Trying to make stock solution of Ammonium Chloride for my 60gallon

reefluvrr

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Hi Randy,

Am I doing something wrong with my math?

I want to make 1L of NH3 where I can dose 5ml into my 60 gallon tank providing about 1.5mg/L (1.5ppm) of NH3 for the tank.

Here is my math:

add 200mg of NH4Cl which has 32% NH3 so 200mg x 0.32=64 g/L

If I put the whole 1L of solution in my tank my NH3 level should be:

64g/L x 1gal/3.785L x 1/60 gallon=0.284g/L or 284mg/L of NH3 in my tank

Now if I divide 284mg/L by 1000mL, I would get 0.284mg/mL concentration.

Does this look right?

For every 1ml of solution I add into my 60 gallon tank, I would be 0.284mg of NH3 in my tank?

Sorry, haven't done this kind of math in over 30 years....

Thank you.
 

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Malcontent

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Thanks Malcontent,

Sorry, I am not sure what to do with the numbers though....

I think you're off by 1000.

1.5 mg NH3/L * 60 gal = 1.5 mg NH3/L * 3.146471 mg NH4Cl / mg NH3 = 4.7197 mg NH4Cl/L
4.7197 mg NH4Cl/L * 227.1247 L = 1071.962 mg NH4Cl
1071.962 mg NH4Cl / 5 mL = 214.392 mg/mL = 214 g/L
 
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reefluvrr

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I think you're off by 1000.

1.5 mg NH3/L * 60 gal = 1.5 mg NH3/L * 3.146471 mg NH4Cl / mg NH3 = 4.7197 mg NH4Cl/L
4.7197 mg NH4Cl/L * 227.1247 L = 1071.962 mg NH4Cl
1071.962 mg NH4Cl / 5 mL = 214.392 mg/mL = 214 g/L
Okay,

Can you help me understand why I want to multiply by 227.1247L (60gal)? If I put 4.7197mg NH4CL into 60 gallons, wouldn't that dilute the total (4.7197/227.1247=0.0278mg/L which means I have 0.0278mg/L of NH4CL in 60 gallons of water?)


Thanks for spending time to help me out on this!
 

Courtney Aldrich

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Here's my calculation.

you want to add 1.5 ppm or 1.5 mg/L of NH3. Your tank is 60 gallons or 227 L, thus you need 1.5 mg/L x 227 L = 341 mg of NH3. Now you want to make a stock solution that adds this amount in 5 mL, so your stock solution should be 341 mg/5 mL = 68 mg/mL. To make 1 L (or 1000 mL), you would add 68 mg/mL x 1000 mL = 68,000 mg = 68 grams of NH3. You have NH4·Cl, so you need 68 grams NH3 x 53.5 g NH4Cl/17.03 g NH3 = 213 grams of NH4Cl. In summary, add 213 grams of ammonium chloride to 1 L of water to prepare a stock solution of 68 mg/mL of NH3.

I would check your tank volume as the listed size might be 60 gallons, but the actual volume minus rocks/sand is likely smaller. For example, I bought a 20 long, but it only holds a little less than 18 gallons of water with nothing else in it, when filled to the brim. I think, a good rule of thumb is to subtract about 15% from your volume to account for the overstated sizes of aquariums and displaced volume from rocks/substrate. If you agree, then you will have to adjust the amounts above. Thus 213 grams x 0.85 = 181 grams of ammonium chloride to 1 L.
 
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reefluvrr

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Here's my calculation.

you want to add 1.5 ppm or 1.5 mg/L of NH3. Your tank is 60 gallons or 227 L, thus you need 1.5 mg/L x 227 L = 341 mg of NH3. Now you want to make a stock solution that adds this amount in 5 mL, so your stock solution should be 341 mg/5 mL = 68 mg/mL. To make 1 L (or 1000 mL), you would add 68 mg/mL x 1000 mL = 68,000 mg = 68 grams of NH3. You have NH4·Cl, so you need 68 grams NH3 x 53.5 g NH4Cl/17.03 g NH3 = 213 grams of NH4Cl. In summary, add 213 grams of ammonium chloride to 1 L of water to prepare a stock solution of 68 mg/mL of NH3.

I would check your tank volume as the listed size might be 60 gallons, but the actual volume minus rocks/sand is likely smaller. For example, I bought a 20 long, but it only holds a little less than 18 gallons of water with nothing else in it, when filled to the brim. I think, a good rule of thumb is to subtract about 15% from your volume to account for the overstated sizes of aquariums and displaced volume from rocks/substrate. If you agree, then you will have to adjust the amounts above. Thus 213 grams x 0.85 = 181 grams of ammonium chloride to 1 L.
Thank you very much for walking through the process with me.

Without helpful people in this Reef2Reef community, I may have never have figured this out. Old brain is definitely rusty!

I agree with you about 15% less volume, so 181g sounds right.
 

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