I am doing a fishless cycle on my 10G. Approximately how much ammonia should I put in to get to the 5ppm. I tried doing a few dilution on my own, but nothing seems to be working. I keep getting very high ammonia levels
ammonia levels
- Thread starter gersheff
- Start date
It's hard to say... there seems to be some variation in the concentration of ammonia available commercially.
For a smallish tank, try 4-6 drops, then test. If it doesn't register, double it and try again. You should be able to creep up to the 4-6 ppm needed for fishless cycling.
Good luck,
Jim
For a smallish tank, try 4-6 drops, then test. If it doesn't register, double it and try again. You should be able to creep up to the 4-6 ppm needed for fishless cycling.
Good luck,
Jim
"PPM" is a very poor unit of measure, desipite its wide use. Most of the time this is a mass/volume measure, but it's never clear. On the test kits, I think it measures mg/L.
If you want 5 ppm NH3 in your tank, let's do a simple calculation.
The density of pure liquid NH3 is 0.682 g/cc, or 682 mg/cc.
10 gallons * 3.7854 L / gallon = ~38 L tank.
38 L * 5 mg NH3/L= 190 mg NH3 in your tank.
Now divide by the density
190 mg / 682 (mg / cc) = 0.2786 cc of pure NH3 in your tank to get to 5 ppm.
Drop size varies, so take some AMMONIA and see how many drops it takes to fill up a teaspoon. Let's say it's 20. Then you have 20 (drops/teaspoon) / 4.928 (mL/teaspoon) = ~ 4 drops/mL.
Now you have 0.2786 mL * 4 drops /mL ~ 1 drop!
This is for pure ammonia. You will have to recalculate if you're using comercially diluted ammonia. Find out the concentration of your ammonia solution. If it's 50%, multiply the number of drops by 2 for a decent approximation.
If you want 5 ppm NH3 in your tank, let's do a simple calculation.
The density of pure liquid NH3 is 0.682 g/cc, or 682 mg/cc.
10 gallons * 3.7854 L / gallon = ~38 L tank.
38 L * 5 mg NH3/L= 190 mg NH3 in your tank.
Now divide by the density
190 mg / 682 (mg / cc) = 0.2786 cc of pure NH3 in your tank to get to 5 ppm.
Drop size varies, so take some AMMONIA and see how many drops it takes to fill up a teaspoon. Let's say it's 20. Then you have 20 (drops/teaspoon) / 4.928 (mL/teaspoon) = ~ 4 drops/mL.
Now you have 0.2786 mL * 4 drops /mL ~ 1 drop!
This is for pure ammonia. You will have to recalculate if you're using comercially diluted ammonia. Find out the concentration of your ammonia solution. If it's 50%, multiply the number of drops by 2 for a decent approximation.
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When I did a fishless cycle on a ten gallon, I used an eydropper and I put in 70 to 80 drops of Great Value brand Clear Ammonia to bring it up to 4 ppm. When I used a more concentrated brand of ammonia, I only put in 30 drops to bring it up to4 ppm. HTH
It looks as if commercially available ammonia is about 1 M, or 17 mg/mL, very very weak stuff.
At 1 M, you'll need 190 mg / 17 mg/ml = 11 mL or about 2 full teaspoons to get it going.
You can see the huge difference between pure ammonia and this comercially available stuff. It's critical to know the concentration. That's why there's such a wide variety of numbers flying on these forums.
At 1 M, you'll need 190 mg / 17 mg/ml = 11 mL or about 2 full teaspoons to get it going.
You can see the huge difference between pure ammonia and this comercially available stuff. It's critical to know the concentration. That's why there's such a wide variety of numbers flying on these forums.
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I am trying to get a hold of the company, with no luck. All i know is that the stuff smells pretty strong.
Ammonium hydroxide is a liquid about the same density of water. It's actually the same thing as ammonia in water since
NH3 + H20 --> NH40H
So if we use 950 mg/mL as the density, at 5% ammonia we have 47.5 mg/mL of NH3, assuming percent by mass is about the same as percent by volume, since the densities aren't too different.
That's 4 mL of your solution to make the 190 mg you need for 5 ppm. (a little less than 1 tsp).
NH3 + H20 --> NH40H
So if we use 950 mg/mL as the density, at 5% ammonia we have 47.5 mg/mL of NH3, assuming percent by mass is about the same as percent by volume, since the densities aren't too different.
That's 4 mL of your solution to make the 190 mg you need for 5 ppm. (a little less than 1 tsp).
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