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Topic: Solutions and dilutions  (Read 2267 times)

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Offline Jojobuller

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Solutions and dilutions
« on: July 11, 2016, 09:11:36 PM »
Hi everyone i'm new here.

My name is joseph i'm 21 and im from australia.

My highest level of schooling is 8th grade due to my mental health. so im trying to teach myself all i can.

I really love fish and aquaria and thats why im coming to you for help. I understand the nitrogen cycle. and i use that principle everday.

What i want to know is, I buy Ammonium Chloride solution (which is VERY expensive). I contacted a wholesaler who makes the ammonum chloride solution i use and they sold me some of the Ammonium Chloride powder.

I would like to know how i make the correct solution from the powder.

On the bottle of powder it says to add 0.5g to 300 litres of water to obtain 4mg/l of ammonia in the aquarium water.

I would like to make a 500ml solution that when you add 1ml to 38 litres it creates 2mg/l of ammonia


Thanks so much in advance. Id also appreciate it if you explained how you work this out.

Offline AWK

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Re: Solutions and dilutions
« Reply #1 on: July 11, 2016, 10:58:37 PM »
Quote
On the bottle of powder it says to add 0.5g to 300 litres of water to obtain 4mg/l of ammonia in the aquarium water
This gives you 1.67 mg of NH4Cl per liter. This means you cannot obtain 4 mg of ammonia per liter.

http://www.chemicalforums.com/index.php?topic=65859.0
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Offline Borek

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Re: Solutions and dilutions
« Reply #2 on: July 12, 2016, 02:55:02 AM »
There are several steps required and several different ideas.

First, you need to calculate how much ammonia is added to the solution when you add 1 g of ammonium chloride. This is relatively easy to calculate using molar masses of ammonia and ammonium chloride.

Second, you need to calculate dilutions. These calculations are basically based on the mass conservation - whatever is put into the solutions, stays there. So, if you need a 38 L of a solution that contains 2 mg/L of ammonia, you need 38 L×2 mg/L = 76 mg of ammonia, and you want it to be present in 1 mL of the solution.
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