[BigJohnny]

I'm getting some ammonium chloride for use with my aquarium, a large 500g bottle of powder.

For obvious reasons (to the chemistry inclined) I think it would be better to premix a batch of liquid solution for adding to the tank, instead of dumping the powder directly into the tank and hoping the crystals dissolve.

I have a few problems however. One of them is the concentration of the solution. According to some information found on the net, 1g is soluble in 3ml of water at room temperature, and 1.4ml at 100C (boiling).
If I mix the solution according to the room temperature formula, then Would that achieve a 100% solution? If I mix it in boiling water, then I would have a supersaturated solution more than 100%? or would that be the formula for making a 100% solution?

Secondly, the data I have found, has been conflicting, the above formulas were found on an elephant care site, and in an actual MSDS it lists the solubility as 29.7g/100g water @ 0C (32F). At least I think its conflicting... how do you convert water into grams?

When making the solution, I would obviously want to use pure water, either distilled or more preferably RO/DI.
I would like to heat the water so I can dissolve more powder in it, But pure water doesn't boil, and adding any "impurities" when the water has reached boiling point, will cause it to explode violently. so how would I add the powder to "boiling", pure water?

Basically I'm aiming for an aqueous ammonium chloride solution that is as pure/strong as possible.
What would be the proper procedure for accomplishing this task?

Thank you.

[enahs]

First, your question about grams-mL.

For water, 1 mL weighs a grams and 1 gram occupies a volume of 1 mL.
Now, that is not exactly 100% true, and it varies slightly with temperature. But unless you have a balance that can weight out to 4 decimal places and a graduated container in the 100ths of mL you are more then fine.


Yes you can make a supersaturated solution by heating it and dissolving more in it, but that would not be advisable. That would not store very well for very long, and you will probably find solid ammonium chloride precipitating out (forming solid in the bottom) over time, or just by accidentally shaking the bottle too much the excess will come out.


It would be best to make your solution at whatever temperature the aquarium will be at.

The best way to make the highest saturated solution at that temperature is to add a lot while stirring, and keep adding a little at a time until you are left with a little solid ammonium chloride that will not disolve (give it 10 minutes with stirring to see if it will disolve). At this point your solution is as saturated under normal conditions that it will be, filter off the unwanted solid (and a couple layers of coffee filters will get it mostly out).


And pure water does boil. Explosive boiling is a physical effect. This can happen with very impure tap water, just by choosing the wrong container. And, you will by no means have pure water, unless you want to spend $200 on a liter of water. And then as soon as you open the container it is not pure anymore!


Your question is then, what is the concentration of your solution, and how much should you add to your aquarium?

[BigJohnny]

I seen on mythbusters about pure water not boiling... then they added a spoon, and it exploded.
anyway, my only concern with that was that if I was going to boil the water and add the AC, it would then cause the water to "explode".
either way, doing it at aquarium temp sounds more stable in the long run. the temp is about 80F.

the pure water would be obtained by my RO/DI unit, which produces water with 0ppm TDS.

Either way, I want the result solution to be as strong as it possibly can be, 100% would be ideal.

The usage is for establishing a colony of nitrifying bacteria in the aquarium, which is started with ammonia in a "fishless cycle". as mentioned, adding the powder doesnt seem like an efficent way to introduce the ammonium chloride into my tank.

Ideally, to start the cycle, you need to have the ammonia level in the aquarium water around 5ppm.

The solution I want to make should be concentrated if possible, so that I dont need to add alot to the water, to get 4-5ppm of ammonia. IE, 4.5ml per 10 gallons of water.
I dont want to have to mix up, and dump a gallon of solution into my tank water to achieve the 5ppm mark.

by adding the powder to water at 80F, how strong can the resulting ammonia solution possibly be?

[enahs]

I seen on mythbusters about pure water not boiling... then they added a spoon, and it exploded.

I saw that one too. And when it comes to anything chemistry related, the Mythbusters are almost always way off, or at least, could do so much more with a solid basic chemistry knowledge. They did not have "pure" water either. It is quite easy to do in your microwave with tap water and certain types of containers; standard cheap ole chemistry beakers can do it quite easily as well.

Anyway.

by adding the powder to water at 80F, how strong can the resulting ammonia solution possibly be?

The solubility for ammonium chloride in water at 80F is ~38.5g/100mL. That is not exact because my CRC handbook listed in at every 10^oC, so I had to extrapolate to the 26 and 2/3 ^oC, but it is fairly close.

That will give you a solution of ~7.06 Mol/L.

That means to get a concentration of 5ppm in 10 gallons of water you would need to take ~0.49mL of your solution, or just say half a milliliter. You could easily destroy the complex chemical environment with too much of this stuff.

[BigJohnny]

Im a little lost, and I dont know what Mol(molarity) is.

Im not too concerned with how much solution I need to make to get 10 gallons to 5ppm . if I make a 100% solution (pure ammonia), then I just need to figure out how much of that solution I need to add to x amount of water to achieve 5ppm.

what complex chemical environment would I be destroying? The point of getting the aquarium to 5ppm of ammonia is to get the bacteria to start growing, and get the nitrification cycle started.
So I want a liquid solution of ammonia to add to the water.

But I need to know the strength of the solution I would be making, which will then allow me to figure out how much of that soloution needs to be added to the aquarium.

there are no fish/living things in the aquarium until the cycle is complete, and the ammonia in the water is broken down into nitrate.

[enahs]

Molarity is moles per liters.
http://dl.clackamas.edu/ch105-04/molarity.htm


I am not sure about the rest of your post. In my previous post I say how much you would need (half a milliliter) for a solution made by dissolving 38.5g Ammonium Chloride in 100mL of water at 80^oF (the roughly theoretical maximum), in order to make 10 gallons of water have 5ppm of Ammonium Chloride.

[JohnPQ]

BigJohnny,

I think what you want to do is impossible. You want to have a solution 100% am I right? But the problem is the maximum
solubility of ammonia chloride at 80 F is  ~38.5g/100mL as enahs had pointed out.

if we approximate that 1mL of water is 1g then 38.5g/100g
so 38.5% by weight is the maxiumum concentration you can make at 80F.

You can convert that to g/gal

38.5g/100 mL = 385g /1000 mL = 385g / L

385g/L * (1/0.2642)L/gal = 1457.22 g/gal

So the maximum solubility of ammonium cloride is roughtly 1457 g/gal at 80F,
it will be slightly less at lower temperature because the solubility of any kind of
salts will decrease at lower temperature.

If you dissolve all your ammonium chloride, you would need:

(500g) / ( 1457g/gal ) = 0.343 gal of water

After you made the solution of 1457 g/gal ,to find out how much you need to add to 10 gal tank
 
use the equation

C*V = C2*V2

C is concentration of solution you have.
V is the volume of concentration you are going to use.

C2 is concentration of new solution you need.
V2 is the volume of the new solution you need.

So in your case you need 5ppm in a 10 gal solution.

5 ppm = 5 mg/L = 0.005 g/L = 0.0189 g/gal

So,
C2 = 0.0189 g/gal (remeber 5ppm = 0.0189g/gal?)
V2 = 10 gal
C = 1457g/gal

we can rearrange the equation as

V = (c2*V2)/c

which come out to 0.0001297 gal, a ridiculous amount, I highly doubt you could accurately measure that, but miraculously if you convert  0.0001297 gal to mL it came out as ~0.49 ml, roughly half a mL as enahs had pointed out.

So I doubt you would want to use concentrated solution at 1457 g/gal it's too easy to add too much. You might want to dilute it further before use.

OK it's 2 in the morning, I don't know if anything I just wrote make any sense. Good luck!

[BigJohnny]

I cant thank you guys enough, And I know I have a BAD habit of making things sound more complicated than they really are.

JohnPQ . You have pretty much answered my entire question, thank you. All of the info in that post of yours was exactly what I was looking for.... I needed some numbers.

I was thinking too large really, I figured you could make like gallons and gallons of solution from the ammonium chloride salts, But with whats been put in front of me I can see now, that is completely wrong.

So with the amounts of solution needed to bring the tank water up, I think I would only need to make about 1L of the solution, not even(If I was going to try and make a gallon, It would require 2 more bottles of ammonium chloride lol)

When I make this stuff up, What is the resulting product? Liquid Ammonium Chloride correct? Or will it just be plain ammonia (same type of stuff you buy at the grocery store for cleaning.... except that they use ammonium hydroxide) either way, im just asking for the sake of knowing what I've got.

Here is some info on the ammonium chloride I've purchased:

Albion Chemicals
Pure laboratory grade chemical reagent
No anti-caking agents
nh4ci=53.49
assay 99%min  (What does this mean??)
nitrate 0.0005%
phosphate 0.001%
sulphate 0.01%
heavy metals 0.0004%


Now for some stupid questions, That I am simply curious about.

Is there any chance that this solution will turn into something else in the presence of other chemicals, sodium thiosulfate for example (a popular compound used in water conditioners for dechlorinating water) or will result in the production of chlorine or other substances from the chloride?

What is the strength of the stuff you would normally buy at a store? Its no 38.5%.. is 38.5% a "big" number in this context?
how would I determine the molarity of this solution I have made?

I cant thank you guys enough for the help, and tolerance to my chemistry n00bness
Im not completely clueless, but I was a little stumped here.. and definetly was using the term "100%" in the wrong context.

[BigJohnny]

Also, about the soluability mentioned in my first post.

Can anyone tell me if this data is conflicting, or if it is the same, just in different scales/measurements.

29.7g/100g water @ 0C
1g/3ml water @ 20C
1g/1.4ml water @ 100C

or really what im asking here is, how many g per 100ml(g) water @ 0C, and what is the forumla used for down/up converting those numbers, so that they are the same scale as the other 2? or 1g/x ml @ 0C ??

[Borek]

if we approximate that 1mL of water is 1g then 38.5g/100g so 38.5% by weight is the maxiumum concentration you can make at 80F

Less than that. 38.5/(100+38.5)

use the equation

C*V = C2*V2

C is concentration of solution you have.

Molar concentration to be precise.

[Borek]

29.7g/100g water @ 0C
1g/3ml water @ 20C
1g/1.4ml water @ 100C

No way to calculate salt solubilities for different temperatures, the only reliable way of checking sulibitiy is by experiment. My handbook gives:

29.4 @ 0C
33.3 @ 10C
37.2 @ 20C
41.4 @ 30C
45.8 @ 40C
50.4 @ 50C
55.2 @ 60C
60.2 @ 70C
65.6 @ 80C
71.3 @ 90C
77.3 @ 100C
(g/100 g H₂O)

Data you listed seems consistent with these numbers (that's assuming 1mL of water is 1 g of water regardless of the temperature - at the precision you need that's pretty good assumption, although it is not correct in general).

[BigJohnny]

What I was trying to get at, is how many ml of water will 1g of ammonium chloride dissolve into, @ 0C ? I think I asked the oppisite of what I really wanted to know.

Also what do you mean is not correct in general.
just for curiousitys sake, how accurate is what I'm doing?

I want to ensure that anything left over to be stored, will be properly labelled with any information I may need/forget in the future.

this includes accuracy, If I label the product as 38.5%, I would like to also include the accuracy in terms of what percentage I may be off by.

your comment about "Less than that. 38.5/(100+38.5)" means what?
If I do that math (rather if I do it correctly... math is not my strong point) then I end up with.
38.5/138.5 = 0.27797833935018050541516245487365
im totally clueless as to what that means

and as I mentioned, I have no idea how to find the molarity of a substance, as well as not even knowing what a mole is.. so I would love to learn a bit more about it.

[BigJohnny]

also, will this solution have any sort of reaction at all, when added to water containing sodium thiosulfate and/or hydrosulfite salts?

[Borek]

What I was trying to get at, is how many ml of water will 1g of ammonium chloride dissolve into, @ 0C?

29.4g of NH₄Cl dissolves in 100 g of water. Assuming 100 g of water is 100 mL, you need 100/29.4 = 2.54 mL.

Also what do you mean is not correct in general. just for curiousitys sake, how accurate is what I'm doing?

Mass of 1mL of water:

 0 deg C - 0.99987 g
 4 deg C - 1.00000 g
10 deg C - 0.99973 g
20 deg C - 0.99913 g
25 deg C - 0.99707 g
40 deg C - 0.99224 g
50 deg C - 0.98807 g
100 deg C - 0.95838 g

As you can see up to 40 deg C error is less than 1%. You need 4 to 5 ppm of ammonia, in other words 4.5 ppm +/- 10% will do. You are making 1% error when tolerance is 10% - ten times more. No problem at all.

your comment about "Less than that. 38.5/(100+38.5)" means what?
If I do that math (rather if I do it correctly... math is not my strong point) then I end up with.
38.5/138.5 = 0.27797833935018050541516245487365
im totally clueless as to what that means

It means that the real concentration is 27.8%. Percentage concentration is (mass of solute/mass of solution)*100% - mass of solute is 38.5, mass of solution is 138.5.

http://www.chembuddy.com/?left=concentration&right=mass-percentage

and as I mentioned, I have no idea how to find the molarity of a substance, as well as not even knowing what a mole is.. so I would love to learn a bit more about it.

http://en.wikipedia.org/wiki/Mole_(unit)

In general - mole is not different from dozen, it just uses differnet number - there are 12 objects in doze, but 6.02*10²³ in mole. You may have dozen marbles, dozen atoms, dozen molecules, as well as mole of atoms, mole of molecules or mole of marbles. Trick is, dozen of marbles fits easily in the pocket, but mole of marbles weight is about the same as Earth mass. On the other hand, mole of molecues of ammonium chloride weight 53.49 g, while mass of dozen of molecules is so low you can't weight them without pretty fancy methods and apparatus.

Molar mass is a mass of mole of substance. Molar concentration is number of moles of substance per 1 L of solution.

http://www.chembuddy.com/?left=concentration&right=molarity

[BigJohnny]

thanks for the response/info.

As you can see up to 40 deg C error is less than 1%. You need 4 to 5 ppm of ammonia, in other words 4.5 ppm +/- 10% will do. You are making 1% error when tolerance is 10% - ten times more. No problem at all.

This sort of confuses me, perhaps its just the wording.
a tolerance of 10% means 10% higher(or lower?) than 4C, where water is 1g?
which means the actual error could be 1% (of weight?), within that 10% temp. tolerance range?

Weight of solution.... 138.5 what? not fully clear on this term.

That molarity stuff seems a bit over my head, mostly just because I have a hard time trying to pick stuff up simply by reading about it. I dont learn well that way.... but anyway, Is it something I even need to bother worrying about with this application anyway?
Or am I just being too anal?