I'm watching the algae and light, and will inject CO2 if it comes to require it. It will be easy to do since I'm using a canister filter.
NH4+ will turn into ammonia in the presence of acidic conditions, but either way I'm not terribly concerned about it because the nitrogen cycle will handle it, and plants are plenty happy to munch on whatever nitrogen source is available. I know there's no ammonia in KNO3, upon re-reading what I'd written in my original post it was slighly unclear what I was trying to do.
That said, I'm ultimately looking for the concentrations of KNO3 and NH4NO3 when in saturated aqueous solutions, so I know how much to safely dose.
At 50 degrees celsius or 122 degrees Fahrenheit, 100 grams of water can dissolve 84.00 grams of pure KNO3. Once that cools to ~24-25 degrees celsius the solution will remain the same. In essence you should have 184 grams of solution. The solution concentration will be 100:84, so for every 1 part water, there will be .84 parts KNO3.
Now with that said, in theory, 100 grams of water should equal around 100.43 ml. according to the density of water. The density of KNO3 is 2.109 grams/ CM^3. ( 1 cm^3 = 1 cc) 84g KNO3 * / 2.109 = 39.83 ml KNO3 (equivalent ). Therefore 100.43 ml h20 + 39.83 ml kno3 = 140.25 ml.
Every 1 ml of this solution, you should have 71.6% water and 28.4% KNO3. That 28.4% will equal .59g of KNO3. In other words 1 CC (ml) of that solution will contain the equivalent of .59G of KNO3. However, that solution will only stay supersaturated IF it is not disturbed, otherwise it will crystallize and become less concentrated.
Please correct me if I'm wrong, but I've checked it several times now and everything checks out. However, you can follow the same steps for a lesser concentration.
Is that confusing enough? 30 minutes later to figure that out.