Converting bicarbonate to meq/L or ppm Alk.

[Earl Karl]

Haven't learned this in chemistry yet, but I plan to go DIY for dosing. What is the formula to convert grams of any bicarbonate source (I'm using sodium acetate and sodium hydroxide) to meq/L or ppm (whichever is simpler)? I'm trying to get 4,000 meq/L of Alk. in a 1L solution for each acetate and hydroxide. Probably not happening with sodium hydroxide because I also plan to put calcium chloride (so basically kalk) in the same solution so I might end up having heavy precipitate, but I know I can achieve 4,000 meq/L of Alk. with sodium acetate and calcium chloride without it precipitating. Just don't know how much sodium acetate to put. From there, I can calculate how much Ca, Mg, and Sr to put. Just an experiment I'm doing.

 

[Randy Holmes-Farley]

You will make a useless mess if you put sodium hydroxide and calcium chloride together.

Sodium acetate and sodium bicarbonate are nearly the same alk potency per unit weight because they have about the same molecular weight.

 

[Randy Holmes-Farley]

If both the sodium acetate and sodium bicarbonate are pure and totally dry, the acetate is 2% more potent.

 

[Earl Karl]

So for anhydrous sodium acetate, how do I know how much to add to get a certain amount of Alk? I'm sure there is a formula for it.

Ok, I'm just going to use regular old Kalkwasser. First off, is it even possible to fully dissolve kalk?

 

[Randy Holmes-Farley]

So for anhydrous sodium acetate, how do I know how much to add to get a certain amount of Alk? I'm sure there is a formula for it.

Ok, I'm just going to use regular old Kalkwasser. First off, is it even possible to fully dissolve kalk?

Use the calculator linked in the stickies, and enter sodium bicarbonate. Subtract 2% if you are being way more exacting that is needed in a Reef context.

Calcium hydroxide will only dissolve to about 2 teaspoons per gallon, which is exactly 40.8 meq/l of alk at 25 deg C. Very little relative to what you wAnt to make.

 

[Earl Karl]

Ah I see ok thank you very much.

 

[Earl Karl]

So I did some calculations to make my own cheap Calcium Acetate solution ($50 per 50lbs bag for each CaCl2 and CH3COONa). I would like you to double check my calculations if you don't mind. Here is what I have as of now to make 1 gallon of solution:
Anhydrous Sodium Acetate 50lbs, using 4699.34 grams to get 15,140 meq/L of Alk for 1 gal.

Briner's Choice Calcium Chloride Anhydrous 50lbs, 94.5% Calcium Chloride. Using 3066.27 grams to get around 276,427 ppm Ca for 1 gal.

BRS Pharma-Grade Magnesium Chloride, 99% Magnesium Chloride. Using 623.1 g to get around 19,745 ppm Mg for 1 gal according to calculator on their website. Unfortunately not anhydrous, but anhydrous MgCl2 is SUPER expensive. Not using MgSO4 due to CaSo4 precipitating. Will dose Two Little Fishies Magnesium Pronto when needed.

Brightwell's Strontium-P, 54% min Sr. 53.24 g to get 7,594 ppm Sr for 1 gal.


The ratio of Ca:Mg:Sr is 91:6.5:2.5. There is 9% more Alk. than Ca. The rest of the 9% I have replaced with 6.5% Mg and 2.5% Sr. Is this ratio fine?

Is this too strong of a solution for 1 gallon? I've seen calcium acetate solutions where they have 70,000 mg/L Ca and 4,000 meq/L Alk. Mine is around 73,000 mg/L Ca, 5,200 mg/L Mg, 2000 mg/L Sr and 4,000 meq/L Alk. I do want my solution strong as possible as demand is fairly high yet have little to no precipitation as possible.

Finally I want to add Potassium Iodide to my solution. How much would you recommend for 1 gallon solution?

I know it's a lot but I hope you could give me advice. This is just an experiment I wanted to test. Thank you.

 

[Earl Karl]

Realized I'm putting many POUNDS of sodium acetate and calcium chloride into my solution. Unless that is normal... which seems strange.

I know some people put 500g of CaCl2 per liter. So in a gallon, that is about 2000g, or about 4 lbs. Which seems a lot for a gallon.

 

[Randy Holmes-Farley]

So I did some calculations to make my own cheap Calcium Acetate solution ($50 per 50lbs bag for each CaCl2 and CH3COONa). I would like you to double check my calculations if you don't mind. Here is what I have as of now to make 1 gallon of solution:
Anhydrous Sodium Acetate 50lbs, using 4699.34 grams to get 15,140 meq/L of Alk for 1 gal.

Briner's Choice Calcium Chloride Anhydrous 50lbs, 94.5% Calcium Chloride. Using 3066.27 grams to get around 276,427 ppm Ca for 1 gal.

BRS Pharma-Grade Magnesium Chloride, 99% Magnesium Chloride. Using 623.1 g to get around 19,745 ppm Mg for 1 gal according to calculator on their website. Unfortunately not anhydrous, but anhydrous MgCl2 is SUPER expensive. Not using MgSO4 due to CaSo4 precipitating. Will dose Two Little Fishies Magnesium Pronto when needed.

Brightwell's Strontium-P, 54% min Sr. 53.24 g to get 7,594 ppm Sr for 1 gal.


The ratio of Ca:Mg:Sr is 91:6.5:2.5. There is 9% more Alk. than Ca. The rest of the 9% I have replaced with 6.5% Mg and 2.5% Sr. Is this ratio fine?

Is this too strong of a solution for 1 gallon? I've seen calcium acetate solutions where they have 70,000 mg/L Ca and 4,000 meq/L Alk. Mine is around 73,000 mg/L Ca, 5,200 mg/L Mg, 2000 mg/L Sr and 4,000 meq/L Alk. I do want my solution strong as possible as demand is fairly high yet have little to no precipitation as possible.

Finally I want to add Potassium Iodide to my solution. How much would you recommend for 1 gallon solution?

I know it's a lot but I hope you could give me advice. This is just an experiment I wanted to test. Thank you.

What is the purpose of the potassium iodide? Iodide, or potassium?

Is there a reason you want a very strong dosing solution? If you make it to match my diy recipe, then you can use the calculators around to determine various things.

 

[Earl Karl]

Really both, but mainly potassium. Iodide sorta helps my chaeto grow ime.

I need strong solution due to high demand in Acro colonies. They suck up everything pretty fast, consuming around 2-3 dkH daily. This may still be too strong but I did compare to other products. While it is stronger probably all of the products, it's not far off from what I see.

 

[Randy Holmes-Farley]

Really both, but mainly potassium. Iodide sorta helps my chaeto grow ime.

I need strong solution due to high demand in Acro colonies. They suck up everything pretty fast, consuming around 2-3 dkH daily. This may still be too strong but I did compare to other products. While it is stronger probably all of the products, it's not far off from what I see.

Potassium iodide is a no go for potassium. You’d have to super overdose iodide to even boost potassium by 1 ppm.

Daily demand is not a great reason to want very high potency, unless you are trying to use some particular dosing pump. A normal two part seems plenty potent.

 

[Earl Karl]

In that case, I'll stick to dosing Lugols and Brightwell Potassium.

I'm using a dosing pump that does 0.1-6,000 mL per minute. I want to continuously dose throughout the day for maximum stability, something like a calcium reactor but without high initial cost. 2 part does not give me the stability I need and I get significantly slower growth. Which is why I am experimenting with Calcium Acetate, see if it makes a difference. I wanted to make my own solution since Calcium Acetate products are generally expensive for what they offer, but I can get things at low cost so I wanted to give it a try on my experiment tank.

If it is too potent, what is the max meq/L of Alk should I do for 1 gallon?

 

[Randy Holmes-Farley]

Just a caution, many dosing pumps are not rated for continuous service 24/7.

The max I’d use, assuming I could dissolve it, would be about the same potency as a two part, which for 3 dKH per day is about 200 mL per day if my two part. That seems like a good volume to dose spread over a day, but if you want a lower volume, more potency can do it.

 

[Earl Karl]

Ok thank you very much! I will lower the potency as the one I stated before is too much.

 

[RobB'z Reef]

Use the calculator linked in the stickies, and enter sodium bicarbonate. Subtract 2% if you are being way more exacting that is needed in a Reef context.

Calcium hydroxide will only dissolve to about 2 teaspoons per gallon, which is exactly 40.8 meq/l of alk at 25 deg C. Very little relative to what you wAnt to make.

hello, old post i know but I'm trying to educate myself on calcium/alk. I want a deeper understanding, not just to follow some recipe. I am struggling a bit with chemistry but I'll get there! What is the weight assumption in mg for 2 US tsp? It must be an approximation but close enough I'm sure. I'm trying to see how you get to 40.8mEq/l/gal mathematically. Near as i can figure mEq = (mass)(valence )/MW => .62 = (23,000mg)(2?)/74,000 => .62mEq/3.78l = not the right answer per gallon haha. I'm missing something fundamental here. Really appreciate your help @Randy Holmes-Farley ! Just wanted to add...I'm assuming all my answers/assumptions are wrong and yours are right

 

[chipmunkofdoom2]

hello, old post i know but I'm trying to educate myself on calcium/alk. I want a deeper understanding, not just to follow some recipe. I am struggling a bit with chemistry but I'll get there! What is the weight assumption in mg for 2 US tsp? It must be an approximation but close enough I'm sure. I'm trying to see how you get to 40.8mEq/l/gal mathematically. Near as i can figure mEq = (mass)(valence )/MW => .62 = (23,000mg)(2?)/74,000 => .62mEq/3.78l = not the right answer per gallon haha. I'm missing something fundamental here. Really appreciate your help @Randy Holmes-Farley ! Just wanted to add...I'm assuming all my answers/assumptions are wrong and yours are right

Have been thinking myself about this as well. Your equation for milliequivalents is correct. I'm not as sure about the calculations.

@Randy Holmes-Farley why is it that we express total alkalinity in meq/L, dKh, or PPM CaCO3 in the context of reef aquaria? With most other chemicals we care to measure, we express concentrations as mg/L or ppm. Calculating the concentration of a solution is very simple if you know the hydration and weight of the compound added. Valence does not seem to be important for these other elements.

 

[Randy Holmes-Farley]

The 40.8 meq/L is for saturated limewater at 25 deg C comes from the known solubility limit of calcium hydroxide in water, and with the knowledge that each mole of calcium hydroxide carries 2 moles of hydroxide, so the alkalinity is twice the molar concentration. It does not come from a teaspoon calculation which is crude and depends on the bulk density (see calcs below).

The solubility is about 1.59 g/L at 25 deg C (see data below)

https://www.lime.org/documents/lime_basics/lime-physical-chemical.pdf

The molecular weight of Ca(OH)2 is 74 g/mole
so 1.59 grams is 1.59 grams / 74 g/mole = 0.021 moles
Since there are two moles of hydroxide per mole of Ca(OH)2, the hydroxide concentration is 2 x 0.021 = 0.042 moles, and hence the alkalinity is 0.042 meq/L.

The teaspoon calculation below very roughly gets there, but is necessarily less accurate due to bulk density variability:

A US teaspoon has a volume of 4.93 mL.

The bulk density of calcium hydroxide varies, but is about 30 pound/cu ft:

https://www.lime.org/documents/lime_basics/lime-physical-chemical.pdf

Since 1 pound per cubic foot = 0.016 g/cm3, the bulk density is about 0.48 g/cm3

Thus, 1 teaspoon contains 4.93 mL x 0.48 g/mL = 2.4 g

The molecular weight of calcium hydroxide is 74 g/mole.

Thus, 2.4 grams = 2.4 g / 74 g/mole = 0.032 moles of calcium hydroxide

If you add two teaspoons to a gallon you are adding twice that (64 mmoles) to a gallon (3.79 L ) --> 64 mmoles/3.79 L = 16.9 mmoles/L

Each mole of calcium hydroxide contains 2 moles fo hydroxide, so the alkalintiy is twice the molarity, and the alk = 2 x 16.9 meq/L = 33.8 meq/L

 

[Randy Holmes-Farley]

Have been thinking myself about this as well. Your equation for milliequivalents is correct. I'm not as sure about the calculations.

@Randy Holmes-Farley why is it that we express total alkalinity in meq/L, dKh, or PPM CaCO3 in the context of reef aquaria? With most other chemicals we care to measure, we express concentrations as mg/L or ppm. Calculating the concentration of a solution is very simple if you know the hydration and weight of the compound added. Valence does not seem to be important for these other elements.

Because alkalinity is not a single thing, but the sum of many things with different molecular weights, weight based concentrations are not suitable. A scientist would use molar concentrations for alkalinity, and old time aquarists used dKH or ppm of an amount of calcium carbonate that would dissolve to give the same alk.

 

[RobB'z Reef]

The 40.8 meq/L is for saturated limewater at 25 deg C comes from the known solubility limit of calcium hydroxide in water, and with the knowledge that each mole of calcium hydroxide carries 2 moles of hydroxide, so the alkalinity is twice the molar concentration. It does not come from a teaspoon calculation which is crude and depends on the bulk density (see calcs below).

The solubility is about 1.59 g/L at 25 deg C (see data below)

https://www.lime.org/documents/lime_basics/lime-physical-chemical.pdf

The molecular weight of Ca(OH)2 is 74 g/mole
so 1.59 grams is 1.59 grams / 74 g/mole = 0.021 moles
Since there are two moles of hydroxide per mole of Ca(OH)2, the hydroxide concentration is 2 x 0.021 = 0.042 moles, and hence the alkalinity is 0.042 meq/L.

The teaspoon calculation below very roughly gets there, but is necessarily less accurate due to bulk density variability:

A US teaspoon has a volume of 4.93 mL.

The bulk density of calcium hydroxide varies, but is about 30 pound/cu ft:

https://www.lime.org/documents/lime_basics/lime-physical-chemical.pdf

Since 1 pound per cubic foot = 0.016 g/cm3, the bulk density is about 0.48 g/cm3

Thus, 1 teaspoon contains 4.93 mL x 0.48 g/mL = 2.4 g

The molecular weight of calcium hydroxide is 74 g/mole.

Thus, 2.4 grams = 2.4 g / 74 g/mole = 0.032 moles of calcium hydroxide

If you add two teaspoons to a gallon you are adding twice that (64 mmoles) to a gallon (3.79 L ) --> 64 mmoles/3.79 L = 16.9 mmoles/L

Each mole of calcium hydroxide contains 2 moles fo hydroxide, so the alkalintiy is twice the molarity, and the alk = 2 x 16.9 meq/L = 33.8 meq/L

thank you so much, the light bulb is starting to come on (slowly haha). I'm gathering mmoles and meq are equal is this context?