Calcium Nitrate question

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I know a few people have used Potassium Nitrate as a nitrate additive. Would Calcium Nitrate fall in the same boat with sodium nitrate, strontium nitrate, and pot nit? Or is there any unknown issues with CaNo3? Randy?
 
All of these nitrates should be fine: sodium, potassium, calcium, magnesium.

I probably wouldn't use strontium nitrate as it may push strontium too high. Not only is strontium by far the lowest of these in seawater, it is the largest percentage metal in the nitrate.
 
Assuming you have a scale, calcium nitrate is 76% nitrate by weight.

Dissolve 10 grams in 1 L of water, and that solution is 10,000 mg * 0.76 / 1 L = 7,600 ppm.

Adding 1 mL of that solution per liter of tank water will boost nitrate by 7,600/1000 = 7.6 ppm.

So 1 ml per 8 L adds about 1 ppm. :)
 
That's an odd one that one intentionally contains ammonia. It wouldn't have been my choice since it isn't labeled for purity and intentionally has ammonia. You can likely use it, but add very small doses at a time. Not more than 0.2 ppm nitrate, IMO.

I would assume a level teaspoon of it is about 5 grams, so for the stock solution I suggested, use 2 level teaspoons per 1 L of water.
 
@Randy Holmes-Farley

Correct me if I'm wrong please

10g to 1000ml rodi = 47ml dose to add 1 ppm in a 100gal

If I reduce to 250ml of rodi and 10g I should be around 11.75 ml to add 1 pmm in 100 g


Using

Screenshot_20201108-223852_Chrome.jpg
 
@Randy Holmes-Farley

Correct me if I'm wrong please

10g to 1000ml rodi = 47ml dose to add 1 ppm in a 100gal

If I reduce to 250ml of rodi and 10g I should be around 11.75 ml to add 1 pmm in 100 g


Using

Screenshot_20201108-223852_Chrome.jpg

No quite.

I do not know if that product is adequately pure or not, but assuming so, it is about 53% nitrate by weight (it is not anhydrous calcium nitrate).

10 grams contains about 5,300 mg of nitrate.

Dissolve that in 250 mL, then it is 5300 mg/250 mL = 21 mg/mL.

100 gallons is 378 L, so to add 1 ppm to 100 gallons, you need to add 378 mg.

378 mg /21 mg/mL = 18 mL.
 

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