ok, the drop thing was a little blasé, but it is a pretty safe (and cheap) chemical.
really, this should be done on molar concentrations, but as it is safe enough and to keep it simple... we'll just stick with ppm.
For every 1 ppm of NH4 you need 5 ppm of H2O2 to completely oxidise it to NO3.
The formula you use to calculate it is a simple one, V1 = C2V2/C1
V1 = Volume of H2O2 required.
C2 = Concentration of H2O2 needed to neutralise the ammonia (5:1 as above)
V2 = Volume of your tank/system.
C1 = Concentration of H2O2 in the bottle.
So, if you have 5ppm NH4, you will need 25ppm of H2O2 to neutralise it.
Say you have a
20L tank.
Say your pharmaceutical/food grade (not cosmetic grade) H2O2 is 3%. (convert % to ppm by multiplying by 10,000, so 3% = 30,000ppm)
V1 = (unknown in L)
C2 = 25ppm
V2 =
20LC1 = 30,000ppm
V1 = 25x20/30,000
V1 =
0.016L (note! that if V2 was in L, V1 will be too! times the answer by 1,000 to convert to mL)
V1 = 16mL
So, to neutralise 5ppm of NH4 in a
25L tank, you need
at least 16mL of 3% H2O2.
I say at least because NH4 is not the only ion or organic matter that will demand (use/deplete) H2O2; bacteria, algae, chlorine, iron...
If all that is too complicated, add a mL at a time till no ammonia is present, it is a rapid reaction, mix the water, test, repeat till gone.
Amquel is sodium formaldehyde bisulphite, it works by a different mechanism, I'm not sure but I think it reacts as such;
2 CH3NaO4S + 2 NH4 = 2 CH3O3SNH4 + 2 Na + O2
At the pH's we typically deal with it's ammonium NH4, however the peroxide still works, as below;
2 NH3 + 9 H2O2 = 2 NO3 + 12 H2O (so you need 4.5 mol of H2O2 to do the job).
I don't know how bioavailable the bound NH4 (or NO2/NO3) will be, my prefererance is to use the peroxide as my aim is to simply rectify the ammonia and allow the biological filtration to deal with the nitrate.