Grams of baking soda required to neutralize 1 ml 88% lactic acid?

[Silver_Is_Money]

In a hypothetical situation involving a mash addition overshoot of 88% lactic acid, for each ml of acid overshoot (lowering the pH below the 5.4 target) how many grams of sodium bicarbonate are required to negate said overshoot and restore the mash pH to 5.4?.

If it has relevance and is necessary to the solution, let's say this mash is comprised of 12 lbs. of malted barley grist and 18 quarts of distilled or high quality RO strike water, with 3 grams of CaCL2-2H2O as the only mineral addition.

In the long run I'm trying to determine how many ppm of alkalinity (as CaCo3) are removed from the mash (assuming the above criteria if relevant) by adding 1 ml of 88% lactic acid, and likewise how many ppm of alkalinity (as CaCo3) are added to the mash by the addition of 1 gram of baking soda.

 

[isomerization]

I'd recommend downloading the Bru'n water spreadsheet: https://sites.google.com/site/brunwater/home/files/Brun Water 1.18a.xls?attredirects=0&d=1

It will allow you to play with all kinds of hypothetical situations

 

[MaltMaker]

1 ml of 88% lactic acid will remove 69 ppm of CaCO3.
1 g of Baking Soda will add 35 ppm of CaCO3.

Though perhaps its not that simple in a complex mash environment one would need to use titration.

 

[Silver_Is_Money]

1 ml of 88% lactic acid will remove 69 ppm of CaCO3.
1 g of Baking Soda will add 35 ppm of CaCO3.

Though perhaps its not that simple in a complex mash environment one would need to use titration.

Thank you! So for all practical purposes 2 grams of added baking soda erases 1 ml of added 88% lactic acid.

 

[Yooper]

Thank you! So for all practical purposes 2 grams of added baking soda erases 1 ml of added 88% lactic acid.

I have a chart cut out and in my brewery that I got from Braukaiser.com. http://braukaiser.com/wiki/index.php?title=Mash_pH_control (read the whole page, but scroll down to see the chart and the "hows and whys":

 

[MaltMaker]

It's not quite as simple as adding/removing the deficit/excess of either acid/alkalinity.

You're dealing with a live complex system that is in a constant state of change.

Overshooting or undershooting the pH is best dealt with by recording the pH and adjusting it on the next batch.

After a pH reading @10-20 minutes - the activity in the mash starts to slow and naturally it tends towards 5.4pH as the mash progresses - IME.

Adding salts and acids to adjust the pH will be fickle at best because it's hard to evenly distribute them in the mash.

IMO, the BrauKaiser chart is useful for experimentation but it's also easy to make it much worse than it is. Having said that if you are off by 0.5 pH or more it may be worth a shot if you can do it quickly and accurately.

 

[Silver_Is_Money]

1 ml of 88% lactic acid will remove 69 ppm of CaCO3.
1 g of Baking Soda will add 35 ppm of CaCO3.

Though perhaps its not that simple in a complex mash environment one would need to use titration.

EZ Water fully agrees with this, but Brewers Friend has 1 ml of 88% lactic acid removing roughly 35 ppm of CaCO3. A huge difference. If one of these mash pH calculators is correct with regard to lactic acid, the other clearly has to be way off.

Edit: The Kaiser Water Calculator also has 1 ml of 88% lactic acid removing roughly 35 ppm of CaCO3. And Kai Troester is involved in both Kaiser and Brewers Friend. The implications of this are staggering if 69 ppm is correct.

 

[MaltMaker]

You're beginning to catch on. Kaiser wrote the Brewers Friend Calculator. I simply gave you the results from the EZWater Calculator. What results does Brun'water give you? Now decide which one you're going to gamble with.

 

[ajdelange]

At and near mash pH lactic acid has a normality of about 11.8. That means that it can provide 11.8 mEq of protons per mL and that it can neutralize 11.8 mEq of alkalinity. Bicarbonate is pure alkalinity. Thus 1 mL of lactic acid will neutralize 11.9 mEq of bicarbonate which, as the equivalence of bicarbonate is 61 mg/mEq is 61*11.8 mg = 719.8 mg/mL. It really is as simple as that. Now if you add 1 mL lactic acid to a mash I can't guarantee that it will neutralize 11.8 mEq (720 mg) of bicarbonate; only that it will neutralize 11.8 mEq of alkalinity some of which might be from the grain and some of which might be bicarbonate. But it doesn't matter which source of alkalinity it neutralizes, the pH shift will be the same and is predictable from the buffering capacity of the entire mash.

Note that this is 720 mg of bicarbonate. You will doubtless add it as sodium bicarbonate which also contains a sodium atom and has molecular weight 23 + 61 = 84. Thus the amount of sodium bicarbonate to neutralize a mL of lactic acid near mash pH is 84*11.8 = 991 mg.

 

[Silver_Is_Money]

At and near mash pH lactic acid has a normality of about 11.8. That means that it can provide 11.8 mEq of protons per mL and that it can neutralize 11.8 mEq of alkalinity. Bicarbonate is pure alkalinity. Thus 1 mL of lactic acid will neutralize 11.9 mEq of bicarbonate which, as the equivalence of bicarbonate is 61 mg/mEq is 61*11.8 mg = 719.8 mg/mL. It really is as simple as that. Now if you add 1 mL lactic acid to a mash I can't guarantee that it will neutralize 11.8 mEq (720 mg) of bicarbonate; only that it will neutralize 11.8 mEq of alkalinity some of which might be from the grain and some of which might be bicarbonate. But it doesn't matter which source of alkalinity it neutralizes, the pH shift will be the same and is predictable from the buffering capacity of the entire mash.

The trick then becomes determining how many mEq of acid and or alkalinity the entire mash contains. That seems to be the difficult part.

 

[Silver_Is_Money]

I have a chart cut out and in my brewery that I got from Braukaiser.com. http://braukaiser.com/wiki/index.php?title=Mash_pH_control (read the whole page, but scroll down to see the chart and the "hows and whys":

View attachment 394395

This chart is quite helpful! Thanks for posting the chart and the link.

 

[mongoose33]

In a hypothetical situation involving a mash addition overshoot of 88% lactic acid, for each ml of acid overshoot (lowering the pH below the 5.4 target) how many grams of sodium bicarbonate are required to negate said overshoot and restore the mash pH to 5.4?.

If it has relevance and is necessary to the solution, let's say this mash is comprised of 12 lbs. of malted barley grist and 18 quarts of distilled or high quality RO strike water, with 3 grams of CaCL2-2H2O as the only mineral addition.

In the long run I'm trying to determine how many ppm of alkalinity (as CaCo3) are removed from the mash (assuming the above criteria if relevant) by adding 1 ml of 88% lactic acid, and likewise how many ppm of alkalinity (as CaCo3) are added to the mash by the addition of 1 gram of baking soda.

At and near mash pH lactic acid has a normality of about 11.8. That means that it can provide 11.8 mEq of protons per mL and that it can neutralize 11.8 mEq of alkalinity. Bicarbonate is pure alkalinity. Thus 1 mL of lactic acid will neutralize 11.9 mEq of bicarbonate which, as the equivalence of bicarbonate is 61 mg/mEq is 61*11.8 mg = 719.8 mg/mL. It really is as simple as that. Now if you add 1 mL lactic acid to a mash I can't guarantee that it will neutralize 11.8 mEq (720 mg) of bicarbonate; only that it will neutralize 11.8 mEq of alkalinity some of which might be from the grain and some of which might be bicarbonate. But it doesn't matter which source of alkalinity it neutralizes, the pH shift will be the same and is predictable from the buffering capacity of the entire mash.

So, AJ, I've highlighted the question above in the original post; what's the answer?

 

[BrewKnurd]

So, AJ, I've highlighted the question above in the original post; what's the answer?

AJs math says that 1 mL of lactic acid will neutralize 720 mg of bicarbonate. As every gram of baking soda contains 730 mg of bicarbonate and 270 mg of sodium, it would be very close to 1 g of baking soda, it appears.

Given the way the question was asked, how much baking soda would be requires to offset a certain overshoot of acid addition, i imagine the makeup of the mash actually isn't important. If one wanted to know how much baking soda to move the mash by a certain pH, then the makeup of the mash would be a critical component.

 

[mongoose33]

AJs math says that 1 mL of lactic acid will neutralize 720 mg of bicarbonate. As every gram of baking soda contains 730 mg of bicarbonate and 270 mg of sodium, it would be very close to 1 g of baking soda, it appears.

Given the way the question was asked, how much baking soda would be requires to offset a certain overshoot of acid addition, i imagine the makeup of the mash actually isn't important. If one wanted to know how much baking soda to move the mash by a certain pH, then the makeup of the mash would be a critical component.

He actually said this: 719.8 mg/mL

I know that's .7198 grams, but per mL?

 

[ajdelange]

We don't work with ppm alkalinity in the mash. We work with the total alkalinity which is the sum of the total water alkalinity with the individual alkalinities of each malt. The water here has no alkalinity so it depends on the alkalinities of the malts but the question cannot be answered without having detailed knowledge of the malts' buffering properties. We can say that this is approximately 45 mEq/kg•pH. If you have 12 lbs malt (5.45) kg in RO/DI water the mash buffering will be approximately 5.45*45 = 245 mEq/pH. If the pH went to 5.2 rather than 5.4 and you wish to restore to 5.4 you will have to neutralize approximately (5.4 - 5.2)*5.45*245 = 27 mEq of acid. To do that requires 27*61 = 1629 mg of bicarbonate or 27*84 = 2268 mg NaHCO3.

If you know that you added x mL too much lactic acid and want to effectively undo that then you must add x*11.8*61 mg bicarbonate or x*11.8*84 mg NaHCO3.

 

[BrewKnurd]

He actually said this: 719.8 mg/mL

I know that's .7198 grams, but per mL?

0.7198g = 719.8 mg = 720 mg for all meaningful purposes. That is the amount he calculated as neutralized by 1 mL of lactic acid, hence the 719.8 mg/mL. from an alkalinity perspective, 720 mg is equivalent to 1 mL of lactic acid.

 

[mongoose33]

We don't work with ppm alkalinity in the mash. We work with the total alkalinity which is the sum of the total water alkalinity with the individual alkalinities of each malt. The water here has no alkalinity so it depends on the alkalinities of the malts but the question cannot be answered without having detailed knowledge of the malts' buffering properties. We can say that this is approximately 45 mEq/kg•pH. If you have 12 lbs malt (5.45) kg in RO/DI water the mash buffering will be approximately 5.45*45 = 245 mEq/pH. If the pH went to 5.2 rather than 5.4 and you wish to restore to 5.4 you will have to neutralize approximately (5.4 - 5.2)*5.45*245 = 27 mEq of acid. To do that requires 27*61 = 1629 mg of bicarbonate or 27*84 = 2268 mg NaHCO3.

If you know that you added x mL too much lactic acid and want to effectively undo that then you must add x*11.8*61 mg bicarbonate or x*11.8*84 mg NaHCO3.

So....why don't you just do the formula and report that? Why are you making this far more complicated than it needs to be?

 

[mabrungard]

Hmm? My calculations with Bru'n Water indicate that it takes about 970 mg of baking soda to neutralize 1 ml of 88% lactic acid. Close enough to AJ's estimate. The 720 mg value is not correct.

 

[ajdelange]

No, the 720 mg is correct but that is the amount of bicarbonate. As the post says...

Note that this is 720 mg of bicarbonate. You will doubtless add it as sodium bicarbonate which also contains a sodium atom and has molecular weight 23 + 61 = 84. Thus the amount of sodium bicarbonate to neutralize a mL of lactic acid near mash pH is 84*11.8 = 991 mg.

I do have to take responsibility for not being absolutely clear on that.

 

[ajdelange]

So....why don't you just do the formula and report that? Why are you making this far more complicated than it needs to be?

First, because the question is worded in such a way that it is clear OP doesn't understand the principles involved fully and second because this is the Brew Science forum where we hope to teach the readers a little brewing science.

 

[mabrungard]

No, the 720 mg is correct but that is the amount of bicarbonate.

Yes, and I was referring to the amount of baking soda required...as requested in the OP.

 

[rayg]

First of all, "88%" lactic acid solutions are not 88%, they vary
between 85-92% depending on which one you buy. One solution is
88-92% in methanol and another is 85-90% in water, another is
labeled "greater than or equal to 85%". There is also a racemic
solution of lactic acid labeled "~90%". Which one is being sold
as "88%" lactic acid? A chemical company like Sigma-Aldrich that sells
to professional chemists accurately labels the substances they
sell. Re-sellers like LD Carlson just slap a convenient label on it.
There are also other impurities some of which may affect the
water chemistry: Sigma-Aldrich's 88-92% lactic acid solution
contains in their words "equal to or less than 100 ppm calcium ion",
while the 85-90% solution contains less than 10ppm calcium ion.

Second, "sodium bicarbonate" is not "bicarbonate". Sodium bicarbonate
is NaHCO3, molecular weight 84, bicarbonate ion is HCO3-, molecular
weight 61, with a single negative charge that I've indicated with
a dash after it.

The op was asking about baking soda, which is NaHCO3, and then confused
the issue by bringing in CaCO3, a different substance with a different
molecular weight, and further confused the issue by mixing absolute
amounts (1 ml of 88% lactic acid) with relative amounts (ppm). ppm is
mg per liter, multiply by the volume of the batch in liters to get
the absolute amount of whatever substance you are talking about.

Lactic acid is monoacidic and has a molecular weight of 90.1, sodium
bicarbonate is monobasic so they react in a 1:1 molar ratio.

If the lactic acid you have happens to be exactly 88% by weight, then if
you know the density of the solution you can calculate the mass of
lactic acid present and then the moles. What's the density? The ones
reported for water solutions of either >85% or ~90% solutions are
1.206gm/mL and 1.209 gm/mL, so 1.21 gm/mL seems a reasonable round-off.

So: 0.88 x 1.21gm/mL x 1mL = 1.065 gm lactic acid. 1.065gm/90.1 gm/mol = 0.0118 moles.
That means we neutralize 0.0118 moles of sodium bicarbonate or carbonate ion.

0.0118 moles x 84 gm/mol = 0.991 gm sodium bicarbonate (baking soda).
0.0118 moles x 61 gm/mol = 0.720 gm bicarbonate ion.

Ray

 

[ajdelange]

Lactic acid is monoacidic and has a molecular weight of 90.1, sodium
bicarbonate is monobasic so they react in a 1:1 molar ratio.

In fact they don't because the pK of lactic acid is not far enough removed from mash pH to bring about full dissociation of the acid. At mash pH 5.2, for example, 4.2% of added HLac is still HLac and has not participated in the neutralization reaction. This is as much as some of the other factors mentioned in the previous post and is why we say 88% lactic acid has normality of about 11.8 near mash pH. This is easily accounted for in brewing software but AFAIK most programs do not take it into account and use the incorrect 1:1 equivalence.

We discourage the use of lactic acid solutions in methanol. In fact we say loud and clear: don't use any. Not that you would need to worry about this if you buy food grade stuff from your brewing supplier.

We do not discuss "racemates" here because of the current PC climate.

 

[isomerization]

We do not discuss "racemates" here because of the current PC climate.

Yes! Too funny

 

[rayg]

In fact they don't because the pK of lactic acid is not far enough removed from mash pH to bring about full dissociation of the acid.

Acetic acid is an order of magnitude less acidic than lactic
acid, yet when you put baking soda into vinegar, the CO2 pours
out immediately. The pKa of lactic acid doesn't matter, what
matters is that the CO2 formed escapes the solution rendering
the reaction irreversible. That's why carbonate is "temporary"
hardness.

CH3CH(OH)CO2H + HCO3- ------> CH3CH(OH)CO- + H2CO3
The equilibrium lies far to the right, but

H2CO3 -------> H2O + CO2 and the CO2 leaves into the air, making the
reaction irreversible.

Here's an experiment for you: take a brown penny and toss into
a bowl with enough vinegar in it to cover it. Nothing much happens.
Then take a pinch of salt and add it, and immediately the penny
lightens. That's because acetic acid reacts with the NaCl to make
a trace amount of HCl, hydrochloric acid:

CH3CO2H + NaCl ------> CH3CO2-Na+ + HCl

The equilibrium lies far, far to the left. But even if the ratio
of acetic acid to HCl is 10 billion to 1, every molecule of HCl
reacts immediately with the copper sulfide in the penny, so that
drives the reaction completely to the right. If you put enough
copper sulfide in the solution, it would consume all of the acetic
acid via reaction with the HCl.

Rayg

 

[ajdelange]

A universe of inorganic, organic, physical, water, environmental, brewing, bio, polymer... chemists accepts the model that you have yourself referenced in several of your posts. This model states that a weak acid is not fully dissociated unless pH of the system is higher than the highest pK of the acid by at least a couple of pH and the Law of Mass Action allows us to determine what fraction is fully dissociated and what fraction(s) are undissociated, monobasic, dibasic... The key to this is the Henderson-Hasselbalch equation which you can look up on Wikipedia or find in any biochemistry text. I really suggest you do that. Or read one of the papers you or I referenced in the thread on nitrogen bubbles.

If you read and comprehend you will find that if you add a mole of lactic acid to a solution of sodium bicarbonate and in so doing reach a pH of 5.2 that 0.96 mole of lactic acid released protons which protons combined with HCO3- converting it to H2CO3 a lot of which decomposed into water and CO2 much of which fizzed off. But 0.04 mole of it did not release a proton which unreleased proton did not react with bicarbonate and thus convert it to CO2 and water. Thus the effective strength of lactic acid in doing a neutralization to pH 5.2 is 96% of its molarity. To pH 6 it's a different story. 6 is well above the pK of lactic acid and the acid is fully dissociated. Each mole of lactic acid removes 1 Eq (or darn close to it) of alkalinity at pH 6.

Acetic acid, as you have noted, is about an order of magnitude weaker than lactic acid so that at pH 5.2 approximately 26% of it is unavailable to fizz baking soda. That does leave 74% which is available for that purpose and that's the part that reacts when you add vinegar to it. If you put 1 mol of baking soda in a beaker and add 1 mol of acetic acid you will not remove all the baking soda. Try it (it's all gone when the alkalinity to pH 4.4 measures 0). To get it all out you would need an extra 0.6 mol and that's why we don't do alkalinity titrations with weak (lactic, acetic) acids but rather strong ones such as sulfuric (closest pK = 1.9).


Pennies in acetic acid: What happens there is that you have hexaaquacopper ions at low pH. When chloride ions are introduced to a solution with hexaaquacopper ions at low pH they can replace some of the water molecules forming tetrachlorocuprate ions and a color change takes place. BTW I don't claim to understand this in depth but it makes sense. What I do know is this has no relevance to the question of partial dissociation of weak acids.

This is meant kindly but if the moderators think it not so let them remove it: I don't know if you have any access to a chemist or even perhaps a colleague who isn't working as a chemist but has had some training as a chemist but if you do perhaps you could run your posts past him. This saves me having to correct them and the risk of other readers becoming misinformed.

 

[MaltMaker]

Yes, that seems obvious but what would happen if I drank the tetrachlorocuprate ions instead?

 

[ajdelange]

Try it (it's all gone when the alkalinity to pH 4.4 measures 0). To get it all out you would need an extra 0.6 mol.

So here's a simple experiment which is relevant to the discussion, is easy to do and drives the last nail into the coffin lid of rayg's thesis (can we call it that?). Weigh out 3.5 grams of NaHCO3. That's 3.5/84 =1/24 mol. Measure out 50 mL of 5% vinegar (that's what most vinegar is and it should say so on the label) that's 2.5 grams of acetic acid which is 2.5/60 = 1/24 mol. Now were acetic acid a strong acid (low pK) 1/24 mole would it would react completely with the sodium bicarbonate. But acetic acid isn't a strong acid. It's pK is 4.76 so much of it is unavailable. According to rayg this doesn't matter but in fact, of course, it does.

If you add the 1/24 mol of acid to the 1/24 mol of baking soda (go slow so it doesn't fizz over and make a mess) it will, of course release some of the CO2 entrained in the Na2HCO3 - but not all of it as rayg claims. The pH of the mix is above 4.4 - the nominal pH at which all carbo has been converted to bicarbonate. The proof that there is still unconverted bicarbonate is simple. Just add more acid. There will be more fizzing.

Note that this is a crude alkalinity titration. We detect end point when fizzing ends which we know to be around 4.4. It is a very crude end point detection method, of course. It take about 80 mL of vinegar. We calculate (in the last post) that at the end point pH it takes about 60% extra acid to neutrailze each mol of bicarb and so calculate that we had (80*.05)/1.6)/60 = 1/24 mol of bicarbonate which we did.

 

[ajdelange]

Withdrawn. See #30.

 

[shoengine]

I'm not going to qualify any statements, but right or wrong headed arguments do not equal a troll, and one can troll even with correct arguments.

Either way, I like this thread. Iron sharpens iron, as the Book says.

 

[ajdelange]

Agreed. I withdrew the post.

 

[ajdelange]

That's why carbonate is "temporary"
hardness.

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.

 

[rayg]

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

 

[MaltMaker]

No wonder there is no chemical reaction software (computer modeling of reactions), it's just not that simple.

 

[rayg]

No wonder there is no chemical reaction software (computer modeling of reactions), it's just not that simple.

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

 

[rayg]

Another error. Carbonate is not temporary hardness. It is alkalinity.

Another error. From a power point from one of our universities, readily available online:

Ray

 

[ajdelange]

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.

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

Lactic acid is monoacidic and has a molecular weight of 90.1, sodium
bicarbonate is monobasic so they react in a 1:1 molar ratio.

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.)

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*.

Thanks. I didn't know that.

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.

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?

But in the case of the reaction with bicarb,
there is something continually removing the [H3O+] from the reaction,
thus *disturbing the equilibrium*.

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.

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 ...

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.

 

[MaltMaker]

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".

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.

 

[ajdelange]

Another error. From a power point from one of our universities, readily available online:

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.

 

[ajdelange]

Writing software to correctly predict chemical reactions is apparently very difficult. The 2013 Nobel Prize went to 3 chemists who wrote such software.

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.