Silver_Is_Money
Larry Sayre, Developer of 'Mash Made Easy'
For our simplified ** procedure example let's assume the following:
4 gallons of sparge water at 100 ppm alkalinity
Desired goal is to neutralize the alkalinity sufficiently to achieve pH 5.4
mEq/mL of 88% Lactic Acid = ~11.5 in the vicinity of pH 5.4
Molecular weight of CaCO3 = 100 g/mole = 100 mg/mmole
Normality of CaCO3 = 2 (such that 1 mole = 2 Eq's or 1 mmole = 2 mEq's)
Normal weight of CaCO3 = 100/2 = 50 g/Eq = 50mg/mEq
Alkalinity in ppm is measured as CaCO3
ppm = mg/L
Initial procedure for neutralization to a final pH of ~4.3:
1) 4 gallons x 3.7854 = 15.142 Liters
2) 100 mg/L alkalinity x 15.142 Liters = 1,514.16 mg alkalinity as CaCo3
3) 1,514.16 mg /100 mg/mmole = 15.1416 mmoles of alkalinity
4) 15.1416 mmoles x 2 mEq/mmole = 30.2832 mEq's of alkalinity
5) 30.2832 mEq/11.5 mEq/mL = 2.633 mL of 88% Lactic Acid to hit 4.3 pH
Final procedure to hit only pH 5.4 (rather than pH 4.3 as above):
1) 2.633 mL x 0.90 = 2.37 mL of 88% Lactic Acid to hit 4.3 pH
Final answer is to add 2.37 mL of 88% Lactic Acid to achieve a pH of 5.4
** Note that this intentionally simplified procedure is only reliably valid for the case of alkaline water within a narrow range of pH's near 7. It should get you sufficiently close to pH 5.4 for alkaline water up to about pH 8.2. A wee pinch of additional 88% lactic acid will be required if your waters pH exceeds 8.2.
++ To solve the above for 10% Phosphoric Acid simply substitute an mEq/mL value of ~1.05 in the above example, rather than using Lactic Acids (~pH 5.4 vicinity) mEq/mL of ~11.5. Or to simplify even further, a ballpark rule of thumb is that 11 x mL of 88% Lactic Acid = mL of 10% Phosphoric Acid
4 gallons of sparge water at 100 ppm alkalinity
Desired goal is to neutralize the alkalinity sufficiently to achieve pH 5.4
mEq/mL of 88% Lactic Acid = ~11.5 in the vicinity of pH 5.4
Molecular weight of CaCO3 = 100 g/mole = 100 mg/mmole
Normality of CaCO3 = 2 (such that 1 mole = 2 Eq's or 1 mmole = 2 mEq's)
Normal weight of CaCO3 = 100/2 = 50 g/Eq = 50mg/mEq
Alkalinity in ppm is measured as CaCO3
ppm = mg/L
Initial procedure for neutralization to a final pH of ~4.3:
1) 4 gallons x 3.7854 = 15.142 Liters
2) 100 mg/L alkalinity x 15.142 Liters = 1,514.16 mg alkalinity as CaCo3
3) 1,514.16 mg /100 mg/mmole = 15.1416 mmoles of alkalinity
4) 15.1416 mmoles x 2 mEq/mmole = 30.2832 mEq's of alkalinity
5) 30.2832 mEq/11.5 mEq/mL = 2.633 mL of 88% Lactic Acid to hit 4.3 pH
Final procedure to hit only pH 5.4 (rather than pH 4.3 as above):
1) 2.633 mL x 0.90 = 2.37 mL of 88% Lactic Acid to hit 4.3 pH
Final answer is to add 2.37 mL of 88% Lactic Acid to achieve a pH of 5.4
** Note that this intentionally simplified procedure is only reliably valid for the case of alkaline water within a narrow range of pH's near 7. It should get you sufficiently close to pH 5.4 for alkaline water up to about pH 8.2. A wee pinch of additional 88% lactic acid will be required if your waters pH exceeds 8.2.
++ To solve the above for 10% Phosphoric Acid simply substitute an mEq/mL value of ~1.05 in the above example, rather than using Lactic Acids (~pH 5.4 vicinity) mEq/mL of ~11.5. Or to simplify even further, a ballpark rule of thumb is that 11 x mL of 88% Lactic Acid = mL of 10% Phosphoric Acid
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