Borisfa
Well-Known Member
Multiple postage. Sorry.
Reducing alkalinity with slaked lime
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Borisfa
Well-Known Member
@ajdelange, and about my calcs, what can you say?
About #3, revisiting my calcs, I have a doubt:
3.1) Would be 1,9 mmol * 74 mg/mmol = 141 mg/L of Lime
or
3.2) Would be 186,5 mg of Alkalinity/L * (74 mg of Lime/mmol) / (200 mg of Chalk/mmol) = 69 mg/L
Ca++ + Ca(OH)2 + 2 HCO3- --> 2 CaCO3 + 2 H2O
MW 40 74 122 200 36
Base y x 186,5
y = 186,5*40/200 = 37,3 mg/L
x = 186,5*74/200 = 69,0 mg/L
But On x we have 37,3 mg/L of Ca++ with the same 37,3 mg/L of Ca++ ion = 74,6 mg/L would be the stocheometric ratio. With 30% recommended on article, 97,0 mg/L of Ca++
For the same reason, stocheometric ratio for Lime indicate 69 mg/L.
What is the right, 3.1 or 3.2?
Thanks a lot for your help.
Fabiano da Mata
Last edited:
Unfortunately, there is no 'right' answer. If you follow the guidelines of the article (with the missing decimal point inserted) looking at the water as is i.e. with its 3,73 mEq/L alkalinity and 3,18 mEq/L calcium hardness you will estimate the lime needed from the temporary hardness which is 3,18 mEq (all the calcium hardness as there is more alkalinity than hardness). The article implies that the reaction
Ca++ + 2HCO3- + (Ca++ + 2(OH)-) ---> 2CaCO3 + 2H2O
goes to completion and as it requires 1 mmol of lime for each 2 mEq of temporary hardness seems to indicate that we would need 3,18/2 = 1,59 mmol of lime (1,59*74.1 = 117.8 grams). The article suggests preparing a slurry of 20% more than that making a total of 141.4 grams for each liter of water to be treated. This slurry is then added incrementally to the water while pH is monitored. All the slurry will not be used. We know that if a decarbonation plant is well operated that we can expect a reduction in the temporary hardness down to about 1 mEq indicating that there will be quite a bit of excess. But we in fact never know exactly what will happen.
We also know in this case that the temporary hardness, and therefore, the amount of alkalinity that can be removed, is limited by the calcium. If we'd like to get more of the alkalinity the answer is to simply increase the calcium making the alkalinity the limiting factor. Any addition of calcium which brings its total to greater than the alkalinity will achieve this. Increasing calcium to >3,73 makes 3,73 mEq/L the temporary hardness and so we calculate that we need 1/2 of 3,73 mEq or 1,86 mmol of Ca(OH)2 the amount we would augment by 20% and prepare as a slurry. But again we don't know exactly what will happen and that is why we use pH meters to control the addition and measure the results. As I said earlier, for a home brewer the boiling process is equally as effective and much simpler to manage. A commercial brewer would not choose boiling as the cost of the lime is much less than the cost of the energy and he has a lab to verify that he is doing the lime treatment optimally.
The other approach is to look at the temporary hardness of 3,18, guess that you are going to reduce that to 1 and calculate the amount of lime based on the 2,18 reduction. You are really flying blind here but I guess this is what a lot of home brewers do.
Ca++ + 2HCO3- + (Ca++ + 2(OH)-) ---> 2CaCO3 + 2H2O
goes to completion and as it requires 1 mmol of lime for each 2 mEq of temporary hardness seems to indicate that we would need 3,18/2 = 1,59 mmol of lime (1,59*74.1 = 117.8 grams). The article suggests preparing a slurry of 20% more than that making a total of 141.4 grams for each liter of water to be treated. This slurry is then added incrementally to the water while pH is monitored. All the slurry will not be used. We know that if a decarbonation plant is well operated that we can expect a reduction in the temporary hardness down to about 1 mEq indicating that there will be quite a bit of excess. But we in fact never know exactly what will happen.
We also know in this case that the temporary hardness, and therefore, the amount of alkalinity that can be removed, is limited by the calcium. If we'd like to get more of the alkalinity the answer is to simply increase the calcium making the alkalinity the limiting factor. Any addition of calcium which brings its total to greater than the alkalinity will achieve this. Increasing calcium to >3,73 makes 3,73 mEq/L the temporary hardness and so we calculate that we need 1/2 of 3,73 mEq or 1,86 mmol of Ca(OH)2 the amount we would augment by 20% and prepare as a slurry. But again we don't know exactly what will happen and that is why we use pH meters to control the addition and measure the results. As I said earlier, for a home brewer the boiling process is equally as effective and much simpler to manage. A commercial brewer would not choose boiling as the cost of the lime is much less than the cost of the energy and he has a lab to verify that he is doing the lime treatment optimally.
The other approach is to look at the temporary hardness of 3,18, guess that you are going to reduce that to 1 and calculate the amount of lime based on the 2,18 reduction. You are really flying blind here but I guess this is what a lot of home brewers do.
Borisfa
Well-Known Member
People, may I add all salts on mash water and not add any salt on sparge water (only adjust pH of sparge water)? What is the potencial benefit (or not) to add all salt only on mash water?
Thanks,
Fabiano da Mata
Thanks,
Fabiano da Mata
Salts added to the mash are to put your pH into the proper range, and the sparge salts are added to contribute more flavor ions, as I understand it. They also help place the finished beer in the proper pH range, although I think I've heard that the beer naturally corrects itself to a certain degree, if possible.
Borisfa
Well-Known Member
People, EZ water and Bru'n Water spreadsheet help us to calculate how many salts we should add to adjust our water.
So, some salts have H2O on their formation.
My doubt is: should I adjust the salt mass to considere that water? Because I suspect that mass is to pure component (CaSO4, MgSO4, CaCl2....) and when I measure, for example 3 g of CaSO4 this mass is not CaSO4, but CaSO4.xH2O.
So, some salts have H2O on their formation.
My doubt is: should I adjust the salt mass to considere that water? Because I suspect that mass is to pure component (CaSO4, MgSO4, CaCl2....) and when I measure, for example 3 g of CaSO4 this mass is not CaSO4, but CaSO4.xH2O.
Borisfa
Well-Known Member
Make sense my doubt? Or epsom, gympsum, CaCl2 labeled on spreadsheats already considere hydrate (water)?
OP
OP
No. The only one that you have to consider is Calcium Chloride (CaCl2) since it will easily absorb water. The answer is to use fresh CaCl2 and keep it tightly closed in a small container and use it quickly.
Water of hydration must be taken into account. Gypsum and epsom salts are pretty stable in this regard and the spreadsheets usually automatically account for the water by assuming 7 for epsom salts and 2 for gypsum. Calcium chloride is trickier. See https://www.homebrewtalk.com/showthread.php?t=501377 for how to manage it.
This is the best thread I've seen on this topic, so I'll necro it to ask a related question. 
Slaked lime absorbs CO2 out of the air and slowly converts to carbonate. Does this need to be taken into account when calculating the amount of lime to add?
Let's say I have a 5 year old bag of Mrs Wage's' Pickling Lime that I bought for making pickles. It's been stored in the original (opened) paper bag in a dry place. How much is still lime and how much is chalk?
Slaked lime absorbs CO2 out of the air and slowly converts to carbonate. Does this need to be taken into account when calculating the amount of lime to add?
Let's say I have a 5 year old bag of Mrs Wage's' Pickling Lime that I bought for making pickles. It's been stored in the original (opened) paper bag in a dry place. How much is still lime and how much is chalk?
Its virtually impossible to know the degree of degradation without significant work. However, it is possible to negate the effect of that degradation by creating a limewater or kalkwasser solution which is a saturated solution of calcium hydroxide. In essence, you add lime to the water and it reaches an asymptotitic pH that typically falls between 12 and 13 at the typical range of room temperatures.
That solution can be relied on to give you the alkalinity that you are seeking. However, I haven't yet worked through the calculations to figure out what that value is with respect to temperature. Sorry.
That solution can be relied on to give you the alkalinity that you are seeking. However, I haven't yet worked through the calculations to figure out what that value is with respect to temperature. Sorry.
Yes and no. If you use Hubert Hanghoffer's method which relies only on pH readings then you don't calculate or measure anything but pH. It goes without saying, however, that you will be well served by having an idea as to how much lime is required before you begin.
Impossible to guess but, fortunately it is pretty simple to determine at least approximately by measuring the alkalinities (note plural) of a 'solution'. Start by putting some of the powder in a test tube or on a plate and drop some acid on it (vinegar will do). If it fizzes then there is appreciable carbonate and you will want to know how much. The presence of the carbonate is by no means detrimental. In fact it is beneficial as a source of nucleation sites.
Measure out 100 mg of the 'lime' you have at hand and dissolve it in cold DI water (RO will do if your TDS is low). Make up to 100 mL. Shake or stir thoroughly. As the solubility of lime is 1730 mg/L at 20 °C the powder should all dissolve eventually, though it may take a lot of stirring, if it is all Ca(OH)2. If it does that is a good indication that there is no CaCO3 (we are assuming no other significant levels of impurities here) and you need go no further.
If undissolved powder remains measure the P alkalinity. This done by adding a bit of phenolpthalein to the beaker or flask and then adding 0.1 N acid until the color disappears. The number of mL added acid is the P alkalinity in mEq/L. What has happened at this point is described by:
xCa(OH)2 +yCaCO3 + (2x+y)H+ ---> (x + y)Ca++ + 2xH2O + yHCO3-
From this it is apparent that
P = 2x + y
where x is the number of millimoles of lime in the powder and y the number of millimoles of calcium carbonate. In words, most of the OH- ions have been protonated to form water and most of the carbonate ions have been protonated to form bicarbonate ions. The 'most of' phrases explain why the answer you get will be approximate. The pH at this point is about 8. Of course, if you have a pH meter you can use it to titrate to pH 8. If you are color blind, as I and 1 in 12 males are (in my whole life I've only met one color blind female and, believe me, that was the least of her problems) the pH meter isn't really an option.
If there was undissolved powder, i.e. carbonate, this will take a looong time to dissolve so you must be patient. The phenolpthalein will clear or the meter will indicate pH 8 but upon standing or stirring the pH will start to creep back up. A colorless but turbid solution indicates that dissolution of the powder is not complete even though the pH is 8 or less. If this happens, start over again with 50 mg of powder or even less. You must have a perfectly clear solution stable at pH 8.
Now add Methyl orange (or whatever indicator your test kit uses for M alkalinity determination or use the meter) and continue titration to the M end point (pH 4.5). The total number of mL of acid added (from the beginning - not from the P end point) is the M alkalinity in mEq/L.
In this step
yH+ + yHCO3- ---> yH2O + yCO2
so the M alkalinity is
M = 2(x+y)
Just knowing y answers your question. As the molecular weight of calcium carbonate is 100 the weight of calcium carbonate in the powder you started with is 100y and dividing this by the mass of the powder you started with gives you the percentage by weight of chalk in your powder but this requires that you have means for measuring weight and volume accurately. If you don't you can still do the test if you don't. In this case you put a 'pinch' of powder in a flask and dissolve with DI (or RO) water. Now add drops of acid of reasonable strength and measure P and M in units of drops.
Now M - P = y and M + P = 4x + 3y from which it is clear that
x = [(M + P) - 3(M - P)]/4 = P - M/2
The weights in mg/L of, respectively, lime and chalk in the powder are thus, respectively
mL = wL(P-M/2)f and mC = wC(M-P)f
where wL is the molecular weight of lime (74.1), wC the molecular weight of chalk (100) and f a constant which converts the measured units of alkalinity to mEq/L. It depends on the strength of the acid and the volume in which the powder is dissolved and the weight of the powder dissolved. If the strength of the acid is 0.1 N, the weight 100 mg and the volume 100 mL then f = 1 and M and P are in units of mL acid.
Then the ratio of lime to chalk in the powder is
rLC = wL(P-M/2)f/wC(M-P)f = (wL/wC)(P-M/2)/(M-P)
and the percentage Ca(OH)2 in the powder is then
% = 100r/(1+r)
Do keep in mind that P and M can be measured in any units such as drops of lactic acid. Diluted hydrochloric acid from the hardware store can also be used (be careful with it!). Vinegar is not a terribly good choice as it is not quite acidic as we would like but will do in a pinch. Adjust the amounts of powder tested, water volume and acid dilution so that M and P are reasonably large numbers. M = 2 and P = 1 drops would not be good alkalinities. Test 10 times more solution or dilute the acid 10 times. Also keep in mind that you do not need to know the weight of powder, volume of water or strength of the acid. Just the drop counts to the M and P pH's.
Going back to the case where there is excess powder for a minute: You can also take the approach that while the undissolved powder may contain some undissolved lime that the solution over it is saturated, look up the strength of saturated solutions of lime at various temperatures and assume that this is the concentration of Ca(OH)2 in the liquid.
In #92 we found that
rLC = (wL/wC)(P-M/2)/(M-P)
A little algebraic fiddling is instructive. Divide numerator and denominator by P and introduce g = M/P to represent the ratio between total (M) and P alkalinities:
rLC = (wL/wC)(1-M/2P)/(M/P-1) = (wL/wC)(1-g/2)/(g-1)
It is clear that 1 < g < 2 as M > P and (1-g/2) must be positive. It is also clear, as one would expect, that the purity of the tested lime depends only on the ratio of the M and P alkalinities and not on their absolute values.
By inserting rLC as calculated from g above into the purity formula, 100r/(1+r), one obtains the percentage formula as a function of g alone and can easily prepare a graph like the one below. For example suppose that a brewer found he needed 100 drops of acid to decolorize the phenolpthalein in a test with some of his picking lime (P = 100 drops) and that 10 more were required to further reduce the pH to 4.5 (M = 110). Then M/P = 1.1 and the graph shows the purity of his lime to be about 77%.
I can't imagine that we haven't been through all this before but that's one of the advantages of getting old: all problems seem novel.
rLC = (wL/wC)(P-M/2)/(M-P)
A little algebraic fiddling is instructive. Divide numerator and denominator by P and introduce g = M/P to represent the ratio between total (M) and P alkalinities:
rLC = (wL/wC)(1-M/2P)/(M/P-1) = (wL/wC)(1-g/2)/(g-1)
It is clear that 1 < g < 2 as M > P and (1-g/2) must be positive. It is also clear, as one would expect, that the purity of the tested lime depends only on the ratio of the M and P alkalinities and not on their absolute values.
By inserting rLC as calculated from g above into the purity formula, 100r/(1+r), one obtains the percentage formula as a function of g alone and can easily prepare a graph like the one below. For example suppose that a brewer found he needed 100 drops of acid to decolorize the phenolpthalein in a test with some of his picking lime (P = 100 drops) and that 10 more were required to further reduce the pH to 4.5 (M = 110). Then M/P = 1.1 and the graph shows the purity of his lime to be about 77%.
I can't imagine that we haven't been through all this before but that's one of the advantages of getting old: all problems seem novel.
Excellent work! The only caution I would have is that we would need to be slow during the assessment of the Total Alkalinity since it might take a few minutes for the chalk to dissolve during the testing. I guess we would titrate down to 4.5 and then monitor the pH for several minutes longer. If the pH rises, a little extra chalk finally dissolved. Just titrate a bit more acid in to bring the pH down to the 4.5 target.
By the way, is 4.5 the proper target? I realize that the target for total alkalinity varies between 4.3 and 4.5. I'm guessing that 4.5 is fine since the chalk should be fully dissolved at that pH...right?
By the way, is 4.5 the proper target? I realize that the target for total alkalinity varies between 4.3 and 4.5. I'm guessing that 4.5 is fine since the chalk should be fully dissolved at that pH...right?
Being sure the end point pH is reached is extremely important but it is with the P alkalinity that it is critical. This is because CaCO3 take so long to dissolve when dissolution involves only conversion to HCO3-. As I described in #92 I've put CaCO3 in a beaker, added DI water, stirred it all up and added acid to pH well below 8 giving me a suspension of chalk milky) with a pH of 5. This is, if course, impossible and so, over time, the pH will climb beyond 8 at which time more acid can be added and the cycle repeats. Eventually the solution clears but even then pH creep takes place. The CaCO3 particles are too small to see but they are still there. Only when all CaCO3 is converted to HCO3- does the pH stay stable at 8.3. To use this method one must have this condition. If he does not then some of the acid added for the M alkalinity mesurement will go to convert CO3-- to HCO3- and then CO2. That will result in M being measured as larger than it should be and P as less than it really is. This can make relatively pure lime read appreciably less pure than it actually is.
It is, of course, important that the M titration be carefully done too but as the HCO3- + H+ --> H2O + CO2 reaction only involves ions (nothing needs to be dissolved) and the reaction of bicarbonate seems to be faster it is not so much of a worry as long as the P titration has completely dissolved the carbonate.
That's debatable. We are making the assumptions that the acid used to get to pH 8 corresponds to 100% (in milimoles) of the carbonate + 100% of the hydroxyl and not all the hydroxy is anihilated. At pH 8 the hydroxyl ion concentration is 1.0E-5 mmol/L, not 0, as we assume, but the error from hydroxy ions is much smaller. The acid used actually amounts to about 102% of the carbonate (titrate to 8.37 with a meter to get to 100% which is not enough to bleach phenolpthalein). Similarly, in going to pH 4.5 from 8 we assume that the acid consumed represents 100% of the bicarbonate which we have assumed is 100% of the carbonate. In other words we assume that M ~ 200% of the carbonate. In fact at pH 4.5 it represents 198.7%. At pH 4.4 it is 198.9% and at pH 4.3 it is 199.1%. Thus there are errors whatever end point we pick but as we go lower in pH it gets a little better. In that sense 4.3 is a better choice but not by that much. This is because the titration curve is flat (if you plot pH as the independent variable) here. It also means that if you have hit 4.5 it is only going to take a wee bit of acid to bring about an appreciable pH drop. So if you are using a buret or digital titrator then go ahead and try to get lower than 4.5 but do it very slowly.
Note that we could get better overall accuracy if we made detailed calculations of the carbo species and hydroxyl iob concentration changes between arbitrary pH's (i.e. we would calculate total and midpoint alkalinities for arbitrary pH's) but that would complicate things appreciably. Here we are looking for something 'quick and dirty'.
