Measuring Calcium Chloride

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The newest Brun Water also includes calculations for CaCl2 solution.

Thanks for the mention. However, I had a brain fart when coding the liquid algorithm. Supporters will be receiving version 3.4 over the next few weeks that corrects the error in liquid CaCl2 calculations. If you will be using a liquid solution before that, be sure to send me a note and I'll get you to the top of the list.
 
Thanks for the mention. However, I had a brain fart when coding the liquid algorithm. Supporters will be receiving version 3.4 over the next few weeks that corrects the error in liquid CaCl2 calculations. If you will be using a liquid solution before that, be sure to send me a note and I'll get you to the top of the list.

I got the updated version but haven't had a chance to enter any calculations. Just wondering, how off was the liquid CaCl2 calculations? Just curious. I have 6 batches in fermentation and wondering what I should expect.
 
Brewed a Pliny clone this weekend. I didn't make up the liquid solution, so I measured the dry stuff and added a bit more to make up for the weight of the moisture it absorbed...

Yeah, I was winging it!
 
Just to make sure I'm doing this right.

I took a 250ml flask, added 20g of CaCl2, and added distilled water to the 200ml mark.

Measured my gravity and it was at 1.070. Enter that into the formula and I get 86.07 g/L.

So if I need 6g of CaCl2 to brew I'd add 6g/86g/L = .069L (69ml), right?
 
This hygroscopic (sucks up water) behavior of calcium chloride is kind of a pain in the rear. Over time, it will draw water out of the atmosphere and attach that water to the solids. We brewers end up not knowing how much calcium and chloride we are actually adding to our brewing water.

How come? Does the water skew the weight measurements by making it heavier?
 
Does my lazy method of keeping it in a sealed containers with a handful of desiccant packets seem reasonable?

I remove it only long enough to weigh some out, and then back in it goes.
 
I keep my calcium chloride in a zip lock bag. I also live in a very humid climate and I've never had an issue with the salt liquefying. I open the ziplock, spoon out what I need, then squeeze out any air from the bag then rezip it.
 
PS: Calcium chloride liquifies when its hydration state is around the hexahydrate stage. That is quite a bit beyond the dihydrate stage. You can't see or tell if your calcium chloride has moved from anhydrous to the dihydrate stage or somewhat beyond. Its not until its way too late that you would notice a problem if liquifying was your indicator.
 
I keep my calcium chloride in a zip lock bag. I also live in a very humid climate and I've never had an issue with the salt liquefying. I open the ziplock, spoon out what I need, then squeeze out any air from the bag then rezip it.
The problem lies not so much in the liquefaction (though that is a mess) as in the weight gain from atmospheric moisture so that if the scale says you have a gram of powder you may have in reality really only 900 or 800 mg of CaCl2 with 100 or 200 mg water. If you have a sensitive balance, one that indicates to the mg, you can put a gram of powder on the pan and watch the numbers increase before your eyes. The early posts here show you how to estimate the water of hydration in your CaCl2 from a hydrometer reading. You can use that technique to determine how much water your material stored in a zip locked bag has picked up and it might be a good idea for you to do that to see whether your method is effective or not.
 
First I will admit that I'm not a 'science' guy so the following may be way out of line. After reading, and rereading this entire thread I came up with the following plan.

Dump all of my CaCl little white round balls into a quart of water. Let them dissolve and then get the mixture to around 77 degrees. Take a Specific Gravity reading and look up the Grams/Liter for that SG on one of the referenced charts.

On brew day take the amount of CaCl I need as calculated in Brun Water and divide that by the Grams/Liter from the chart and multiply by 1000. That gives me the amount of the liquid I need, in mL, to put in my kettle.

The next time I brew, I measure the SG to see if it has changed. If it has, look up the new G/L in the chart and repeat.

No percentages to figure out, no weighing.

As I said I'm not 'sciency'. Will this work?
 
Yes. And the simplicity of measurement is a lot of the appeal. You shouldn't need to recheck the SG as the solution, while it will exchange water with the air trapped in the bottle when you open it, doesn't do so nearly as dramatically as the powder.

Use a plastic bottle as the solution will corrode a metal closure.
 
Yes. And the simplicity of measurement is a lot of the appeal. You shouldn't need to recheck the SG as the solution, while it will exchange water with the air trapped in the bottle when you open it, doesn't do so nearly as dramatically as the powder.

Use a plastic bottle as the solution will corrode a metal closure.

Exchange water with the air trapped in the bottle when you open it? Isn't it already fully liquified?
 
Would anyone advocate that calcium chloride is one way to save an over-attenuated beer?

Mines just finished at 12 when I wanted 18 so I thought by increasing the chloride from 50 to 80 ppm this would promote the body and sweetness of the beer.
 
Exchange water with the air trapped in the bottle when you open it? Isn't it already fully liquified?

If the chemical potential of water (proportional to the log of the partial pressure of water vapor) in the gas over the liquid isn't equal to the chemical potential of water in the solution (proportional to the log of the mole fraction of water) water will flow into or out of the solution in an attempt to acheive balance. This should be a minor effect.
 
Yes. And the simplicity of measurement is a lot of the appeal. You shouldn't need to recheck the SG as the solution, while it will exchange water with the air trapped in the bottle when you open it, doesn't do so nearly as dramatically as the powder.

Use a plastic bottle as the solution will corrode a metal closure.

Thank you and a plastic bottle it is.
 
I should probably sit down and get my CaCl straightened out. I've been assuming it was heavier than normal for a while now, and just adding about 10-20% more for each addition.

Is it really that critical to the average homebrew recipe?
 
Is it really that critical to the average homebrew recipe?

I'd be interested on its impact as well, my main curiosity being whether I could increase it to balance a beer that has over-attenuated and finished too dry and too high in ABV as it supposedly increases the maltier aspects such as sweetness, flavour and perceived body/mouthfeel.
 
Hey everyone, just wanted your opinion on how to get a reliable CaCl2 concentration using a ppm meter. First let me say I do not own narrow range hydrometers, but I do have a ppm meter. This question of how hydrated my CaCl2 has become itched at my inner OCD far too long. So here is my math. I added 2.912 grams of my CaCl2 to 800mL of distilled water. I take a ppm measurement and receive a reading 3000ppm dissolved solute. Therefore I took the equation

(grams solute)
--------------------------------- X 1,000,000 = Concentration in ppm
(grams solvent+grams solute)

and solved for grams of solute

(Grams of solute) = (3000 ppm / 1,000,000) x (800g+2.912g)

Grams of solute = 2.408 meaning the remaining 0.504 grams was made up of water molecules...Aka I have a 82.6% sample of CaCl2 at my disposal. I could also use portions of my 800 ml since the concentration is now known ala if I add 10ml it is the same as adding 0.029g of CaCl2

Sound ok? Thanks everyone!

-Bobby
 
Your ppm meter is really a conductivity meter most probably scaled for sodium chloride. 3000 ppm NaCl means 100*3000/1000000 = 0.3%. An 0.3% solution of NaCl has conductivity of 4.543 mS/cm so we assume that this is the conductivity of your CaCl2 solution. A CaCl2 solution with conductivity 4.543 mS/cm is 0.3043 % CaCl2 w/w. You thus have
0.003043*(800*0.998203 + 2.912) = 2.43889 grams of CaCl2 in your mix and the purity of the salt is apparently 100*2.43889/2.912 = 83.7531% (many more decimal places here than justified). So, yes, assuming that your 'ppm' meter is calibrated for NaCl your method is indeed valid and rather clever too! There is one caveat and that is that the conductivities of NaCl and CaCl2 solutions are very close as long as the solutions are weak. As long as you stay below 5% w/w you should be OK.
 
I think this idea is important enough to propose a procedure for making CaCl2 solutions of known strength. Here is a suggested one.

Put 50 grams of CaCl2 in a 1 liter container. Add about 3/4 of a liter of DI water and stir until completely dissolved. Allow to cool. Make up to the mark (1L). You have a solution of nominal concentration 50/1000 = 5%. Making sure the temperature is 20 °C measure the conductivity. It should be around 67.4 mS/cm. If you are using a TDS meter you should read around 48138 ppm which you would convert to conductivity by the use of the formula
G = 15.22*pct -0.25315*pct*pct
where pct = ppm/10000. Insert the conductivity in the formula

pct = ( 15.026 -sqrt(15.026^2 - 4*0.30933*G))/(2*0.30933)

to get the actual strength of the solution in percent w/w (grams of CaCl2 per 100 grams of solution). For purposes of illustration lets say you read 60 mS/cm. Then
( 15.026 -sqrt(15.026^2 - 4*0.30933*60))/(2*0.30933) = 4.38978% w/w

Now plug that into the formula

((0.0010909*pct -0.08074)*pct + 10.01)*pct = grams/L

in our example

((0.0010909*4.38978 -0.08074)*4.38978 + 10.01)*4.38978 = 42.4781 g/L

As you put 50 grams of powder in the liter of solution you know that the purity of your CaCl2 is 100*42.478/50 = 84.956% and that each mL of the solution contains 42.5 mg CaCl2.0H2O.

Another approach is to add the 50 mg of salt to 1 L of water (998.203 grams at 20 °C). Let's say that we again measure 60 mS/cm so that we have, again, made a 4.38978% w/w solution. Now we know that we have 0.0438978*(50 + 998.203) = 46.0138 grams of CaCl2.0H2O our solution and since we used 50 g of salt to make it the purity is 100* 46.0138/50 = 92.0276% (we have added more water this time so clearly the salt would have to be purer to give the same conductivity). Since the grams CaCl2 per liter depends only on the conductivity the w/v strength of this solution is the same as before: 42.5 mg/mL
 
I just wanted to say I wish I had a tenth of aj's mind...lol. Thanks man for putting up with all of us dummies. We're lucky to have you on here
 
Thanks for the mention. However, I had a brain fart when coding the liquid algorithm. Supporters will be receiving version 3.4 over the next few weeks that corrects the error in liquid CaCl2 calculations. If you will be using a liquid solution before that, be sure to send me a note and I'll get you to the top of the list.

When using liquid CaCl2 in the latest version, are the adjustments done in grams per gallon or mL per gallon?
 
If I needed 5 grams CaCl2 and had a solution of SG 1.050 I'd use the formula in No. 1 to calculate that the solution contained 61 grams per liter and measure out 5/61 = 0.082 L (82 mL) of that solution but if you wanted to you could go further to calculate that the 82 mL of this solution weighs 82*1.050*0.998203 = 85.9453 grams, tare a container and weigh out 85.9453 grams of the solution. Seems like extra trouble to me and not justified by the accuracy required of this application. And it introduces the risk of accidentally dripping or pouring CaCl2 solution into your balance. For the program to be really useful it should take in the SG of the solution and convert any CaCl2.0H20 weight to cc of that solution. It could, of course, also calculate the weight of that solution.
 
If I needed 5 grams CaCl2 and had a solution of SG 1.050 I'd use the formula in No. 1 to calculate that the solution contained 61 grams per liter and measure out 5/61 = 0.082 L (82 mL) of that solution but if you wanted to you could go further to calculate that the 82 mL of this solution weighs 82*1.050*0.998203 = 85.9453 grams, tare a container and weigh out 85.9453 grams of the solution. Seems like extra trouble to me and not justified by the accuracy required of this application. And it introduces the risk of accidentally dripping or pouring CaCl2 solution into your balance. For the program to be really useful it should take in the SG of the solution and convert any CaCl2.0H20 weight to cc of that solution. It could, of course, also calculate the weight of that solution.

i think 100g would be fine. This isn't rocket surgery.
 
Started playing around with some solution today and need a reality check!

My solution has a SG of 1.060, or 6.7% w/w according to Bru'n Water. The Pale Ale profile suggests 55 ppm of Chloride and to get there, starting with RO water, I need 37 grams of CaCI2. This comes out to ~500ml of solution assuming I want to get to the number using CaCI2 alone.

I know I can use canning salt to add chloride but was just playing with different ways to "skin the cat".

Does this seem right? For those of you who use a solution, what is your typical strength and how much do you make at a time?
 
I'm pretty sure the 37 grams you mention is the weight of the solution, not the CaCl2! At least that's the way I interpret the Bru'n Water calc.

My solution is around 17% IIRC, and I use a dropper to weigh out the solution into a little stainless steel cup and then drop and rinse that in with the mash water.
 
My solution has a SG of 1.060, or 6.7% w/w according to Bru'n Water. The Pale Ale profile suggests 55 ppm of Chloride and to get there, starting with RO water, I need 37 grams of CaCI2. This comes out to ~500ml of solution assuming I want to get to the number using CaCI2 alone.

This is why the formula is given in g/L of solution:

g/L = -684.57 + 175.12*SG + 509.45*SG*SG

If your SG is 1.060 you have

-684.57 + 175.12*1.06 + 509.45*1.06^2= 73.4752 grams per liter and if you want 37 grams you then need 1000*37/73.4752 = 503 mL

That is a lot of CaCl2. Are you doing a couple of barrels?

BTW as a liter of your solution weighs 1.06*998.203 grams its strength is
100*( -684.57 + 175.12*1.06 + 509.45*1.06^2)/(1.06*998.203)=6.9441% w/w so there is apparently a small problem in Bru'n water but there is no need to worry about w/w strength if you use the w/v formula as given in the sticky.


Does this seem right?
Yes. 503 mL vs 500 mL is 10*log(503/500) = 0.0259 dB. I don't worry until error get to a dB or more.


For those of you who use a solution, what is your typical strength and how much do you make at a time?
To be honest I don't worry about it. I just assume my CaCl2 is about 80%. If it's actual strength is 100 % of as little as 60% that's only ±1.2 dB and it really doesn't matter. From time to time I check my CaCl2 stock to see what the CaCl2 concentration actually is and adjust the 80% assumption accordingly.
 
This is why the formula is given in g/L of solution:

g/L = -684.57 + 175.12*SG + 509.45*SG*SG

If your SG is 1.060 you have

-684.57 + 175.12*1.06 + 509.45*1.06^2= 73.4752 grams per liter and if you want 37 grams you then need 1000*37/73.4752 = 503 mL

That is a lot of CaCl2. Are you doing a couple of barrels?

BTW as a liter of your solution weighs 1.06*998.203 grams its strength is
100*( -684.57 + 175.12*1.06 + 509.45*1.06^2)/(1.06*998.203)=6.9441% w/w so there is apparently a small problem in Bru'n water but there is no need to worry about w/w strength if you use the w/v formula as given in the sticky.


Yes. 503 mL vs 500 mL is 10*log(503/500) = 0.0259 dB. I don't worry until error get to a dB or more.


To be honest I don't worry about it. I just assume my CaCl2 is about 80%. If it's actual strength is 100 % of as little as 60% that's only ±1.2 dB and it really doesn't matter. From time to time I check my CaCl2 stock to see what the CaCl2 concentration actually is and adjust the 80% assumption accordingly.

Thanks for the response. After reading your post I went back and not only did I "mis-use" Bru'N Water, I also screwed up a formula!

If you're going to make one mistake, you might as well make a few more and pile on the problem, I guess!

That's why I originally posted, I just couldn't get my head around using that much CaCI2 and didn't see my mistake(s).

Thanks again!
 
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