Lactic acid calculations

[jonboris]

I think ions are very very small so probably it's a lot. I'm curious why it's important to know how many.

I want to know the amount of ions so i can fine tune the recipe and dont contribute with more than whats detectable, in the water book it says 400ppm is a threshold but some can taste it even sooner.
I want to know dag nabit....

Your brewing water spreadsheets should be telling you how to contribute enough to correct for pH, your palate should tell you what's detectable. ppm isn't a measure of ions. I can assure you you are not contributing enough to taste at 1ml in your sparge and mash water.

If you feel you can taste it, switch to phosphoric acid. It's a lower concentration but is vastly more undetectable flavorwise as a pH adjuster. Vastly.

[SweBrew]

Thanks all, but i know how much i´ll need to correct for mash pH and at 1-2ml for a 5gal batch its not detectable, but it would be nice to know how much it contributes. and when it´s noticeable.

[cclarkk]

I came across this thread while looking for an answer to a question that I had. I know it's an old question, but maybe I can shed some light on this for anyone else wondering.

Short answer: 5.35*10^21 molecules, each contributing one (H+) and one anion (CH3CHOHCOO-).

Long answer:
First off, it's simply a lot easier to use the same units for everything. Using non-standard units causes most of the head scratching for everyone. Water reports are in ppm, acid in % by dry weight per volume, equivalents, etc. It's easier for the agencies, manufacturers, or whoever to use what they use, so that's why we are always scratching our heads.

As was stated in a previous reply, acid percentage is grams of dry acid per 100mL of solvent, in this case water.
For an 80% solution of lactic acid, you have 80g lactic acid powder added to 100mL water. Using the molar mass of lactic acid, 90.08 g/mole, you can divide into your 80g starting weight and find your solution to contain 0.888 moles. Multiply this by Avogadro's number, 6.022*10^26, and you will find that you have approximately 5.35*10^23 molecules per 100 mL of solution. I say approximately because adding the acid to the solution increases the volume. If you really wanted to know exactly what you had, you could measure the density and use that to calculate the exact amount.

So, for 1mL, you have 5.35*10^23 / 100 = ~5.35*10^21 molecules. The formula for lactic acid is CH3CHOHCOOH, and it dissociates into CH3CHOHCOO- and H+, so you get one ion of each contributed to your solution for each molecule that you add. In a highly concentrated solution, you do not get complete dissociation because it is a weak acid. However, as you are adding it to a 5 gallon batch, you should see complete dissociation.

Keep in mind that 1 ppm of liquid is not 1 molecule of chemical X per million molecules of solvent Y. It is one milligram per liter of solvent, usually water. So, for 400 ppm, you would have 0.400g of X per liter of water. You still need to multiply by total volume and Avogadro's number, then divide by the molar mass of X to find the number of molecules or ions present.

If you were feeling froggy, you could even calculate the expected pH of your strike water. And if you knew what the statistical contribution was from the components of your mash, you could calculate your pH mid mash.

Hope this helps.

[ajdelange]

For an 80% solution of lactic acid, you have 80g lactic acid powder added to 100mL water.

Not so. When the strength of a solution is expressed 80% 'by weight', often written 80% w/w it means, as the abbreviation suggests, that the dissolved substance is 80% of the weight of the solution. Thus to find the weight of lactic acid in 1 mL of the solution you must know its density. This is, for lactic acid, density = (-0.0000032247*W + 0.0026771)*W +0.99464 grams/mL where W is the w/w strength expressed as a percentage (i.e. 80). The number of grams of lactic acid is then (W/100)*density per mL and the number of millimoles 1000*(W/100)*density. The 'strength' is the number of protons emitted by the acid which is found by determining the fraction, f, of acid molecules which dissociate. This is found by computing r = 10^(pH - 3.86), where 3.86 is the pK of lactic acid and then f = r/(1+r). pH is the target pH for whatever it is that is being acidified. Because lactic acid isn't quite a strong acid f is close to but not equal to 1 e.g. something like 0.97 for typical mash target pH. The strength of the acid is then f*1000*(W/100)*density. This is the number of millimoles of protons (usually called milliequivalents) delivered by 1 mL of the acid in transitioning to the target pH. We call this the effective normality of the acid. As an example, 80% lactic acid delivers 10.3 mEq protons per mL to pH 5.5 and is thus said to be 10.3 N. If you wish to mash at pH 5.5 and have a proton deficit of X then the number of mL of 80% w/w acid you need are X/10.3.

The normality of lactic acid of strength W% w/w is well approximated by the simple polynomial -12.4962+ 0.0668348*W+4.57878*pH + 0.000231738*W*W + 0.00787279*W*pH -0.4193*pH*pH