water.

Whereas with Na+OH-, the initial conversion to H2O does not happen, as it does in the case of ammonium hydroxide, as the sodium is not protic and cannot donate an H+. So as you heat it past its melting point, all you have is a puddle of ions... which are involatile.

Is that the long and short of it?

Offline AWK

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Re: Volatility of NaOH
« Reply #8 on: August 09, 2012, 03:42:11 PM »
solid ammonium hydroxide
As far as I knot such solid doe'nt exist. You have solution containing up to about 25 % ammonia at which ~1 % of ammonia exists as NH4+ + OH-.
The rest is OK
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Offline Omega Glory

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Re: Volatility of NaOH
« Reply #9 on: August 09, 2012, 03:51:13 PM »
Is there a reason why such a solid cannot exist? Why can NH4OH not form a crystal lattice when NaOH can?

Wikipedia lists a pair of boiling points for ammonium hydroxide, as well as a pair of melting points  http://en.wikipedia.org/wiki/Ammonium_hydroxide ... What is the interpretation of this information as it applies to the real world? Is it the temperature at which an aqueous solution of 25%/32% NH4OH melts and boils?

If so, the temperatures they give, e.g. 37.7C for 25% NH4OH -- does that signify that at 37.7C, all of the residual NH4+ will have been converted to NH3, and all of the NH3 will have boiled off? Clearly it cannot mean that all of the water will have boiled away as well, because that would not happen until 100C

Offline vmelkon

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Re: Volatility of NaOH
« Reply #10 on: August 09, 2012, 08:54:50 PM »
My guess is that when you heat the ammonia solution at that temperature, it boils. The bubbles would be NH3.
Now obviously, as it boils, the conc of NH3 goes down and so the bp rises. As the conc of NH3 goes to 0, the bp approaches the bp of pure water.

"Is there a reason why such a solid cannot exist? Why can NH4OH not form a crystal lattice when NaOH can?"

I don't know. Good question.

Offline Dan

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Re: Volatility of NaOH
« Reply #11 on: August 10, 2012, 03:13:40 AM »
Is there a reason why such a solid cannot exist? Why can NH4OH not form a crystal lattice when NaOH can?

Hydroxide is a much stronger base than ammonia, so the equilibrium:

NH4+-OH  :requil: NH3 + H2O

Heavily favours the right hand side (hydroxide easily deprotonates ammonium).

I myself have never liked the term ammonium hydroxide, it's just aqueous ammonia.
My research: Google Scholar and Researchgate

Offline Omega Glory

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Re: Volatility of NaOH
« Reply #12 on: August 10, 2012, 09:20:00 AM »
What do the following encyclopedia values signify, then?

Ammonium hydroxide

Melting point    

−57.5 °C (25%)
−91.5 °C (32%)

Boiling point    

37.7 °C (25%)
24.7 °C (32%)

This from http://en.wikipedia.org/wiki/Ammonium_hydroxide

Quote
Is there a reason why such a solid cannot exist? Why can NH4OH not form a crystal lattice when NaOH can?

Hydroxide is a much stronger base than ammonia, so the equilibrium:

NH4+-OH  :requil: NH3 + H2O

Heavily favours the right hand side (hydroxide easily deprotonates ammonium).

I myself have never liked the term ammonium hydroxide, it's just aqueous ammonia.

Is this to say that any time you have two ions together joined by an ionic bond, and one of them is substantially more acidic or basic than the other (as in the case of Na+ vs. OH-), that difference in affinity completely circumvents the ability of ions to bond together and solidify?

Offline Dan

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Re: Volatility of NaOH
« Reply #13 on: August 10, 2012, 09:31:44 AM »
Quote
What do the following encyclopedia values signify, then?

Ammonium hydroxide

Melting point   

−57.5 °C (25%)
−91.5 °C (32%)

Boiling point   

37.7 °C (25%)
24.7 °C (32%)

This from http://en.wikipedia.org/wiki/Ammonium_hydroxide

I expect it refers to the melting/boiling points of 25% NH3(aq) and 32% NH3(aq) solutions.

Quote
Is this to say that any time you have two ions together joined by an ionic bond, and one of them is substantially more acidic or basic than the other (as in the case of Na+ vs. OH-), that difference in affinity completely circumvents the ability of ions to bond together and solidify?

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.
« Last Edit: August 10, 2012, 09:45:14 AM by Dan »
My research: Google Scholar and Researchgate

Offline Omega Glory

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Re: Volatility of NaOH
« Reply #14 on: August 10, 2012, 09:40:15 AM »
I expect they are the melting and boiling points of 25% NH3(aq) and 32% NH3(aq) solutions.

Again, though, according to this post...

My guess is that when you heat the ammonia solution at that temperature, it boils. The bubbles would be NH3.
Now obviously, as it boils, the conc of NH3 goes down and so the bp rises. As the conc of NH3 goes to 0, the bp approaches the bp of pure water.

"Is there a reason why such a solid cannot exist? Why can NH4OH not form a crystal lattice when NaOH can?"

I don't know. Good question.

...my interpretation is, as the temperature rises, the solution will reach the boiling point of ammonia, and the NH3 will evaporate off, until there is no more NH3 left in solution, and all you have is pure water sans any impurities. Once the NH3 is gone, the liquid ceases to boil. Then, as the temperature is further ramped up, you come to 100C, and the water itself begins to boil.

What, then, do the encyclopedia values signify? Are they the temperatures at which the NH3 begins to boil? Or are they the temperatures at which there is no more NH3 left in solution? What, exactly, is their physical significance?

Also, if it has already been established that there is no such thing as solid ammonium hydroxide, then what does the melting point refer to? Does this refer to the melting point of a block of ice that includes 25% NH3 by weight as in impurity?

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