pKa dissociation constants and weak acid mEq strength at targeted pH

[Silver_Is_Money]

How do we determine a weak acids mEq/mL strength at a given (or targeted) pH?

Lets begin this excercise with a "nominally*" monoprotic acid, which by popular demand will be 88% Lactic Acid.
(*nominally monoprotic because it is diprotic if one considers that it has a pKa2 at a pH related value of 15.1, which is generally and safely ignored)

88% Lactic Acid:
pKa1 = 3.86, pKa2 = 15.10 (but we can for any practical purposes ignore pKa2, since we are nowhere near pH 15.10 when mashing, etc...)
Density = 1.206 g/CC
Valence = 2, but since we are ignoring pKa2 we will presume it to be 1 (and hence monoprotic)
Molecular Weight = MW = 90.078 g/mole
Concentration = 0.88

Step 1) Calculate grams of Lactic Acid per mL:
0.88 x 1.206 = 1.06128 g./mL

Step 2) Calculate Molarity (M):
1000(1.06128 g./mL) = 1061.28 g/L
1061.28 g/L / 90.078 g/mole = 11.7818 moles = 11.7818M = 11.7818 mmole/mL (presumes full H+ dissociation)

Step3) Calculate Normality (N)
N = M x Valence
N = 11.7818 x 1* = 11.7818N = 11.7818 mEq/mL (*Valence is actually 2, but we are ignoring pKa2, so for us Valence = 1)

Step 4) Establish a target pH:
Target pH = pH_T (whereby within the mash most typically 5.40 pH is targeted)

Step 5) Determining the pKa1 Percent dissociation (or liberation) of the H+ (acid) ion at pH_T
Percent Freely Dissociated H+ =1-1/(1+10^(pH_T - pKa1))

Example for pH_T = 5.40
Percent Freely Dissociated H+ =1-1/(1+10^(pH_T - pKa1))
Percent Freely Dissociated H+ =1-1/(1+10^(5.40 - 3.86))
Percent Freely Dissociated H+ = 0.971968
(where by "Percent", I'm intending percent as a fraction, or fractional percent)

Step 6) Determine mEq/mL Acid strength
11.7818 mEq/mL for full dissociation x 0.971968 dissociation = 11.4515 mEq/mL acid strength at pH 5.40

Answer for the case where pH_T = 5.40 = 11.4515 mEq/mL

HINT: By definition the H+ dissociation at any given pKa = 50% (or 0.50) when pH_T = pKa, so you can test your math by setting pH_T = pKa1. If you get 0.50 for dissociation, you are doing this correctly.

​

 

[Silver_Is_Money]

The pKa's for triprotic (meaning 3 H+ ions available for dissociation) Phosphoric Acid (H3PO4) are:
pKa1 = 2.16
pKa2 = 7.21
pKa3 = 12.32

Since we mash at ballpark pH 5.40, and this lies between pKa1 and pKa2, we can not ignore pKa2 for H3PO4 acid.

Anyone want to take a stab at calculating the mEq/mL strength of 85% Phosphoric Acid at a target pH of 6.00?
(note that pH 6.00 is yet closer to, and thereby more influenced by pKa2)

Hint: Sum the H+ dissociation fractions computed for each pKa, and divide the sum by H3PO4's valence, to get overall H+ dissociation at pH 6.00. H3PO4's valence = 3, since there are 3 H+ ions per H3PO4 molecule, and the valence of each individual H+ is 1, whereby 3 x 1 = 3.

 

[Silver_Is_Money]

How do you know when you are dealing with a "weak acid", whereby the above is relevant to the determination of relative acid strength (relative here meaning relative to a targeted pH)? If it is not one of the below listed "strong" acids, it is a "weak" acid:

List of strong acids:

1. Hydrochloric acid (denoted by the chemical formula HCl)
2. Hydrobromic acid (denoted by the chemical formula HBr)
3. Hydroiodic acid or hydriodic acid (denoted by the chemical formula HI)
4. Sulfuric acid (denoted by the chemical formula H2SO4)
5. Nitric acid (denoted by the chemical formula HNO3)
6. Chloric acid (denoted by the chemical formula HClO3)
7. Perchloric acid (denoted by the chemical formula HClO4)

A strong acid dissociates all of its H+ ions at nigh on any pH target, so its strength is effectively always simply its normality expressed as mEq/mL

A weak acid does not readily give up into solution (or dissociate) its H+ ions, and therefore the fraction of H+ which is dissociated with respect to any given pH must be calculated whereby to determine its strength relative to a targeted pH.

 

[Silver_Is_Money]

No takers for the question of 85% Phosphoric Acids "relative" mEq/mL strength at pH 6? Lets go through the same steps as for Lactic Acid, albeit apply them to Phosphoric Acid:

85% Phosphoric Acid:
pKa1 = 2.16, pKa2 = 7.21, pKa3 = 12.32
Density = 1.689 g/CC
Valence = 3
Molecular Weight = MW = 97.995 g/mole
Concentration = 0.85

Step 1) Calculate grams of Phosphoric Acid per mL:
0.85 x 1.689 = 1.43565 g./mL

Step 2) Calculate Molarity (M):
1000(1.43565 g./mL) = 1435.65 g/L
1435.65 g/L / 97.995 g/mole = 14.6502 moles = 14.6502M = 14.6502 mmole/mL (presumes full H+ dissociation)

Step3) Calculate Normality (N)
N = M x Valence
N = 14.6502 x 3 = 43.9507N = 43.9507 mEq/mL

Step 4) Establish a target pH:
Target pH = pH_T (whereby for our example 6.00 pH is targeted)

Step 5) Determine pKa1, pKa2, and pKa3 Percent dissociation (or liberation) of the 3 x H+ (acid) ions at pH_T

Percent Freely Dissociated H+ = 1-1/(1+10^(pH_T - pKa))

Example for pH_T = 6.00
pKa1 Percent Freely Dissociated 1st H+ ion =1-1/(1+10^(6.00 - 2.16)) = 0.999855477
pKa2 Percent Freely Dissociated 2nd H+ ion =1-1/(1+10^(6.00 - 7.21)) = 0.058078414
pKa3 Percent Freely Dissociated 3rd H+ ion =1-1/(1+10^(6.00 - 12.32)) = 0.000000479

Step 6) Determine mEq/mL Acid strength
Sum of 0.999855477 + 0.058078414 + 0.000000479 = 1.05793437
1.05793437/3 = 0.35264479 (meaning only 35.264% of the three H+ ions within H3PO4 are liberated at pH 6)
43.9507 mEq/mL (full dissociation) x 0.352645 "overall" H+ dissociation = 15.4989 mEq/mL strength at pH 6.00

Relative Acid Strength answer for the case where pH_T = 6.00 = 15.50 mEq/mL (rounded)

 

[Silver_Is_Money]

What if we added 1 mL of 85% Phosphoric Acid to 250 mL of fresh DI water, and then to this we added a squirt of phenolphthalein solution, and then we wanted to titrate this solution with 1N sodium hydroxide (NaOH), which is a "strong" base. How many mL of 1N NaOH should it take whereby to neutralize the acidified water and then raise its pH to the 8.3 pH level required for phenolphthalein to turn from clear to first 'stable' faint pink?

Example for pH_T = 8.30
pKa1 Percent Freely Dissociated 1st H+ ion =1-1/(1+10^(8.30 - 2.16)) = 0.999999276
pKa2 Percent Freely Dissociated 2nd H+ ion =1-1/(1+10^(8.30 - 7.21)) = 0.924827221
pKa3 Percent Freely Dissociated 3rd H+ ion =1-1/(1+10^(8.30 - 12.32)) = 0.00009549

Step 6) Determine mEq/mL Acid strength
Sum of 0.999999276 + 0.924827221 + 0.00009549 = 1.924921987
1.924921987/3 = 0.641640662 (meaning 64.164% of the three H+ ions within H3PO4 are being liberated at pH 8.3)
43.9507 mEq/mL (for full dissociation) x 0.641640662 "overall" H+ dissociation = 28.2006 mEq/mL acid strength at pH 8.3

28.2 mEq/mL of liberated H+ ions at pH 8.3 require 28.2 mEq/mL of OH- ions whereby to raise the solution to pH 8.3, and since we know that 1 mL of a 1N concentration "strong" base solution = 1mEq of base at all pH's, the calculated answer (mathematical prediction) is:

28.2 mL of 1N NaOH are required whereby to hit 8.3 pH and turn the phenolphthalein indicator to a 'first lasting' faint pink

 

[Silver_Is_Money]

Who wants to put post #5 to the test?

10 grams of NaOH (sodium hydroxide, or lye) made up to 250 total mL with DI or good distilled water = 1N NaOH