I have a small problem (I am not a chemist) - for various reasons we ended up with a 3,000L water tank that has 114kg of ammonia dissolved in it (ammonia as anhydrous before it got dumped into the tank). It is a pain in the ass because no legitimate hazardous waste disposal company is willing to touch it as is. It is no good for agricultural use either because it may contain some oil from a refrigeration system as well. My initial thought is to neutralize it with either CO2 or Citric Acid. CO2 is not great because I think the reaction will create an insoluble precipitate (ammonium bicarbonate), which would muck up the bottom of the tank. I wanted to look into citric acid, but I need to double check how much I might need if we go down this route and if it was an okay idea. I think the salt of the reaction would be soluble, which is good for us.Assuming 114 kg of NH3Molecular Weight: 17.0367g/mol
Kb = 1.78E-5
114,000g * 1/17.0367 * 1/3000 = 2.23mol/L
x = [OH-]
Kb = x^2/(2.23-x)
Assume x is fairly small
Kb = x^2/2.23
x = 0.0063 mol/L
pOH = -log(0.0063) = 2.2
pH - 14 - 2.2 = 11.8 -> too high for hazardous waste
I have 2.23 * 3000 = 6,691 moles of NH3
Citric acid can donate three H+
3NH3 + CA = H2O + Salt
So I need 6,691 mol / 3 = 2,230 moles of citric acid
Citric acid molecular weight = 192.124 g/mol
2,230 * 192.124 = 428.5 kg of pure citric acid?
Sorry about the lack of correct significant figures...
I didn't find the right solution from the internet.