Yttrium2
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Azeotropes
Can someone clue me in as to azeotropes?
Does 91% isopropanol form an azeotropes with water?
How dilute can the mixture be while still firming an azeotropes?
At first the idea of distillation seemed so simple. Now I rarely know what will distill, and what won't because of azeotropes
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walruslover69
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Isopropanol forms an azeotrope with water. The vapor that is distilled is 91% isopropanol and 9% water. Azeotropes form at all concentrations. This is
also assuming you have a proper fractionation, otherwise the vapor will have even more water in it. Distillation seems very simple of the surface but
there is actually quite a lot too it.
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DavidJR
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Some examples
Example 1: Zeotropic mixtures
Consider a mixture of methanol (b.p. 65c) and water (b.p. 100c), which do not form an azeotrope. You can be certain that the boiling point of any such
mixture is between the boiling points of the components, which covers the full range of %methanol. More importantly, the vapour produced by boiling
such a mixture will always be richer in the lower-boiling component (methanol) than the mixture you started with. This means you can easily separate
them by fractional distillation.
Example 2: Minimum-boiling (positive) azeotropic mixtures
Consider a mixture of isopropanol (b.p. 83c) and water (b.p. 100c). Isopropanol and water form an azeotrope at 91% v/v isopropanol, which boils at
~80c.
2a: If our mixture contains 50% v/v isopropanol and we put it in a simple distillation apparatus, with the boiling flask heated by a
water/oil/sand bath at 85c, the mixture will boil producing a vapour richer in isopropanol than the starting mixture. Once the mixture stops boiling
(since temp only 85c we won't drive over all of the water) we can empty out the boiling flask, and put the distillate in the boiling flask for a
second run. This time we'll still increase the % isopropanol in the distillate, but not by as much as we did before. We can repeat this, but with
diminishing returns. Essentially a fractionating column would do this work for you as the vapour has the opportunity to recondense/reevaporate many
times. The most concentrated you'll ever get is 91% v/v because at this point we have an azeotrope, in other words, the %isopropanol in the vapour
produced by boiling the mixture is the same as in the liquid.
2b: If instead our starting mixture contained 95% isopropanol (above the azeotrope) then boiling this would produce vapour containing less
isopropanol (though still above the azeotrope). So, if we were to fractionally distill our 95% IPA, the distillate would be the 91% azeotrope.
However, if we distill until all of the water has been carried over as the azeotrope, then the residue left in the boiling flask will be ~100%
isopropanol.
Example 3: Maximum-boiling (negative) azeotropic mixtures
Consider a mixture of nitric acid (b.p. 83c) and water (b.p. 100c). Nitric acid and water form an azeotrope at 68% w/w HNO3, which boils at
121c.
3a: If our mixture contains 50% w/w HNO3, then we have more water than the azeotrope. So if we perform a fractional distillation,
then the excess water will come over first at 100c, and then the azeotrope will come over at 121c.
3b: If instead our starting mixture contained 85% w/w HNO3, then we would have less water than the azeotrope. So if we perform a
fractional distillation, then the excess HNO3 will come over first at 83c, and after that, the azeotrope will come over at 121c.
N.B. for simplicity of explanation i've assumed in these examples that your fractionating column is impossibly good
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ninhydric1
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Quote: Originally posted by DavidJR  |
2b: If instead our starting mixture contained 95% isopropanol (above the azeotrope) then boiling this would produce vapour containing less
isopropanol (though still above the azeotrope). So, if we were to fractionally distill our 95% IPA, the distillate would be the 91% azeotrope.
However, if we distill until all of the water has been carried over as the azeotrope, then the residue left in the boiling flask will be ~100%
isopropanol.
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Wait, so if I, theoretically, distill my 99% isopropanol, I'm able to get 100% isopropanol in my distilling flask and get a small amount of 91%
isopropanol in my receiving flask? I'm assuming I'd need a good fractionating column though.
The philosophy of one century is the common sense of the next.
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DavidJR
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| Quote: Originally posted by ninhydric1 |
Wait, so if I, theoretically, distill my 99% isopropanol, I'm able to get 100% isopropanol in my distilling flask and get a small amount of 91%
isopropanol in my receiving flask? I'm assuming I'd need a good fractionating column though. |
Theoretically, yes. But you'd probably be better off using a drying agent.
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unionised
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Theoretically, yes.
But you would need to remove the drying agent (most are at least slightly soluble).
The most practical way to do that is by distillation.
And, unless you have a dry box or something, while you filter off the drying agent and transfer the stuff to a flask, it will pick up moisture from
the air.
So the distillation will end up removing the last traces of water anyway.
:-)
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DavidJR
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| Quote: Originally posted by unionised |
Theoretically, yes.
But you would need to remove the drying agent (most are at least slightly soluble).
The most practical way to do that is by distillation.
And, unless you have a dry box or something, while you filter off the drying agent and transfer the stuff to a flask, it will pick up moisture from
the air.
So the distillation will end up removing the last traces of water anyway.
:-)
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Sure, but if you have 99% isopropanol and you want to get 100% then you'd distill off the azeotrope and the residue in the boiling flask would be your
~100% isopropanol. If you used a drying agent then distilled, you'd keep the distillate, not the residue. So that method has the advantage of
eliminating any potential non-volatile impurities.
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