I'm trying to make a solution of KNO3 from Ca(NO3)2 and K2SO4. I've only ever done high school chemistry, and that was five years ago so I really need someone to check this in case I screwed up horribly
K2SO4 + Ca(NO3)2 --> CaSO4 + 2KNO3
Room temperature here is always around 28-30*C so solubility for each is:
K2SO4: 0.13 g/mL
Ca(NO3)2: 1.52 g/mL
CaSO4: 0.00264 g/mL
KNO3: 0.48 g/mL
I used this site
to figure out how many grams of the reagents I need to get 50 g of KNO3 in solution.
K2SO4: 43.0894 g; needs at least 331.4572 mL to dissolve
Ca(NO3)2: 40.5743 g; needs at least 26.6936 mL to dissolve
So I'm thinking I'll just dissolve them in 400 mL of distilled water.
What I should get, theoretically, is 50 g of KNO3 in 400 mL (after I filter out the CaSO4). Which is 125000 mg/L, or 0.125 g/mL. Since 50 g of KNO3 only needs around 104 mL of water to dissolve, I guess I could evaporate it off to around 110 mL? (I wonder, would evaporation to dryness work with KNO3?) If I can evaporate it off it'd be 454545.45 mg/L or 0.454545 g/mL. But since I don't have any proper equipment to make sure I'm evaporating the right amt of water I'll just assume I won't evaporate it off for the rest of this:
My tank will have around 34-36 gallons of water in it. According to this calculator
, for EI and with 36 gal I should add 2.1 g of KNO3 per dose. Which should be 16.8 mL of the solution.
Is this right? Idk if dosing in solution works this way