OP
OP
Randy Holmes-Farley
Reef Chemist
Staff member
Super Moderator
Excellence Award
Expert Contributor
Article Contributor
R2R Research
My Tank Thread
- Joined
- Sep 5, 2014
- Messages
- 67,349
- Reaction score
- 63,689
Let's look at magnesium incorporation, and how that might impact the answer. We'll start with a "worst case" analysis, usign the organisms that incorporates the most magnesium that I am aware of:
Calcareous alga Corallina pilulifera
Assessment of calcareous alga Corallina pilulifera as elemental provider
https://www.sciencedirect.com/science/article/pii/S0961953498000841
"The major elemental composition in C. pilulifera is as follows: sodium 0.13%, chloride 1.75%, magnesium 4.37%, calcium 18.4%, iron 0.31%, and carbonate 28.5%."
Assume we add 1000 mg/L of CaCl2. That doses 639 mg/L of chloride and 361 mg/l calcium.
If we incorporate all of that into calcium carbonate, then by the above paper, we would also incorporate 4.375/18.4 x 361 mg/l = 85.8 mg/L magnesium.
To balance the Ca (361 mg/L, 9.01 mmolar) and magnesium (85.8 mg/L, 3.53 mmolar) we need 9.01+ 3.53 = 12.54 mmolar carbonate, and hence, 2 x 12.54 = 25.08 mmolar sodium (since there are two sodium ions per carbonate in sodium carbonate). That equals 577 mg/L sodium.
I chose to base all the subsequent calculations on the amount of sodium added because it is the limiting element. No additional sodium is needed in any possible NaCl free salt mix, while, as noted earlier, you do need more chloride and cannot base calcualtions off the chloride added.
So we have 577 mg/L of sodium added. How much magnesium and how much sulfate are needed to exactly offset that sodium addition to leave a seawater ratio?
NSW has 10,800 mg/L sodium, 2700 mg/l sulfate, and 1280 mg/L magnesium (roughly). We will use this ratio.
Using 577 mg/L sodium, and matching that ratio, we must add
577/10,800 x 2700 mg/L of sulfate = 144 mg/L of sulfate.
577/10,800 x 1280 mg/L of magnesium = 68.4 mg/L of magnesium.
That, of course, is the answer when no magneisum incorporation is account for: more sulfate is needed by about a factor of two.
BUT, we also incorporated magnesium into the calcium carbonate, depleting it from the water and leaving a less than NSW ratio of magnesium. If we really want a NSW residue, we must add more of it, and we calculated above we have to add and ADDITIONAL 85.8 mg/l of magnesium.
That incorporation means that to provide a true NSW ratio, we need to add a total of 85.8 mg/L plus 68.4 mg/L = 154.2 mg/L of magnesium.
That makes it the largest component on the list of ions that I posed in this question.
in a real reef tank, is the magnesium incorporation that high? Almost certainly not. Unless nearly all incorporation is by coralline algae, the ratio is likely to be much lower. Sulfate is still likely to big winner here in a reef application, but it shows that the intent of the product could impact the answer. [emoji4]
Calcareous alga Corallina pilulifera
Assessment of calcareous alga Corallina pilulifera as elemental provider
https://www.sciencedirect.com/science/article/pii/S0961953498000841
"The major elemental composition in C. pilulifera is as follows: sodium 0.13%, chloride 1.75%, magnesium 4.37%, calcium 18.4%, iron 0.31%, and carbonate 28.5%."
Assume we add 1000 mg/L of CaCl2. That doses 639 mg/L of chloride and 361 mg/l calcium.
If we incorporate all of that into calcium carbonate, then by the above paper, we would also incorporate 4.375/18.4 x 361 mg/l = 85.8 mg/L magnesium.
To balance the Ca (361 mg/L, 9.01 mmolar) and magnesium (85.8 mg/L, 3.53 mmolar) we need 9.01+ 3.53 = 12.54 mmolar carbonate, and hence, 2 x 12.54 = 25.08 mmolar sodium (since there are two sodium ions per carbonate in sodium carbonate). That equals 577 mg/L sodium.
I chose to base all the subsequent calculations on the amount of sodium added because it is the limiting element. No additional sodium is needed in any possible NaCl free salt mix, while, as noted earlier, you do need more chloride and cannot base calcualtions off the chloride added.
So we have 577 mg/L of sodium added. How much magnesium and how much sulfate are needed to exactly offset that sodium addition to leave a seawater ratio?
NSW has 10,800 mg/L sodium, 2700 mg/l sulfate, and 1280 mg/L magnesium (roughly). We will use this ratio.
Using 577 mg/L sodium, and matching that ratio, we must add
577/10,800 x 2700 mg/L of sulfate = 144 mg/L of sulfate.
577/10,800 x 1280 mg/L of magnesium = 68.4 mg/L of magnesium.
That, of course, is the answer when no magneisum incorporation is account for: more sulfate is needed by about a factor of two.
BUT, we also incorporated magnesium into the calcium carbonate, depleting it from the water and leaving a less than NSW ratio of magnesium. If we really want a NSW residue, we must add more of it, and we calculated above we have to add and ADDITIONAL 85.8 mg/l of magnesium.
That incorporation means that to provide a true NSW ratio, we need to add a total of 85.8 mg/L plus 68.4 mg/L = 154.2 mg/L of magnesium.
That makes it the largest component on the list of ions that I posed in this question.
in a real reef tank, is the magnesium incorporation that high? Almost certainly not. Unless nearly all incorporation is by coralline algae, the ratio is likely to be much lower. Sulfate is still likely to big winner here in a reef application, but it shows that the intent of the product could impact the answer. [emoji4]
Last edited:
