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Old 08-20-2009, 04:03 AM   #1
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Default Variability of hydration in Calcium Chloride

Now that I'm finally delving into water adjustment, something jumped out at me when looking at the salts being used. Most of the salts seem to be fairly stable with a certain number of hydrates, but it seems that CaCl2 is a very hygroscopic species and can (and will) absorb moisture readily from the air.

Does anyone try to account for this when using CaCl2 to treat water?

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Old 08-20-2009, 04:51 AM   #2
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In Homebrewing Volume I by Al Korzonas he lists the ppm from CaCl2 for both anhydrous and the dihydrate form. Beersmith uses the dihydrate form in its calculations so that is what I am going by.

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Old 08-20-2009, 03:11 PM   #3
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If you look in Wikipedia's entry for CaCl2 it lists anhydrous and four hydrates (mono-, di-, tetra-, hexa-). In brewing literature (most notably Palmer's spreadsheet) it's listed in it's dihydrate form. One reason for this could be that the hygroscopic nature of the CaCl2 decreases enough in the dihydrate state that it can remain stable in typical storage conditions. Although, it is definitely still hygroscopic, because if you leave it in open air it will actually turn to a solution.

I'd love more clarification on this, as the difference between anhydrous and dihydrate, or the other higher hydrates, is pretty significant.

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Old 08-20-2009, 10:05 PM   #4
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Quote:
Originally Posted by JKoravos View Post

I'd love more clarification on this, as the difference between anhydrous and dihydrate, or the other higher hydrates, is pretty significant.
Anhydrous simply refers to a compound that is free of water, but because many compounds such as CaCl2 are highly hydroscopic compounds know as ionic salts they are hard to maintain in an anhydrous state, and are much more likely to be found in a hydrated state.

di-hydrous refers to the state of having two water molecules per salt compound or
2H2O + Ca Cl2

So when you are calculating the amount of CaCl2 needed for a dilution, you have to account for the two molecules of water that are present for every unit of CaCl2.

If you look at the molecular weight of CaCl2 it is Ca(40 g/mol) + 2xCl (35.43g/mol)=111g/mol of CaCl2. but when you look at the di-hydrous state you have to add the 2 water H2O (36g/mol) you have the formula weight of 147g/mol.

so if you used the molecular weight to do your calculations instead of the formula weight, you would be 25% short.

I'm not sure the significance of the other states, regarding CaCl2, but the di-hydrate state is commonly commercially available. So I think it is the most relevant...
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Old 08-20-2009, 11:38 PM   #5
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Quote:
Originally Posted by Bsquared View Post
Anhydrous simply refers to a compound that is free of water, but because many compounds such as CaCl2 are highly hydroscopic compounds know as ionic salts they are hard to maintain in an anhydrous state, and are much more likely to be found in a hydrated state.

di-hydrous refers to the state of having two water molecules per salt compound or
2H2O + Ca Cl2

So when you are calculating the amount of CaCl2 needed for a dilution, you have to account for the two molecules of water that are present for every unit of CaCl2.

If you look at the molecular weight of CaCl2 it is Ca(40 g/mol) + 2xCl (35.43g/mol)=111g/mol of CaCl2. but when you look at the di-hydrous state you have to add the 2 water H2O (36g/mol) you have the formula weight of 147g/mol.

so if you used the molecular weight to do your calculations instead of the formula weight, you would be 25% short.

I'm not sure the significance of the other states, regarding CaCl2, but the di-hydrate state is commonly commercially available. So I think it is the most relevant...

The question I have is, how do you know you have the dihydrate? If the material is hygroscopic it will absorb water from the atmosphere. Even if you may have bought a package that specified dihydrate, you may have a higher hydrate due to water being adsorbed from the atmosphere.
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Old 08-21-2009, 02:27 AM   #6
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Okay, I see what you are after. I what back in time to my Chemistry books,and my CRC handbook and that CaCl2 is most stable in its CaCl2•6H20, but this form has a melting temp of -55ºC. So like you said if you leave it out in the air it turns to a liquid, and that is the CaCl2•6H20 form, I could not find any data for the 4H2O, so I'd assume that that these are rare transitory forms.

If your CaCl2 is still crystallized, it is dihydride . if it's a liquid its CaCl2•6H20.

Hope that helps

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Old 08-21-2009, 02:30 AM   #7
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Quote:
Originally Posted by Bsquared View Post
Okay, I see what you are after. I what back in time to my Chemistry books,and my CRC handbook and that CaCl2 is most stable in its CaCl2•6H20, but this form has a melting temp of -55ºC. So like you said if you leave it out in the air it turns to a liquid, and that is the CaCl2•6H20 form, I could not find any data for the 4H2O, so I'd assume that that these are rare transitory forms.

If your CaCl2 is still crystallized, it is dihydride . if it's a liquid its CaCl2•6H20.

Hope that helps

THAT is what I was looking for. Awesome. Thank you.
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Old 08-21-2009, 11:29 PM   #8
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I bought some CaCl2 at the LHBS last night. I brought it to work today to weigh out the salt for a batch Monday. By the time I took it out of my car it was getting mushy. There was still enough dry salt to weigh out, but it was so hot and humid that any crystal that was exposed to the atmosphere for more than a few seconds turned into solution.

I think I'm going to try to bake out the water, then store it in the freezer.

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Old 08-22-2009, 08:23 AM   #9
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When I was learning about water chemistry, we used the 5H20 form for calculation - not sure why, but there you go. Be sure to keep it in an airtight environment, and weigh it just before you use it, otherwise it turns into sludge.

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