The hydration of CaCl2 is quite exothermic. The solution gets hot. One needs to let it cool in order to be able to get an accurate hydrometer reading.
Measuring Calcium Chloride
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The hydration of CaCl2 is quite exothermic. The solution gets hot. One needs to let it cool in order to be able to get an accurate hydrometer reading.
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 I meant 77 degrees F since that's what the chart used as a reference temp.
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?
Brew_Meister_General
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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.
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Very easy to find out. Just put a bit in a glass of the beer in question.
LtCornuBrew
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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
(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.
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
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
orangehero
Well-Known Member
When using liquid CaCl2 in the latest version, are the adjustments done in grams per gallon or mL per gallon?
Its grams. Mass measurement is still superior to volumetric measurement.
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.
Norselord
Well-Known Member
i think 100g would be fine. This isn't rocket surgery.
Or brain science either. How about 80 mL?
Murphys_Law
Well-Known Member
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?
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 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.
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.
Murphys_Law
Well-Known Member
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!
Bigdaddyale
Well-Known Member
sub
Silver_Is_Money
Larry Sayre, Developer of 'Mash Made Easy'
I just finished making up one liter of CaCl2 solution. I added 109.74 grams and carefully brought it up to 1 liter in an Erlenmeyer flask, and when it cooled it measured in at a final SG of 1.077. From this I believe that my "fresh' CaCl2 was about 86.6% by weight as CaCl2, or somewhere between the anhydrite and the dihydrate form.
If I'm calculating this correctly, then 7.95 ml of my solution (call it 8 ml) is the equivalent of 1 gram of the most commonly assumed dihydrate form. Does this sound correct?
If I'm calculating this correctly, then 7.95 ml of my solution (call it 8 ml) is the equivalent of 1 gram of the most commonly assumed dihydrate form. Does this sound correct?
Without a precision hydrometer, could I use a refractometer instead?
I just tested my 5 month old bottle of LD Carlson Calcium Chloride.
I had about half a bottle (26.8g). I heated 500mL of water (which I weighed out since my flask is not exactly precision), dissolved the CaCl2 into it, and cooled to room temp.
I used a regular calibrated hydrometer to measure it and it looked like it was 1.039. I used my refractometer to double-check it but that only showed 1.029. I'm guessing there's some reason a refractometer doesn't work for this purpose since nobody else mentioned it.
I plugged that SG into your equation and got 46.86g/L.
Since my original g/L was 53.6 it looks like my CaCl2 is ~86%.
Sound about right?
I just tested my 5 month old bottle of LD Carlson Calcium Chloride.
I had about half a bottle (26.8g). I heated 500mL of water (which I weighed out since my flask is not exactly precision), dissolved the CaCl2 into it, and cooled to room temp.
I used a regular calibrated hydrometer to measure it and it looked like it was 1.039. I used my refractometer to double-check it but that only showed 1.029. I'm guessing there's some reason a refractometer doesn't work for this purpose since nobody else mentioned it.
I plugged that SG into your equation and got 46.86g/L.
Since my original g/L was 53.6 it looks like my CaCl2 is ~86%.
Sound about right?
A refractometer works just fine for this purpose but you cannot use the Brix scale. You must reduce the Brix reading to a refractive index reading and then look up the calcium chloride concentration corresponding to that refractive index. You can find tables of Brix vs RI and CaCl2 concentration vs RI on line. From them you can obviously prepare a table of Brix reading vs CaCl2 concentration.
For those looking for the chart referenced in Martin's second post of this thread, it has been changed to http://www.osi-univers.org/IMG/pdf/CalciumChloridHandbook-2.pdf.
Unfortunately the Dow Handbook does not list RI vs concentration. I have found such tables on the net but can't remember where (if I did I'd tell you).
Royski
Well-Known Member
And it tells us that the data are in the CRC Handbook. Good to know.
While we are at it one can use conductivity as well but in this case one needs to know the conductivity of CaCl2 solutions and the scale that one's particular conductivity meter uses.
While we are at it one can use conductivity as well but in this case one needs to know the conductivity of CaCl2 solutions and the scale that one's particular conductivity meter uses.
PartagasD4
Well-Known Member
Can someone help please. Twice now I have made a 1.090 solution with store bought distilled water. After about a month I am getting a lot of calcium, I think it's calcium, dropping out of the solution. When swirling the jar it looks like a snow globe. Is this something to worry about? Am I making the solution to strong? What am I doing wrong?
Take some of the precipitate (filter it out through a paper towel) and drip vinegar on it. Does it fizz?
