Silver_Is_Money
Larry Sayre, Developer of 'Mash Made Easy'
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.
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.
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