Lactic Acid Alkalinity Reduction

[Dim]

Hi,

I am trying to square the circle between theory and reality and get my maths right. I have gone through many, many previous threads but nothing seems to be able to explain the delta that I am seeing.

Looking at the technical product sheet of Murphy & Son's Lactic Acid (80%), I can see the following statement:

6 ml of LACTIC ACID per hl reduces the alkalinity by 16 mg/litre (ppm)
​

According my calculations, adding 6 ml of 80% lactic acid to 100L of water should reduce (HCO3) alkalinity by ~39 ppm. Here's how I would calculate this:

1. The density of 80% lactic is ~1.2 g/mL
2. The pure acid weight can be calculated as: 6 ml × 80% × 1.2 g/mL = 5.76g
   
3. Lactic has an equivalent weight of 90.08g.
4. 5.76 g ÷ 90.08 g/Eq ÷ 100L = 0.000639 Eq/L
5. HCO3 has an equivalent weight of 61.02g.
6. The amount of HCO3 that can be neutralised is: 0.000639 Eq/L × 61.02 g/Eq ~ 0.039 g/L ≙ 39 ppm

How come my results are so vastly different from the spec? I assume they refer to alkalinity as HCO3, but it would be equally relevant if they meant CaCO3. Sorry, I am not a chemist and I would appreciate your input.

 

[Silver_Is_Money]

Due to incomplete dissociation, the nominal mEq/mL strength of any weak acid is totally dependent upon a pH target.

That said, looking at it your way:
-------------------------------------
The density of 80% Lactic Acid is 1.187 g/mL
80% x 1.187 g/mL = 0.9496 g. of pure lactic acid per mL
0.9496 g/mL /90.08 g/Eq = 0.01046 Eq/mL = 10.46 mEq/mL (for the case of 100% dissociation, and also for the case of being monoprotic)
[but as merely a technicality, lactic acid is not monoprotic, it is actually polyprotic, being capable of liberating 2 H+ ions rather than 1]
Lactic Acids pKa's are: pKa1 = 3.86, and pKa2 = 15.1, so at pH's of concern to brewers it is close to being monoprotic.

16 ppm Alkalinity (as CaCO3): Note here that alkalinity is not HCO3-, as that is bicarbonate.
-------------------------------------------------------------------------------------------------------
16 mg/L / 50.04345 mg/mEq = 0.31972 mEq/L alkalinity (as CaCO3)
0.31972 mEq/L * 100L = 31.972 mEq of alkalinity

31.972 mEq / 10.46 mEq/mL = 3.06 mL of 80% lactic acid required

So for the ideal case of 100% dissociation for a monoprotic lactic acid the answer is 3.06 mL to neutralize the alkalinity of 1 HL of 16 ppm alkalinity water.

In the real world however, due to incomplete dissociation, 80% Lactic Acid does not have a strength of 10.46 mEq/mL. At a target pH of 5.5 (for but one target pH example) it actually has a strength of only 10.3059 mEq/mL

Therefore, to hit pH 5.5:

31.972 mEq alkalinity / 10.3059 mEq/[email protected] = 3.10 mL of 80% Lactic Acid required.

But there is yet another catch. To hit zero extant bicarbonate species sufficient acid must be added to hit pH 4.3.

And at pH 4.3, the mEq/mL strength of 80% Lactic Acid is only 7.734 mEq/mL

Thus:

31.972 mEq alkalinity / 7.734 mEq/mL = 4.13 mL of 80% lactic acid to hit pH 4.3 and finally be able to definitively declare alkalinity to be at zero ppm.

Still not 6 mL as per Murphy....

Bottom line: I believe you would be far better served with the pH 5.5 target answer of 3.1 mL of 80% Lactic Acid. But as stated, it depends upon what pH you want to hit as your target goal.

The "relative" (or "effective") strengths which I calculate for 80% lactic acid for a given target pH are as follows over the range of pH's that are of most concern to brewers:

For 5.0 pH, 80% lactic acids effective "acid strength" = 9.83 mEq/mL
For 5.1 pH, 80% lactic acids effective "acid strength" = 9.97 mEq/mL
For 5.2 pH, 80% lactic acids effective "acid strength" = 10.08 mEq/mL
For 5.3 pH, 80% lactic acids effective "acid strength" = 10.17 mEq/mL
For 5.4 pH, 80% lactic acids effective "acid strength" = 10.25 mEq/mL
For 5.5 pH, 80% lactic acids effective "acid strength" = 10.31 mEq/mL
For 5.6 pH, 80% lactic acids effective "acid strength" = 10.35 mEq/mL
For 5.7 pH, 80% lactic acids effective "acid strength" = 10.39 mEq/mL
For 5.8 pH, 80% lactic acids effective "acid strength" = 10.42 mEq/mL
For 5.9 pH, 80% lactic acids effective "acid strength" = 10.45 mEq/mL

Notice that it is only for the case whereby the pH target is fully 2 points greater than pKa1 whereby the acid strength is essentially fully representative of the monoprotic ideal acid strength initially calculated at 10.46 mEq/mL. When the pH is 2 full points away from pKa1 nearly all of the first (of potentially two) H+ ions per molecule are being liberated.

 

[Silver_Is_Money]

A similar error appears on Murphy's 75% phosphoric acid technical sheet. It also indicates nearly twice as much acid addition as is required.

 

[mabrungard]

A proper acid/base water chemistry calculator can provide a better answer when dealing with weak acids. Bru’n Water has a Water Acidification calculator that is what is needed.

 

[Silver_Is_Money]

Here's a shocker. I finally found my "Water" book, which had gone missing for roughly 3 years, and in Appendix 'B' A.J. deLange explains (presuming that I'm reading him correctly) that reducing alkalinity in sparge water to hit a specifically targeted pH such as 5.5, etc... only requires somewhere between 80% and 90% of the straight up calculated acid requirement, such as would be seen for my examples of hitting pH 5.5 and 4.3 in post #2 above. And a quick check of both the MpH 3.0 calculator and Bru'n Water shows that for the case of targeting pH 5.5 their calculators only call for about 86%-87% of the calculated requisite acid addition which I computed in post #2 above. I'm presently both flummoxed and dumbfounded by this. Can anyone guide me out of my massive confusion here?

PS: The Kaiser water calculator breaks ranks with MpH 3.0 and Bru's Water and calculates the full acid addition with no reduction, in good agreement with my calculated acid additions as seen in post #2 above for the case of alkalinity only added to sparge water. But oddly enough, if minerals are added to the sparge water within the Kaiser calculator, it's acid volumes required to adequately de-alkalize sparge water plummet to valuations even lower than those witnessed within MpH 3.0 and Bru'n Water.

 

[Silver_Is_Money]

In continuing with the post above, if (for the specific case of targeting 5.5 pH) you add only ~87% of the acid calculated (or anywhere from 80% to 90% as per AJ, who is looking at a broader range spanning targets from ~pH 5.3 to ~pH 5.7 or 5.8), you leave behind in the water 13% of its alkalinity (or 10% to 20% as per AJ's broader span). This may not matter all that much if you start with ~75 ppm alkalinity and remove all but ~10 ppm, but if you start with ~300 ppm and leave behind ~39 ppm of alkalinity it could easily make a huge difference.

And on top of that, if you are shooting for 5.5 pH, and the zero alkalinity point is definitively achieved only at 4.3 pH, then at 5.5 pH there must still remain some ppm quantity of alkalinity that has not been eliminated (due to not hitting pH 4.3), so factually something more than 10 ppm or 39 ppm is actually being left behind. This is why I'm flummoxed.

 

[Silver_Is_Money]

It turns out to be the case that the Kaiser Water Calculator only calculates sufficient acid addition to sparge water to rid it of its Residual Alkalinity, and not its factually present Total Alkalinity. For the impossible case of water with alkalinity only, residual alkalinity and total factual alkalinity are equal. For the case of using the Ward Labs quantified analyticals for my very high alkalinity well water, the Kaiser calculator calculates only enough acid to rid sparge water derived from this source of only ~66% of its alkalinity, thus leaving behind well more than 100 ppm alkalinity.

Something appears fishy in all of this.

 

[Dim]

It turns out to be the case that the Kaiser Water Calculator only calculates sufficient acid addition to sparge water to rid it of its Residual Alkalinity, and not its factually present Total Alkalinity. For the impossible case of water with alkalinity only, residual alkalinity and total factual alkalinity are equal. For the case of using the Ward Labs quantified analyticals for my very high alkalinity well water, the Kaiser calculator calculates only enough acid to rid sparge water derived from this source of only ~66% of its alkalinity, thus leaving behind well more than 100 ppm alkalinity.

Something appears fishy in all of this.

I just wanted to say thanks for the detailed explanation and the subsequent thoughts. I have gone through various acid calculations myself and found not much common ground between them which made me curious.

I referenced Murphy and Son's spec because I assumed they (or their manufacturer) would have done some empirical analysis to determine their numbers and I wanted to understand how that can be replicated in theory. I still have a hard time to believe that they are wrong by such a huge margin. To demonstrate the vast difference, I used a simplified calculation where I omitted adjustments for pH (relatively negligible @ 5.5 pH target as you have demonstrated) and the density of 1.2 as an approximation.

 

[Silver_Is_Money]

I think I can see where the ~86% to 87% (and AJ's 80% to 90%) may come from. See the chart below. 5.5 pH seems to cross horizontally to the left and hit the 'Y' axis at just about 0.13 or 0.14, or 13% to 14% (corresponding to 87% and 86% acid additions respectively). 80% (or 0.20) crosses over and down to ~pH 5.7-5.8, and 90% (or 0.10) crosses over and down to ~ph 5.3.

But to me it appears that there may be double dipping occurring here. If one initially compensates for the acids relative mEq/mL "acid strength" for a given targeted pH, and then on top of that applies the chart derived 'Y' Axis "corrective" multiplicative factors spanning from 80% to 90% (pH 5.7 or perhaps pH 5.8, to pH 5.3), one is in doing so double dipping on the compensation end of things, no?

Perhaps @ajdelange, @mabrungard, and @dmr will assist here and lift the confusion.

 

[Silver_Is_Money]

Page 96 of the "Water" book offers a chart which is a bit easier to follow than the one I posted above as to garnering what I'm perceiving to be the requisite(???) corrective 'acid scale back' factor to apply for any given target pH. I applied a simple and straightforward "times 0.87" multiplicative correction factor to a modified Mash Made Easy 8.10 test version of my spreadsheet which targets sparge water acidification to exclusively 5.5 pH and it appears in admittedly limited testing to be giving output results for sparge water acidification additions that are essentially splitting the narrow output difference found between MpH and BW. I can also at least tentatively offer that neither BW or MpH appear to be basing their acidity advice upon Residual Alkalinity or extant sparge water mineralization levels as for the Kaiser calculator. But as stated in one of my above posts, this approach leaves a measurable and high load of remaining ppm's of alkalinity for the case of high alkalinity initial source water, and this somehow seems to be highly intuitively wrong. Still awaiting an explanation as to why it is not wrong.

 

[Dim]

@mabrungard I have been using Bru’n Water for years (thanks for all the work that went into it, btw) but my actual pH with Lactic 80% was consistently above the prediction. I always attributed this to inconsistencies between the water report and my actual tap water.

As mentioned above, the numbers from the Murphy & Son's spec are most likely based on some empirical evidence and also much more in line with my own experience. It could obviously be a fluke but I'm interested in the subject and that's why I'm asking for expert advice/opinion on how/why this would be.

 

[Silver_Is_Money]

OK, after thoroughly going over Palmer and Kaminski's book titled "Water" on pages 93-96, I have finally modeled what AJ is doing therein. The below rather lengthy non-linear regression formula (wherein x = water pH) closely mirrors and duplicates the chart found on page 96 of "Water" to a respectably high R value of confidence specifically for water pH's spanning the pH range as seen on the chart. From there it's merely a matter of following along with the given formulas on the preceding pages to compute the actual mEq's of alkalinity to be removed via an acid addition for a given volume of sparge water with a given ppm alkalinity (as CaCO3, and when titrated to pH 4.3, or alternately when titrated to the newer ISO standard of pH 4.5) to hit a desired sparge water target pH for the effective neutralization of its alkalinity.

Water Charge (chart on page 96) ~=-780.132225013302+930.675267920368*x-468.985903380742*x^2+129.285941580041*x^3-21.0407808980981*x^4+2.02073321523162*x^5-0.106055330905865*x^6+0.00234788488449278*x^7

 

[Dim]

Me again! Sorry, I am not quite ready to give up yet, as I still struggle to understand seemingly basic water calculations. I have picked out another example where Kaiser, Palmer and Brungard all (sort-of) agree, but - to me - they all seem to have made the same mistake in their calculations. I have to assume that I am wrong and they are obviously right, but could someone PLEASE explain to me why?

Here's the example:

• Water Profile (for completeness): Ca: 100, Mg: 5, Na: 30, SO4: 50, Cl: 50, HCO3: 245, CaCO3: 200
• Acid addition: 4ml of 75% phosphoric acid to 20L
  
• Kaiser:
  
  • assumes 4.81g (pure acid) x 1000 ÷ 98g/Eq ~ 49.1 mEq
  • result: HCO3: 95, CaCO3: 78, RA: 4
• Palmer:
  
  • assumes 96 g/Eq
    
  • result: CaCO3: 77, RA: 3
• Brungard:
  • calculations protected, cannot verify assumptions
  • result: HCO3: 96, CaCO3: 79

As you can see, all three are reasonably close, but they all seem to assume 98 g/Eq for H3PO4. To my knowledge H3PO4 has a molar mass 98g/mol which needs to be divided by 3 to get the equivalent weight - https://www.quora.com/What-is-the-gram-equivalent-weight-of-phosphorous-acid-H3PO4.

When I do my 'napkin math', I get: 4.7g x 1000 ÷ 32.7g/Eq ~ 143.7 mEq

Any thoughts/ideas? I do appreciate your feedback!

 

[Silver_Is_Money]

Although Phosphoric Acid is polyprotic and has 3 protons (H+ ions) it can release, at a pH of 5.5 the picture looks like this:

1st proton: 99.9543121% released
2nd proton: 1.9125528% released
3rd proton: 0.0000151% released

For 75% Phosphoric Acid at pH 5.5, its effective acid strength (by my calculation) is 12.31 mEq/mL. (rounded)

It is only about 1.0187 times stronger at pH 5.5 than if it was merely monoprotic.

 

[Dim]

Although Phosphoric Acid is polyprotic and has 3 protons (H+ ions) it can release, at a pH of 5.5 the picture looks like this:

1st proton: 99.9543121% released
2nd proton: 1.9125528% released
3rd proton: 0.0000151% released

For 75% Phosphoric Acid at pH 5.5, its effective acid strength (by my calculation) is 12.31 mEq/mL. (rounded)

It is only about 1.0187 times stronger at pH 5.5 than if it was merely monoprotic.

OK, thanks (so much!), I think I now got it. At pH 5.5, the calculation would be then something along the lines of:

r1 = 10^(5.5 - 2.14)
r2 = 10^(5.5 - 7.2)
r3 = 10^(5.5 - 12.37)
m0 = 1 / (1 + r1 + r1 * r2 + r1 * r2 * r3)
m1 = m0 * r1 ~ 0.98
m2 = m1 * r2 ~ 0.02
m3 = m2 * r3 ~ 0.00
sf = m1 + 2 * m2 + 3 * m3 ~ 1.019
strength = 4.7 * 1000 / 98 * sf ~ 48.8 mEq/L

 

[Silver_Is_Money]

4 mL x 12.306 mEq/mL = 49.22 mEq's (for pH 5.5)

I use 2.16, 7.21, and 12.32 as my pKa's.

 

[Jocky]

I stumbled across this thread after having some similar confusion over lactic acid's ability to neutralise alkalinity.

While I can't contribute to the chemistry discussion above, I can contribute that Murphy's have conflicting data on their website. Going through their resources their latest Lactic Acid data sheet says...

"3 ml of Lactic Acid per hl reduces the alkalinity by 16 mg/litre (ppm)."

From https://www.murphyandson.co.uk/wp-content/uploads/2018/10/Lactic-acid-Rev-4.pdf
via https://www.murphyandson.co.uk/resources/datasheets/#tech-datasheet-L


I've also seen a water analysis sent to a friend back in 2015 that recommends double that dosing rate.

 

[Silver_Is_Money]

"3 ml of Lactic Acid per hl reduces the alkalinity by 16 mg/litre (ppm)."

I've also seen a water analysis sent to a friend back in 2015 that recommends double that dosing rate.

For 80% Lactic Acid the Murphy's advice should bring 100 Liters of ~pH 7.5 water with an initial 19 ppm of Alkalinity down to about pH 5.7 and 3 ppm of remaining Alkalinity, so for this case I see it as decently reliable information. Doubling it would be a mistake.

 

[Jocky]

For 80% Lactic Acid the Murphy's advice should bring 100 Liters of ~pH 7.5 water with an initial 19 ppm of Alkalinity down to about pH 5.7 and 3 ppm of remaining Alkalinity, so for this case I see it as decently reliable information. Doubling it would be a mistake.

That was also my thought on reading my friend's water analysis. It suggests adding 17ml of 80% lactic acid to every 25 litres of liquor, in order to reduce 212ppm alkalinity as CaCO3 down to 40ppm.

It was only today having seen the product sheet linked in the first post of this thread that things clicked together and I realised that Murphy's had previously got things very wrong at some time in the past.

I wonder how many of their clients realised this?

 

[Silver_Is_Money]

That was also my thought on reading my friend's water analysis. It suggests adding 17ml of 80% lactic acid to every 25 litres of liquor, in order to reduce 212ppm alkalinity as CaCO3 down to 40ppm.

I calculate ~8.3-8.4 mL to accomplish this (initial water pH dependent).

With 80% Lactic Acid Tech Sheet revision #4 Murphy's has corrected the earlier error.

 

[Silver_Is_Money]

FWIW, I recently came across a peer reviewed industry level brewing document which stated that alkalinity within sparge water needs only to be reduced to a maximum of 50 ppm (as CaCO3), as at that juncture and below it does not seem to induce tannins, astringency, etc... All are free to take or leave this information, or provide comment as to their thoughts on it.