Agreed. I withdrew the post.
Grams of baking soda required to neutralize 1 ml 88% lactic acid?
- Thread starter Silver_Is_Money
- Start date
Help Support Homebrew Talk:
- Status
Another error. Carbonate is not temporary hardness. It is alkalinity. Calcium or magnesium content in the water that is equal to or less than the alkalinity, equivalent for equivalent, is temporary hardness because it can be precipitated out by either heating the water or adding lime to it:
Ca++ + 2HCO3- <---> CO2 + CaCO3 + H2O
Ca++ + Ca(OH)2 + 2HCO3- <---> 2CaCO3 + 2H2O
Calcium or magnesium that is greater than the alkalinity is termed permanent hardness because it cannot be removed by either of these simple treatments.
As the equations show, in reduction of temporary hardness by heating half the alkalinity removed is removed through ejection of CO2 and the other half by precipitation. In the case of lime softening all alkalinity removed is removed by precipitation.
1. Zymotechnia Fundamentalis is there because I like history, brewing
historic brews, and because it sounds cool and refers to "Fundamental
Yeast Technology".
2. Lactic acids reacts 1:1 with bicarbonate ion, it doesn't matter
what the pKa is. If you have 100 molecules of lactic acid, but only
one reacts with bicarbonate, it reacts with one molecule of bicarbonate.
3. pKa is the negative log of the Ka, which is ratio that tells you
the relative amounts of substances in solution when the solution
*is at equilibrium*. If I place 1 mole of lactic acid in water solution
such that the total volume is 1 liter, it dissociates:
CH3CH(OH)CO2H -------> CH3CH(OH)CO2- + H3O+
The ratio [CH3CH(OH)CO2-][H3O+]/[CH3CH(OH)CO2H], where the square
brackets mean concentration of the substance in the brackets in moles/liter,
is equal to 1.4 x 10^-4, which is a pretty small number, typical for
carboxylic acids, and is called the Ka. The negative log of this number
is 3.85 and it is called the pKa. However, this ratio does not tell you the absolute
amount of any of these substances. I began with 1 mole lactic acid, that
means x amount dissociates, and since the amount of lactate has to equal
the amount of H3O+, [x][x]/[1-x] = 1.4 x 10^-4. Since 1 mole is much
greater than 1.4x10^-4, you can ignore the x in the denominator and
get a close answer by just writing x^2 = 1.4x10-4, or x = 1.18 x 10^-2.
That means that there is about 1-0.018 = 0.982 moles of lactic acid
remaining *at equilibrium*. But in the case of the reaction with bicarb,
there is something continually removing the [H3O+] from the reaction,
thus *disturbing the equilibrium*.
For now, let's call it The Leprechaun (and hope that he's not a racist
leprechaun...). Let's say The Leprechaun removes all of the [H30+] and
replaces it with Na+, that means we are back to the beginning, but we are
starting with 0.982 moles of lactic acid instead of 1.0. So now the equation
is [x][x]/[0.982-x] = 1.4 x 10^-4, so now x^2 = (0.982)(1.4 x 10^-4), so
x = 0.0117, and we have 0.982-0.0117 = 0.970 moles of lactic acid remaining.
Do you see what happens? The Ka value never changes, but the lactic acid is
consumed until The Leprechaun stops taking the H3O+. In this case, "The Leprechaun"
is the loss of CO2 to the atmosphere. At the elevated temperatures of mashing
and then boiling wort, for all intents and purposes all of the CO2 is lost, and
all of the lactic acid would be consumed if an equal molar amount of bicarbonate
were present, *even if the pKa was 14 instead of 3.85*.
4. This is just an example of Le Chatelier's principle, explained in any
general chemistry text, usually as the last topic in the chapter on
equilibria. If you want to prevent the lactic acid from completely reacting
and rely on the pKa, you would put the system under pressure and prevent the
CO2 from escaping. I learned this from Masterton + Slowinski, I've taught it
from Zumdahl and Chang, but really any intro chem text should have this information.
5. You are confusing the chemistry that occurs in closed aqueous systems and
acids and bases that don't form gases that escape, such as phosphate ions, with
the CO2 system, where the system is open and some of the reactants/products can escape.
Ray
historic brews, and because it sounds cool and refers to "Fundamental
Yeast Technology".
2. Lactic acids reacts 1:1 with bicarbonate ion, it doesn't matter
what the pKa is. If you have 100 molecules of lactic acid, but only
one reacts with bicarbonate, it reacts with one molecule of bicarbonate.
3. pKa is the negative log of the Ka, which is ratio that tells you
the relative amounts of substances in solution when the solution
*is at equilibrium*. If I place 1 mole of lactic acid in water solution
such that the total volume is 1 liter, it dissociates:
CH3CH(OH)CO2H -------> CH3CH(OH)CO2- + H3O+
The ratio [CH3CH(OH)CO2-][H3O+]/[CH3CH(OH)CO2H], where the square
brackets mean concentration of the substance in the brackets in moles/liter,
is equal to 1.4 x 10^-4, which is a pretty small number, typical for
carboxylic acids, and is called the Ka. The negative log of this number
is 3.85 and it is called the pKa. However, this ratio does not tell you the absolute
amount of any of these substances. I began with 1 mole lactic acid, that
means x amount dissociates, and since the amount of lactate has to equal
the amount of H3O+, [x][x]/[1-x] = 1.4 x 10^-4. Since 1 mole is much
greater than 1.4x10^-4, you can ignore the x in the denominator and
get a close answer by just writing x^2 = 1.4x10-4, or x = 1.18 x 10^-2.
That means that there is about 1-0.018 = 0.982 moles of lactic acid
remaining *at equilibrium*. But in the case of the reaction with bicarb,
there is something continually removing the [H3O+] from the reaction,
thus *disturbing the equilibrium*.
For now, let's call it The Leprechaun (and hope that he's not a racist
leprechaun...). Let's say The Leprechaun removes all of the [H30+] and
replaces it with Na+, that means we are back to the beginning, but we are
starting with 0.982 moles of lactic acid instead of 1.0. So now the equation
is [x][x]/[0.982-x] = 1.4 x 10^-4, so now x^2 = (0.982)(1.4 x 10^-4), so
x = 0.0117, and we have 0.982-0.0117 = 0.970 moles of lactic acid remaining.
Do you see what happens? The Ka value never changes, but the lactic acid is
consumed until The Leprechaun stops taking the H3O+. In this case, "The Leprechaun"
is the loss of CO2 to the atmosphere. At the elevated temperatures of mashing
and then boiling wort, for all intents and purposes all of the CO2 is lost, and
all of the lactic acid would be consumed if an equal molar amount of bicarbonate
were present, *even if the pKa was 14 instead of 3.85*.
4. This is just an example of Le Chatelier's principle, explained in any
general chemistry text, usually as the last topic in the chapter on
equilibria. If you want to prevent the lactic acid from completely reacting
and rely on the pKa, you would put the system under pressure and prevent the
CO2 from escaping. I learned this from Masterton + Slowinski, I've taught it
from Zumdahl and Chang, but really any intro chem text should have this information.
5. You are confusing the chemistry that occurs in closed aqueous systems and
acids and bases that don't form gases that escape, such as phosphate ions, with
the CO2 system, where the system is open and some of the reactants/products can escape.
Ray
Don't need chemical reaction modeling software, all you need is a pencil, paper and calculator. It looks hard because you've never seen it before probably, but this is taught in every general chem class. Go to the local library and get a general chem text, it's there in the acid-base equilibria chapter, and using the ICE table (Initial concentration, Change, at Equilbrium). It's even in the index under "ICE" or "ICE method".
Ray
Another error. From a power point from one of our universities, readily available online:
Ray
![Craft A Brew - Safale BE-256 Yeast - Fermentis - Belgian Ale Dry Yeast - For Belgian & Strong Ales - Ingredients for Home Brewing - Beer Making Supplies - [3 Pack]](https://m.media-amazon.com/images/I/51bcKEwQmWL._SL500_.jpg)
Again you miss the point. Of course one molecule of lactic acid reacts with a molecule of bicarbonate. How could it be otherwise. The point is that if you have added 100 molecules of lactic acid to a solution containing bicarbonate and that solution is at pH 5.2, four of those molecules stay as lactic acid and do not react with anything. Thus your statement
is untrue as anyone with knowledge of basic chemistry beyond the phlogiston theory knows and as you can verify by the simple experiment I posted earlier. I know you are not going to do it because you will not like the outcome but should you decide to open your mind use acetic acid rather than lactic because the ratio depends on the difference between pK and pH and it is more dramatic with acetic (2.8:1 molar ratio rather than 1.04 with lactic.)
Thanks. I didn't know that.
Yes, actually it does. If you put C mole of lactic acid in a liter of pure water its fairly easy to figure out how much HLac, Lac-, H+ and OH- will be found. For example, if you add 0.1 mmol the pH will go to 4.18, there will be 1000*10^-4.18 mmol of H+, 1000*10^(4.18 - 14.17) mmol of (OH)-, 0.0327 mmol of HLac (un dissociated lactic acid) and 0.067 mmol of Lac- (Lactate ion). How do I know this. By applying basic chemistry. It's in lots and lots of texts. Again I ask why you don't get one and read it?
And that's why we don't approach the problem the way you do. The question asked was "If I have x mol of bicarbonate how much lactic acid do I need to neutralize it?" First you have to understand what neutralize means. It means bring the water containing the bicarbonate to mash pH because that's what we as brewers seek to do. To answer this question you suppose that you had to put y mol (really we are interested in mmol here but that's OK for now) HLac into the water to reach pH 5.2 (or whatever you choose) at equilibrium, calculate the distribution of HLac and Lac-, realize that you have f0*y HLac, f1*y HLac-, realize that each HLac- is the result of the reaction HLac + HCO3- ---> CO2 + H2O so that f1*y = x (simple conservation of mass) and solve for y = x/f1. You then add y. I don't know why you have so much trouble understanding this. I suspect it's because you are afraid that you will see the light if you ponder it long enough and I don't think you want to see the light. Anyway, the formulas for calculating f0 and f1 are simple (I've posted them before in this thread)
r1 = 10^(pH-pK)
f0 = 1/(1 + r1)
f1 = r1*f0
These fall directly out of the mass action equation. Stick in 5.2 for mash pH and 3.86 for lactic acids pK and you'll get y = 1.033*x. Thus the answer to the question asked is "You need 3.3% more moles of lactic acid than you have moles of bicarbonate to neutralize. If you want to do the vinegar experiment recognizing that all CO2 is gone at pH 4.5 stick in 4.5 for the pH and 4.76 and you'll see that you need 2.8*x acetic acid.
You are critiquing an approach that I am not taking even though I have explained several times the approach I am taking and though I swore I would waste no more time on this I always get caught up in the hopes that there is yet a clearer way to explain something. So this post represents one final attempt on my part
If you are responsible for teaching young minds you really owe it to them and yourself to understand it.
Checking on Zumdhl and Chang (of whom I have never heard) it appears that this is a highschool chemistry text. No, you won't find Henderson Hasselbalch in a high school text. You will need a college text or one written for industry. So maybe if they are highschool kids it's not so much for them but you should understand it as you are misleading the readers here with assertions that x mol of a weak acid neutralize x mol of bicarbonate.
Writing software to correctly predict chemical reactions is apparently very difficult. The 2013 Nobel Prize went to 3 chemists who wrote such software.
https://www.scientificamerican.com/article/2013-chemistry-nobel-for-molecule-computer-models/
http://chemistry.stackexchange.com/questions/8351/software-for-predicting-chemical-reactions
Working the mechanics of the reaction by hand is comparatively trivial, though there even appears to be some disagreement about which approach is best even for that... thus the difficulty in writing software.
Yes, you are right. The slide is wrong. It isn't even consistent within itself. In the last major bullet it it says hardness can be 'caused by Ca, Mg, Fe' and then in the first sub-bullet it says temporary hardness is 'due to carbonate and bicarbonate' and in the second that permanent hardness is 'due to Cl, SO4'.
It's obvious that the author wrote 'due to' when he meant 'paired with'. Most people, at least those familiar with water chemistry, seeing that would shrug the error off but an error it is. As they say in the first bullet it is the cations that react with soap giving the 'hard' curds.
As well it should. Combining quantum mechanics with the non relativistic models (wave equations) is certainly a monumental task but when we are looking at a simple question like the one posed here the math is pretty trivial:
r1 = 10^(pH - pK)
f0 = 1/(1 + r1)
f1 = r1*f0
y = x/f1
is easily put into a spreadsheet or figured out with a scientific calculator. I did write to the King of Sweden but they turned me down.
It occurred to me this morning that perhaps another way to explain why lactic acid doesn't neutralized bicarbonate 1:1 might be to use the scheme I came up with for Palmer in his book on water in order to explain the acid/base chemistry of the mash. Its also the basis of the TQ paper on mash pH estimation. The basic concept is to let the titration curve of the materials involved tell the story. In this case the material in question is lactic acid (the other material is carbonic acid). To get the titration curve for lactic acid we would put a millimole of it in a beaker with a liter of water and add strong base in increments recording the pH after each addition. If we plotted the mEq of strong base added against the y axis and the resulting pH against the x axis we would get a plot that looks like the one below. This is similar to the carbonic acid plot in John's book except that it has two steps and one riser (with the pK at the inflection point of the riser) and the carbonic acid plot has three steps and two risers (with the two pK's at their inflection points). A plot for phosphoric acid (in the book), with its three protons, has 4 steps and 3 risers with the 3 pK's at their inflection points).
The numbers on the y axis represent the number of milliequivalents of base added and the total charge on the lactate (Lac-) ion in the water at equilibrium after the base has been added and has reacted with the acid:
HLac + Na+ + (OH)- ---> Na+ + H2O + Lac-
A reacting acid (HLac) molecule gives up its proton to the base (OH)- forming water and acquiring a negative charge in the process. IOW it becomes an Lac- ion. Not all acid molecules do this. The fraction that make the sacrifice depends on the final pH hence the curve. The fraction that do is given by the magnitude of the charge. Thus we see that if the final pH > 6.5 or so nearly all the lactic molecules have transitioned and Lactic acid is, in such cases, a strong acid. Conversely if pH < 1 very few molecules do and Lactic acid is a weak acid.
At mash pH we are almost to the point where we can call lactic acid strong as the y axis values are close to -1 in that region. At the same time it is clear from the curve that if we neutralize bicarbonate (or any other form of alklinity) to pH 5 with lactic acid that only 94% of the acid we added has reacted and we had better add a bit extra if we want to get rid of a specified amount of alkalinity. The amount is
mmol_acid_needed = mEq_alkalinity_to_be_removed/reading_from_graph
From yesterday's post it is probably clear that reading_from_graph = f1(pH) and that is indeed the case which also reveals how the graph was prepared. Put another way, the graph reading gives you the fraction of the added acid that reacts at a given pH and 1 minus the reading, f0(pH), gives you the fraction that didn't react.
I really hope that makes it clear how this works. Note that the same issue arises with phosphoric acid. Mash pH lies in the middle of the step between the first two pK's of phosphoric acid and thus the charge is ~-1. John and I debated this one quite a bit and he finally decided that he was just going to say that 1 mmol of phosphoric acid would wipe out 1 mEq of alkalinity but just as that isn't the case with lactic, it isn't the case with phosphoric. If you want 1:1 exactly you'll have to go to a strong acid (WRT mash pH) such as sulfuric acid with one negative pK and the other 1.92 or hydrochloric with one negative pK.
The numbers on the y axis represent the number of milliequivalents of base added and the total charge on the lactate (Lac-) ion in the water at equilibrium after the base has been added and has reacted with the acid:
HLac + Na+ + (OH)- ---> Na+ + H2O + Lac-
A reacting acid (HLac) molecule gives up its proton to the base (OH)- forming water and acquiring a negative charge in the process. IOW it becomes an Lac- ion. Not all acid molecules do this. The fraction that make the sacrifice depends on the final pH hence the curve. The fraction that do is given by the magnitude of the charge. Thus we see that if the final pH > 6.5 or so nearly all the lactic molecules have transitioned and Lactic acid is, in such cases, a strong acid. Conversely if pH < 1 very few molecules do and Lactic acid is a weak acid.
At mash pH we are almost to the point where we can call lactic acid strong as the y axis values are close to -1 in that region. At the same time it is clear from the curve that if we neutralize bicarbonate (or any other form of alklinity) to pH 5 with lactic acid that only 94% of the acid we added has reacted and we had better add a bit extra if we want to get rid of a specified amount of alkalinity. The amount is
mmol_acid_needed = mEq_alkalinity_to_be_removed/reading_from_graph
From yesterday's post it is probably clear that reading_from_graph = f1(pH) and that is indeed the case which also reveals how the graph was prepared. Put another way, the graph reading gives you the fraction of the added acid that reacts at a given pH and 1 minus the reading, f0(pH), gives you the fraction that didn't react.
I really hope that makes it clear how this works. Note that the same issue arises with phosphoric acid. Mash pH lies in the middle of the step between the first two pK's of phosphoric acid and thus the charge is ~-1. John and I debated this one quite a bit and he finally decided that he was just going to say that 1 mmol of phosphoric acid would wipe out 1 mEq of alkalinity but just as that isn't the case with lactic, it isn't the case with phosphoric. If you want 1:1 exactly you'll have to go to a strong acid (WRT mash pH) such as sulfuric acid with one negative pK and the other 1.92 or hydrochloric with one negative pK.
Neutralizing lactic acid with a strong base has nothing to do with neutralizing it with a base
like bicarbonate which reacts to form H2CO3 and then CO2 + H20. Also, this plot has nothing
to do with the elevated temperatures during mash and wort boiling, which help to drive off the
CO2. I'm not wasting anymore time with this, I suggest you stop trying to decipher brewing
books written by non-chemists and start with a general chemistry textbook.
Ray
like bicarbonate which reacts to form H2CO3 and then CO2 + H20. Also, this plot has nothing
to do with the elevated temperatures during mash and wort boiling, which help to drive off the
CO2. I'm not wasting anymore time with this, I suggest you stop trying to decipher brewing
books written by non-chemists and start with a general chemistry textbook.
Ray
You refuse to understand that bicarbonate is not phosphate and that it reacts with acid to form CO2 which escapes.
Both Zumdahl and Chang are university level general chemistry texts, I have used them to teach at the community college level. Here are my copies:
If you search the MIT library at libraries.mit.edu you will find both in the catalog. As I said in the other post, this is a waste of time until you learn some chemistry.
Ray
Evidently you have yet again failed to read the post or have misunderstood it. The comments regarding strong base had to do with how we would construct the curve in the lab. We would probably use a strong base to do the titration but we could very well do it with a weak base (bicarbonate) too. It doesn't matter what the acid protonates. Just that it protonates something which can be any partially deprotonated acid with a higher pK. I find the fact that you teach chemistry and are unaware of this frankly shocking. In fact the curve was not calculated based on a titration with either NaOH or NaHCO3 but calculated from the pK of Lactic acid which is why the pK point goes right through 0.5.
Doesn't matter again. If one wants to work at higher temperature he just slides the curve to right or left depending on the pK at the temperature of interest.
I think that would be wise. You keep punching me in the fist with your face. It must be getting painful by now. Plus it keeps me from having to refute your misinformation.
I'm afraid this chemistry isn't found in textbooks aimed at the high-school/junior college/community college level. Perhaps that is why you are unaware of it. You would have to go to a university level text, probably not freshman but certainly not necessarily senior or graduate level either. Most of the sources I consult fit the latter description. Also books written for the water treatment industry, any good brewing text with a chapter on water, and biochemistry texts have it. I'd suggest you get one of those, take an advanced chemistry course or, as I have suggested before, find a colleague who knows some chemistry beyond the high school level. I have in other posts suggested several of these texts to you so I won't repeat other than to say that Stumm and Morgan is probably the best (but I doubt you would be able to read it at first - with some persistence though you should be OK). I'll add Levine's Physical Chemistry to the list. But you shouldn't have to consult a text to figure this out as it all falls out of law of mass action (I assume you've heard of that). I've shown you how to do this several times (and it's in a Sticky above).
Of course I do. The failure in understanding here is on your part. The escape of CO2 has nothing to do with it as we check pH after all the CO2 has escaped.
And I'm sure my highschool and freshman college texts are too. Actually just checked and Sienko and Plane is still for sale at Amazon. Two things I shall never forget about that book. One is that Prof Plane (who was head of the Chem dept.) had a bookcase behind his desk that covered the entire wall. It was full of copies of that book - no two in the same language. The other is the animation exhibited by Prof. Sienko (usually a most taciturn man) during his lecture on fermentation.
We did mass action and equilibria (hated it) in that course but not Henderson-Hasselbalch or partition of species (that I can recall).
Though one never knows for sure I would think it manifestly clear to most who knows more chemistry here, at least on this particular subject, so I think your decision wise.
If you decide to open your mind as by, for example, measuring the alkalinity of a millimolar solution of sodium bicabonate (alkalinity 50 ppm as CaCO3 = 1 mEq/L) with sulfuric acid, hydrochloric acid and lactic acid you will find it takes half a millimole/L sulfuric, 1 mmol/L hydrochloric and 1.23 mmol/L lactic to reach the ISO endpoint of pH 4.5 clearly proving that 1 mmol of lactic acid does not remove 1 mmol bicarbonate. If you aren't sure what alkalinity is or how to measure it we can fill you in. And, when you have your results and are wondering why they don't match your theories we can go back to trying to get you straightened out.
The reason for mentioning this again is that in refuting it we gain another insight and that is that the technique we are really using here, referred to as monitoring the 'proton condition' by Stumm and Morgan and 'tracking the protons' by me and which really means just recognizing that protons are conserved is the same very powerful technique we use to come up with estimates of mash pH. I didn't think this was advanced chemistry but apparently it is though anyone with basic chemistry should be able to understand it.
If one adds lactic acid to brewing water or mash or anything else he is coming it at the left of the curve. Pure lactic (or sulfuric or phosphoric...) acid has 0 charge and very low pH. If there is any things in the water (OH- ions, HCO3- ions, pale malt) which can absorb protons the lactic acid will transfer protons to them and the pH will rise from what I call the 'intrinsic pH' of the lactic acid (very low) to a new pH and the number of protons transferred is found by reading the curve at the new pH and multiplying by the number moles of lactic acid added as the curve. If a proton is transferred to a bicarbonate ion that becomes a carbonic acid molecule and if the mash is hot at this point that carbonic acid molecule will decompose into water and CO2 and the CO2 will fly off. The proton stays in the mash. If the proton lands on an OH- ion that converts to water and the proton stays in the mash. If it lands on an ion from malt that ion becomes protonated and the proton stays in the mash but the pH of the system is reduced relative to the intrinsic pH of the malt. The major points here are
1)It doesn't matter where the protons go in the mash they stay in the mash and are thus accounted for.
2)Those that land on HCO3- ions give rise to CO2 and thus the amount of CO2 produced is easily calculated and is thus accounted for.
3)We can extend this same technique to mash pH prediction and control by considering separate titration curves for the various malts and acids we may add to a mash and, of course, the titration curve for carbonic acid if the water has any alkalinity. As I have written extensively on this elsewhere I will not go into further detail here (unless someone wants me to).
Ok, I think we've all said enough.
- Status
