So you used 1/2 tsp each for gypsum and calcium chloride?
Measuring Calcium Chloride
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So you used 1/2 tsp each for gypsum and calcium chloride?
Wait, I have probably been reading the original primer wrong. It says baseline is 1 tbs of calcium chloride. Then it says for IPAs add 1 tbs gypsum and 1 tbs of calcium chloride. Is it saying for an IPA the total should be 1 tbs of gypsum and 2 tbs of calcium chloride? Or a total of 1 tbs of gypsum and 1 tbs of calcium chloride?
Dcpcooks
Well-Known Member
Yes 1/2 of CaCL and 1/2 of gypsum per five gallons of liquor inclusive of sparge water. Don't forget the acid malt at 2% of the grain bill. That's needed to bring your ph down. It's also relative to your water source, I'm assuming your using RO water with very little if any Carmel malt and you don't have a ph meter.
Dcpcooks
Well-Known Member
The original primer recommendation for IPA was 1 tsp of each.
The newer recommendation is half that amount.
Yes indeed. Applying your calculations i have a difference of 0.7 gr/L on my measurements between hydrometer and refractometer. That is great!
Great indeed! Tells me I got the math about right.
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Thanks!!
JDL
Well-Known Member
I did the above and kept the solution in a mason jar sealed up and have used it for 2 brews in the past 2 months. I am about to brew again tomorrow and went and looked at my jar and sediment has formed on the bottom. I figure I have three choices and wanted to see what others would do or hopefully ajdelange sees this in time.
1) Take liquid off the top and test it for strength and adjust as needed.
2) Do something to mix the solution again i.e. shake or heat it up and then test the strength.
3) Not use it as an addition and just have the water chemistry be a little off and order more for the next brew day.
Thanks in advance,
Jeff
This has been discussed previously... can't remember if this thread or another. It is almost certainly calcium carbonate.
1. You should test the strength, but i suspect you'd see barely any change.
2. Calcium carbonate is insoluble, so it won't mix back in easily.
Something you can try is to decant off the liquid and pour vinegar on the (dried) solids. If they fizz then you've got calcium carbonate.
Does the amount of CaCl and amount of water matter? Or are we just looking to make a solution of any random amount of both and taking that SG whatever it may be? In my head it would make sense that the amount of CaCl and water used doesn't matter. We are just looking for the SG.
Well at least that is what I'm hoping because that is what I did. My CaCl was old so I figured I'd do it anyway. The SG of the solution is 1.090. I couldn't tell you how much CaCl or water I added. So if I'm doing this right, I get g/L = 111.5883. And if I need 2g of CaCl, I'll need to use 17.923mL of the solution.
Well at least that is what I'm hoping because that is what I did. My CaCl was old so I figured I'd do it anyway. The SG of the solution is 1.090. I couldn't tell you how much CaCl or water I added. So if I'm doing this right, I get g/L = 111.5883. And if I need 2g of CaCl, I'll need to use 17.923mL of the solution.
As I understand it, you are correct if you are using BnW or otherwise have some way of figuring out the conversion of SG of CaCL solution to CaCL pellet weight.
The beginning of this thread talks about how you can determine the strength of your CaCL pellets but to your point, in practice it doesn't matter...just make a solution, measure it, and be done with it.
Please correct me if I'm wrong because this has been my working theory since I read this thread.
I read it again and it seems like what I did was correct. Took a "large" amount of CaCl and add enough water to take an SG reading. No specific amounts are given. Plug the SG into the formula to get g/L. Take the grams needed for the recipe divided by the g/L amount and multiple by 1000 to get mL.
My CaCl was really old. It wasn't getting clumpy but it wasn't air tight. I figured I'd just make the solution with the entire CaCl and keep it in a flask. From what I have read, this solution shouldn't go bad or change SG since it has been fully hydrated. Is that correct?
Yes, your surmise is correct. The theory is that all CaCl2 in the powder will dissolve releasing bound water into the solution as well while any CaCO3 will precipitate. You measure SG and determine the g/L dissolved CaCl2 from the formula in the OP and that tells you how many mL of that solution you would need to deliver X grams of CaCl2 (X/g/L). But it also tells you how much CaCl2 was in the dissolved powder. That's simply the volume of the solution (in liters) multiplied by the g/L. That number can be used to determine the purity of the original powder if you weighed it out. This should be pretty accurate unless there is a layer of undissolved CaCO3 on the bottom of the container in which case you should decant the liquid, wash the residue with a little DI water and decant that too.
The solution should be pretty stable, certainly more stable than the powder, but, if the container is left open it will exchange water with the air losing water if the humidity is low and the solution weak or taking some on if the humidity is high and the solution strong.
The solution should be pretty stable, certainly more stable than the powder, but, if the container is left open it will exchange water with the air losing water if the humidity is low and the solution weak or taking some on if the humidity is high and the solution strong.
Silver_Is_Money
Larry Sayre, Developer of 'Mash Made Easy'
17.923 mL agrees with a spreadsheet that I made based upon AJ's formula. This either confirms that you are on the right track, or it confirms that both of us are making the exact same calculating error. I vote for the former.
In reading back over this thread I note a lot of concern about the precipitation of flakes of calcium carbonate over time to the point that some have abandoned the scheme for management of calcium chloride proposed in this thread. Early in the thread concern seems to have been focused on the fact that exactly what this material is was unknown an secondarily on that clearly as something is coming out of solution something must be being lost. Common sense says that the material is calcium carbonate produced by carbon dioxide in the air and experiments by several indicate that this is the case as the material fizzes when exposed to acid. I had originally opined that the reaction involved is
Ca++ + 2Cl- + CO2 + H2O ---> CaCO3 + 2H+ + 2Cl-
but we note the release of hydrogen ions implying a reduction in pH. This reaction can indeed tale place but once the pH has been lowered to 8.2 no more CO2 will dissolve as atmospheric CO2 (assumed to be at 0.0004 bar) will be in equilibrium with the precipitated chalk so we doubt this is the responsible mechanism. A little investigation revealed that most CaCl2 produced is as a byproduct of the Solvay process in which limestone and salt (NaCl) are feedstocks for the production of soda ash (NaCO3). This process is carried out at high pH and as a consequence it is not surprising that the CaCl2 is 'contaminated' with some (OH)- ions. We also found a specification for FCC grade CaCl2 (what we buy from the LHBS) which promised alkalinity of less than 0.3% w/w as Ca(OH2). Thus our CaCl2 solution is likely to contain some 'lime water' and we all remember from high school chemistry that lime water is used to detect CO2:
Ca++ + 2(OH)- + CO2 ---> CaCO3 + H2O
It is thus quite reasonable for us to expect that a strong solution of FCC CaCl2 is, if, as inevitably it will be, it is exposed to air, going to precipitate some chalk.
Some may want to know how much of this will precipitate. To start lets look at 100 grams of 'CaCl2' powder. If the hydroxyl contamination level is 0.3 % that means it is equivalent to 300 mg. As the molecular weight is 74.093 that implies the equivalent of 300/74.093 = 4.05 mmol of Ca(OH)2. or 8.1 mEq of (OH)- ions in the solution which, according to the equation above, will precipitate 4.05 mmol of calcium. The molecular weight of anhydrous calcium chloride is 110.98 so nominally 100 grams of that is 1000*100./110.98 = 901.063 mmol and thus our calcium loss would be 4.05/901.063 = 0.45% or 1.5 times the specified alkalinity of the product if that alkalinity is expressed w/w as Ca(OH)2 which in all liklihood it is going to be though in fact not all the hydroxyl ions are going to be paired with calcium though, of course, most of them will be. Given the feedstocks for the Solvay process there will doubtless be magnesium, sodium, potassium and probably other cations present as well.
As, in general, we don't have the alkalinity for the product we bought from the LHBS there may be those that want to be able to obtain some indication as to how much hydroxyl contamination is actually present the the product they are holding in their hands. There is a simple test which tells us how much (OH)- is in a sample and that is one with which we are well familiar: the alkalinity test. Most here are aware that in that test acid of known strength is added to a solution and the pH monitored until the pH is reduced to 4.5 (ISO test). The alkalinity of the solution is simply the number of mEq of acid required to bring a liter of the solution to that pH. The number, called the M alkalinity (or sometimes T for 'total') is often multiplied by 50 to give units of 'ppm as CaCO3' but we won't do that here. What some may not realize is that the test is actually done in two steps with the first being a titration to pH 8.3. The amount of acid required to get to pH 8.3 is called the P alkalinity. By making a solution of the product in question and determining its alkalinity, P and M, we can determine (OH)- concentration.
Let's assume we start, as suggested above, with 100 grams of 'CaCl2' in liter of DI water. We've calculated that for 0.3% alkalinity we would have 8.1 mEq of (OH)- ions in the solution. This implies that the pH of such a solution is going to be 14 + log(0.0081) = 11.91. Thus the first clue that your solution is going to eventually precipitate chalk is a high pH like this. Were the alkalinity 0.15% the (OH)- concentration would be 4.1 mmol/L and the pH =14 +log(0.0041) = 11.61 i.e. it drops by 0.3 for each halving of the alkalinity. Were the alkalinity 0.03% as opposed to 0.3% the pH would drop 1 unit to 10.91 i.e. by 1 unit for each factor of 10 reduction in the alkalinity. So for 0.003% contamination pH would be 9.91 etc. so that it is pretty clear that a pH < 10 means insignificant loss of calcium to precipitation. Thus one really does not really need to do a titration. A simple pH measurement is really sufficient to tell you whether you need to worry about precipitation or not or indeed whether you can expect to even have it.
But as this is the Brewing Science forum, here goes:
The amount of acid required to move a liter of solution from pHs (its sample pH) to pH = 8.3 is the P alkalinity and is, approximately
P = [(OH)-] + Ct*(-1 - Q(pHs)) = [(OH-] - Ct*(1 + Q(pHs))
P is in units of mEq acid added per liter to reach pH = 8.3
[(OH)-] is the concnetration of hydroxyl ions in mEq/L
Ct is the total concentration of carbon atoms in CO3--, HCO3- and CO2 in solution in units of mmol/L
Ct is found from the approximation
Ct ~ (M - P)
i.e. the difference between the M and P alkalinities. It should be small or 0 and is there only if the CaCl2 is contaminated with chalk which, again, given the feedstocks, is reasonable to assume but as it is so insoluble there shouldn't be much.
Q(pHs) can be found from the curve in the Palmer Kaminsky book but is easily calculated from
r1 = 10^(pHs - 6.38)
r2 = 10^(pHs - 10.38)
f0 = 1/(1 + r1 + r1*r2)
f1 = r1*f0
f2 = r2*f1
Q = -(f1 + 2*f2)
Typical value: Q(10.38) = -1.5
Then
[(OH-] = P + Ct*(1 + Q(pHs))
If M and P are close, as we would expect
[(OH-] ~ P
but you can also, in this case, calculate it from
[(OH-] = 1000*10^(pHs - 14)
To sumarize: the stuff is chalk (CaCO3). It is harmless. Very little of it precipitates thus very little calcium is lost. If the pH of a solution of 100 grams/L is below 10 the loss is likely to be less than 0.0045% of the dissolved calcium ion. The solution SG will not be appreciably effected.
Ca++ + 2Cl- + CO2 + H2O ---> CaCO3 + 2H+ + 2Cl-
but we note the release of hydrogen ions implying a reduction in pH. This reaction can indeed tale place but once the pH has been lowered to 8.2 no more CO2 will dissolve as atmospheric CO2 (assumed to be at 0.0004 bar) will be in equilibrium with the precipitated chalk so we doubt this is the responsible mechanism. A little investigation revealed that most CaCl2 produced is as a byproduct of the Solvay process in which limestone and salt (NaCl) are feedstocks for the production of soda ash (NaCO3). This process is carried out at high pH and as a consequence it is not surprising that the CaCl2 is 'contaminated' with some (OH)- ions. We also found a specification for FCC grade CaCl2 (what we buy from the LHBS) which promised alkalinity of less than 0.3% w/w as Ca(OH2). Thus our CaCl2 solution is likely to contain some 'lime water' and we all remember from high school chemistry that lime water is used to detect CO2:
Ca++ + 2(OH)- + CO2 ---> CaCO3 + H2O
It is thus quite reasonable for us to expect that a strong solution of FCC CaCl2 is, if, as inevitably it will be, it is exposed to air, going to precipitate some chalk.
Some may want to know how much of this will precipitate. To start lets look at 100 grams of 'CaCl2' powder. If the hydroxyl contamination level is 0.3 % that means it is equivalent to 300 mg. As the molecular weight is 74.093 that implies the equivalent of 300/74.093 = 4.05 mmol of Ca(OH)2. or 8.1 mEq of (OH)- ions in the solution which, according to the equation above, will precipitate 4.05 mmol of calcium. The molecular weight of anhydrous calcium chloride is 110.98 so nominally 100 grams of that is 1000*100./110.98 = 901.063 mmol and thus our calcium loss would be 4.05/901.063 = 0.45% or 1.5 times the specified alkalinity of the product if that alkalinity is expressed w/w as Ca(OH)2 which in all liklihood it is going to be though in fact not all the hydroxyl ions are going to be paired with calcium though, of course, most of them will be. Given the feedstocks for the Solvay process there will doubtless be magnesium, sodium, potassium and probably other cations present as well.
As, in general, we don't have the alkalinity for the product we bought from the LHBS there may be those that want to be able to obtain some indication as to how much hydroxyl contamination is actually present the the product they are holding in their hands. There is a simple test which tells us how much (OH)- is in a sample and that is one with which we are well familiar: the alkalinity test. Most here are aware that in that test acid of known strength is added to a solution and the pH monitored until the pH is reduced to 4.5 (ISO test). The alkalinity of the solution is simply the number of mEq of acid required to bring a liter of the solution to that pH. The number, called the M alkalinity (or sometimes T for 'total') is often multiplied by 50 to give units of 'ppm as CaCO3' but we won't do that here. What some may not realize is that the test is actually done in two steps with the first being a titration to pH 8.3. The amount of acid required to get to pH 8.3 is called the P alkalinity. By making a solution of the product in question and determining its alkalinity, P and M, we can determine (OH)- concentration.
Let's assume we start, as suggested above, with 100 grams of 'CaCl2' in liter of DI water. We've calculated that for 0.3% alkalinity we would have 8.1 mEq of (OH)- ions in the solution. This implies that the pH of such a solution is going to be 14 + log(0.0081) = 11.91. Thus the first clue that your solution is going to eventually precipitate chalk is a high pH like this. Were the alkalinity 0.15% the (OH)- concentration would be 4.1 mmol/L and the pH =14 +log(0.0041) = 11.61 i.e. it drops by 0.3 for each halving of the alkalinity. Were the alkalinity 0.03% as opposed to 0.3% the pH would drop 1 unit to 10.91 i.e. by 1 unit for each factor of 10 reduction in the alkalinity. So for 0.003% contamination pH would be 9.91 etc. so that it is pretty clear that a pH < 10 means insignificant loss of calcium to precipitation. Thus one really does not really need to do a titration. A simple pH measurement is really sufficient to tell you whether you need to worry about precipitation or not or indeed whether you can expect to even have it.
But as this is the Brewing Science forum, here goes:
The amount of acid required to move a liter of solution from pHs (its sample pH) to pH = 8.3 is the P alkalinity and is, approximately
P = [(OH)-] + Ct*(-1 - Q(pHs)) = [(OH-] - Ct*(1 + Q(pHs))
P is in units of mEq acid added per liter to reach pH = 8.3
[(OH)-] is the concnetration of hydroxyl ions in mEq/L
Ct is the total concentration of carbon atoms in CO3--, HCO3- and CO2 in solution in units of mmol/L
Ct is found from the approximation
Ct ~ (M - P)
i.e. the difference between the M and P alkalinities. It should be small or 0 and is there only if the CaCl2 is contaminated with chalk which, again, given the feedstocks, is reasonable to assume but as it is so insoluble there shouldn't be much.
Q(pHs) can be found from the curve in the Palmer Kaminsky book but is easily calculated from
r1 = 10^(pHs - 6.38)
r2 = 10^(pHs - 10.38)
f0 = 1/(1 + r1 + r1*r2)
f1 = r1*f0
f2 = r2*f1
Q = -(f1 + 2*f2)
Typical value: Q(10.38) = -1.5
Then
[(OH-] = P + Ct*(1 + Q(pHs))
If M and P are close, as we would expect
[(OH-] ~ P
but you can also, in this case, calculate it from
[(OH-] = 1000*10^(pHs - 14)
To sumarize: the stuff is chalk (CaCO3). It is harmless. Very little of it precipitates thus very little calcium is lost. If the pH of a solution of 100 grams/L is below 10 the loss is likely to be less than 0.0045% of the dissolved calcium ion. The solution SG will not be appreciably effected.
Silver_Is_Money
Larry Sayre, Developer of 'Mash Made Easy'
Amazing! Thanks A.J.
Note that in this technique we use the two alkalinities to determine the carbo (Ct) and (OH)- concentrations. That is, of course, exactly what we want to do in determining the extent to which lime Ca(OH)2 has been contaminated with aerial CO2 and so we measure P and M for lime purity as well.
Silver_Is_Money
Larry Sayre, Developer of 'Mash Made Easy'
If I stick my pH meter into my CaCl2 solution is there any chance of damaging the electrode by doing this test?
Only if you bang the bulb against the container.
I would think that would be OK. Such containers are designed not to leach into the food products they contain. I can't imagine that they would leach into your CaCl2 solution either.
Bigdaddyale
Well-Known Member
little help, please.I have a brix of 26.2 @ 73.5 F
Per No. 116:
grams CaCl2/Liter solution = 5.9392*Bx + 0.050098*Bx^2 -0.00018856*Bx^3 + 4.7399e-06*Bx^4 = 188.839 for Bx = 26.2 as read from a refractometer
Bigdaddyale
Well-Known Member
Bigdaddyale
Well-Known Member
I have no idea what this means, but I have hope if I read the earlier threads it will all make sense.
Means that your solution has 188.8 grams of calcium chloride per liter and you will need 5,3 ml of that solution to get 1 gr of calcium chloride (or 5,6 gr if you wish to weigh the solution).
I made the solution in my 2L flask and let it sit. Now after two PBW soaks I can't remove these white stains. I assume these stains are related to the CaCl solution but not sure why they won't come off. Any ideas?
Bigdaddyale
Well-Known Member
Try using a Star San solution or vinegar
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^+1 @Bigdaddyale is right, StarSan should do it.
