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What do I get when I add Ammonium Chloride to water?

Edward Shave

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27 Aug 2018
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Kirk Langley, Derbyshire
So I have purchased a quantity of 99.5% Ammonium Chloride. I'm thinking I will end up with a total ammonia solution if I add some to distilled water.
Now it comes to doing it I thought I might just check here first as I'm no chemist.
I want to use the resulting ammonia to test Seachem's Ammonia Alert.
 
Hi all,
It dissociates into its two component ions: you get ammonium NH4+ and chloride Cl-
It is back to the <"solubility rules chart">, nearly all chlorides are soluble. Whether the ammonium ion (NH4+) remains in solution, or becomes de-ionised to ammonia gas (NH3) <"is dependent on pH">.

=NH3-NH4_equlibrium.png

cheers Darrel
 
It dissociates into its two component ions: you get ammonium NH4+ and chloride Cl- :)
I saw mention of chloride when googling and stupidly misread it as chlorine so thought it would gas out leaving just the ammonia.
Can I ignore the chloride content or will it distort my ammonia test kit results?

Just to be clear I'm carrying out these experiments in just a tank of water. no fish are involved.
 
Can I ignore the chloride content or will it distort my ammonia test kit results?
Might be better to get some household ammonia and use that ?

Just check ingredients to confirm only ammonia and its strength.

I have some from Homebase ammonia, put dirty oven shelves in a poly bag, add ammonia, shake, leave overnight outside and by the morning the grease and burnt on gunk just falls off.
 
Hi all,
Can I ignore the chloride content or will it distort my ammonia test kit results?
You can ignore it in terms of the test, but you need to take it into account when you work out the percentage of ammonia in the ammonium chloride (NH4Cl).
Might be better to get some household ammonia and use that ?
You can use either, you just need to know the % of ammonia/ammonium.

Ammonium chloride is 33. 6% NH4
(The workings are the RAMs of N = 14, H = 1 and Cl = 35.5 and 14 + (4*1) + 35.5 = 53.5 and NH4 = 18 and 18/53.3 = 0.336)

I want to use the resulting ammonia to test Seachem's Ammonia Alert.
Because you want a very weak solution of ammonia/ammonium you are best using <"serial dilution"> to get the right strength of solution to test.

Therefore when you've got 2.97g of NH4Cl you have 1g of NH4+ (1/0.336). The units "mg/L" and "ppm" are equivalent and there are 100mg in 1g, so if you make up 2.97g of NH4Cl to 1000g with RO water you have a 1000ppm solution of NH4+ / NH3 (dependent upon pH). This is your "stock solution".

If you take 100 g of the stock solution and make this up to 1000g (1000 mL) you have a 100 ppm solution ("solution 1") and you can dilute "solution 1" further to give solution strengths more applicable to aquariums.

cheers Darrel
 
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