Determining Acid Equivalents

[BigGutHeavyD]

Suppose 1 kg of malt is crushed, added to DI water and a pH reading is taken. From that point a solution of .1N HCl is added until the pH reads 5.0.

pH ml of .1N HCl
5.7 0 (Starting DI pH)
5.6 1
5.5 2
5.4 4
5.3 6
5.2 9
5.1 11
5.0 15

It can now be reasonably assumed that 4ml of .1N HCl would be needed per Kg of malt to reach a mash pH of 5.4.

The problem for me is that when actually brewing I'm use 88% Lactic Acid or 10% Phosphoric Acid solutions and not .1N HCl.

Without doing the titration with the 88% Lactic or 10% Phosphoric acid, how does one determine their equivalents to the .1N HCl.

In other words, "How much 88% Lactic Acid is equivalent to 4ml of .1N HCl?" or "How much 10% Phosphoric Acid is equivalent to 4ml of .1N HCl?"

Can someone please explain the math involved in solving these problems?

(I'm not a chemist, scientist or mathematician so please go easy, explain and simplify as much as possible, would be much appreciated. Also .1N HCl is just an example, could also be 0.5 N H2SO4, or some other acid solution and concentration.)

 

[ajdelange]

Lactic and phosphoric acids are weak acids and so as a first step we have to determine the charge on 1 mole of each as a function of pH. I won't go into the details as it's fairly involved (falls out of Henderson-Hasselbalch) but I can give the algorithm for doing the actual calculations in the space available here.

First you need the ratios

r1 = 10^(pH - pK1)
r2 =10^(pH - pK2)
r3 = 10^(pH - pK3)

where pH is the pH you are working at. IOW if you want to calculate the amount of acid needed to get to mash pH 5.4 then pH = 5.4. r1 is the ratio of the concentration of the acid less one proton to the concentration of the acid which hasn't lost any (undissociated). H3PO4 <--> H+ + H2PO4-

The pKs are properties of the acid. Lactic acid has only one proton to lose and so has only one pK = 3.86. Phosphoric acid has 3 with associated pKs pK1 = 2.148, pK2 = 7.198 and pK3 = 12.319 (Wikipedia - temperature not specified).

The acid will, according to the pH, distribute itself among the available species i.e. the undissociated acid, acid less one proton (monobasic ion), acid less two protons (dibasic ion) and acid less three protons (tribasic ion). The fraction (0 - 1) of undissociated ions is

f0 = 1/(1 + r1 + r1*r2 + r1*r2*r3).

As the ratio of monobasic to undissociated ions is r1

f1 = r1*f0

and, similarly

f2 = r2*f1
f3 = r3*f2

We call undissociated fraction 'f0' because the charge on the undissociated acid molecules (H3PO4) is 0. We call the monobasic fraction 'f1' because the charge on the monobasic ion (H2PO4-) is -1, the dibasic (HPO4--) fraction is f2 because the charge on the dibasic ion is -2 and so on for the tribasic ion. Thus if we add uncharged acid to a mash and the resultant pH is is 5.4 the charge on the acid derived molecules becomes

Q = -Pt*(0*f0 + 1*f1 + 2*f2 + 3*f3)

in which Pt is the total number of moles of acid added. There is a plot of Q vs pH in Palmers Water book for 1 mol phosphoric acid.

This describes phosphoric acid which has three protons. To make the same math (in a spreadsheet) work we pretend lactic acid also has 3 protons too but that they are so closely bound they never get released which we model by using pK2 = pK3 = 40. This results in r2 = r3 = 0 and f2 = f3 = 0.

Each unit of Q represents a unit of protons released and a unit of base neutralized so the strength of the acid added when Pt moles are supplied is Q mEq.

Because pK1 for lactic acid is close to mash pH and because mash pH falls about half way between the first two pKs of phosphoric acid Q, in each case, is about 1 mEq per 1 mmol of the acid.

This leaves us to determine the number of mmol of lactic or phosphoric acid in a mL of stock acid of specified strength and this depends on the density of the acid as a function of its strength.

Putting all of this together we can calculate that 88% lactic acid has strength, depending on the end pH of
pH mEq/L
5.2 11.26 mEq/L = 11.26 N
5.3 11.36
5.4 11.45
5.5 11.51
5.6 11.56
5.7 11.61

If a solution provides 1 Eq/L = 1 mEq/mL it is said to be normal ( 1 N). Thus the numbers of mEq/mL are the normalities of the solutions.

Thus, the higher the pH you are shooting for the stronger the acid will be as it is more dissociated at higher pH. Similarly for 10% phosphoric

pH N
5.2 1.08
5.3 1.08
5.4 1.09
5.5 1.09
5.6 1.10
5.7 1.10
5.8 1.11

And for 80% phosphoric acid

pH N
5.2 13.42
5.3 13.45
5.4 13.50
5.5 13.55
5.6 13.61
5.7 13.69
5.8 13.79

 

[ajdelange]

Suppose 1 kg of malt is crushed, added to DI water and a pH reading is taken. From that point a solution of .1N HCl is added until the pH reads 5.0.

pH ml of .1N HCl
5.7 0 (Starting DI pH)
5.6 1
5.5 2
5.4 4
5.3 6
5.2 9
5.1 11
5.0 15

We would, based on these data, conclude that the malts alkalinity vs. pH is

mEq/kg = -7.0268*(pH - 5.7) + 20.225*(pH - 5.7)^2 - 0.013101*(pH - 5.7)^3

This isn't very realistic as the first coefficient is usually at least -60 and at most around -30 mEq/kg•pH

 

[BigGutHeavyD]

You are, of course, correct as the data was made up on a whim.

Thank you for the explanation, I'll have to take some time and digest it.

 

[BigGutHeavyD]

How much 88% Lactic Acid is equivalent to 4ml of .1N HCl?

1.) The density of 88% lactic acid is 1.206 g/cc
2.) 1 mL of 88% lactic acid contains 0.88 * 1.206 = 1.06 grams of lactic acid
3.) Lactic acid has a gram equivalent weight of 90.08 g
4.) 1.06 grams of lactic acid delivers (1.06 / 90.08) * 1000 = 11.8 mEq protons

5.) The density of a 0.1N - 37% hydrochloric acid is 0.12 g/ml
6.) 1 mL of 0.1N - 37% HCl contains 0.37 * 0.12 = 0.0444 grams of hydrochloric acid
7.) HCl acid has a gram equivalent weight of 36.461 g
8.) 0.0444 grams of HCl delivers (0.0444 / 36.461) * 1000 = 1.22 mEq protons
9.) 4 mL of .1N HCl delivers (4 * 1.22) = 4.88 mEq protons

9.) (X / 90.08) * 1000 = 4.88 mEq protons (X = unknown grams of lactic acid)
10.) X = 0.4395904 grams of lactic acid delivers 4.88 mEq protons
11.) 0.4395904 grams of lactic acid / 1.06 grams of lactic acid per 1 ml of 88% lactic acid solution =

12.) 0.415 ml of 88% lactic acid solution = 4ml of .1N HCl solution

Not being well versed in chemistry this may well be wrong, please correct me so that I may learn.

Of course that does not account for gram equivalent weight change as pH changes.

I believe, Mr. Delange, that you explained it fairly well at the end of this thread:

https://www.homebrewtalk.com/showthread.php?t=422296

 

[ajdelange]

The problem is in your step 4. One mmol of lactic. Acid does not deliver 1 mEq of protons at mash pH but slightly less than that with the actual amount being determined by the pH. The number is close enough to 1 that using 1 is doubtless good enough for home brewing as is the case for phosphoric so your result is close enough for government work.