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### Topic: Making an Ammonium Chloride Solution. How?  (Read 40989 times)

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#### JohnPQ

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #15 on: July 15, 2007, 06:41:55 PM »

Less than that. 38.5/(100+38.5)

Molar concentration to be precise.

The C*V = C2*V2 equation should apply to any kind of concentration, not just molar concentration.

#### Borek

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #16 on: July 15, 2007, 06:43:35 PM »
a tolerance of 10% means 10% higher(or lower?)

It means you need 4.5 ppm solution of ammonia, but if its concentration is 10% higher or 10% lower it is still OK.

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which means the actual error could be 1% (of weight?), within that 10% temp. tolerance range?

Between 0 and 40 deg C assumption that 1 mL of water weights 1 g never gives error larger than 1%. SO, if you calculate amount of water needed in this temparture range using this assumption, you may be off by 1% - and you can be off by 10% and you will be still OK. Thus this assumption is pretty safe to use.

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

When you dissolve 38.5 g of ammonium chloride in 100 g of water total weight of the solution is 100+38.5=138.5 g.

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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?

IMHO you may safely ignore it, just don't use formulas that use molar concentrations.

WHen it comes to dilutions and solution preparation you may take a look at my program CASC, although it will be most likely way too powerfull for your needs. However, I know that at least one aquarium proffesional bought it, but I believe he is dealing with many tanks and many solutions on the daily basis.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### Borek

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #17 on: July 15, 2007, 06:53:15 PM »

Molar concentration to be precise.

The C*V = C2*V2 equation should apply to any kind of concentration, not just molar concentration.

Let's assume you have 100 mL of 50% acetic acid and you dilute it to 1000 mL. What is the final concentration? Using C*V = C2*V2 you get 5%, in reality it is 5.25%. That's because density is concentration dependent. In most cases difference is negligible, but in general this approach is incorrect.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### BigJohnny

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #18 on: July 15, 2007, 06:53:52 PM »
do you have a trial version of your software? id love to give it a whirl, but cant afford $64 right now... plus im in canada, so its like$70

nevermind.. just found it lol.

#### Borek

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #19 on: July 15, 2007, 07:27:51 PM »
Just remember that trial doesn't have density table for NH4Cl. But it doesn't matter (much) - even assuiming density of all solutions is always 1g/mL you will be still below 10% error.
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#### JohnPQ

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #20 on: July 15, 2007, 07:31:37 PM »

Let's assume you have 100 mL of 50% acetic acid and you dilute it to 1000 mL. What is the final concentration? Using C*V = C2*V2 you get 5%, in reality it is 5.25%. That's because density is concentration dependent. In most cases difference is negligible, but in general this approach is incorrect.

OK this is little bit off-topic,

If you have 100 mL of 50% acetic acid, that mean you have 50g of acetic acid in solution?
if you have 1000mL of 5.25% acetic acid, then you have 5.25g of acetic acid in solution?

Where does the extra 0.25g came from???
That's all caused by change in density?

Do you mind explain concentration dependent of density? I have no clue what does that mean.

Thanks!

#### Borek

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #21 on: July 15, 2007, 07:50:16 PM »
If you have 100 mL of 50% acetic acid, that mean you have 50g of acetic acid in solution?

Nope. 50% acetic acid has density of 1.0575 g/mL, so mass of 100 mL is 105.75 g, 50% of that is acetic acid - 52.8 g.

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if you have 1000mL of 5.25% acetic acid, then you have 5.25g of acetic acid in solution?

5.25% acetic acid has density of 1.0059 g/mL, mass of 1000 mL is 1005.9 g, 5.25% of that is 52.8 g.

Note that exact calculation give error of about 0.07 g, that's from the rounding errors. Probably correct concentration should be 5.24 or 5.26 - or something like that, it is almost 2 a.m. here and I am too tired to play with numbers to find out.

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Where does the extra 0.25g came from???

From the assumption that solution density is constant regardless of the acetic acid concentration. It is not.

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That's all caused by change in density?

Yep.

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Do you mind explain concentration dependent of density? I have no clue what does that mean.

Every solution density depends on its concentration - in most cases knowing concentration you can find out (using tables) density, knowing denisty you may find out concentration.

For some substances it is not that easy. Acetic acidy density vs concentration curve has maximum at about 78%, and for example 60% and 92% solutions have the same density - knowing concentration you can find out density in tables, but knowing density you can't tell whether you have 60% or 92% solution.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

#### BigJohnny

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##### Re: Making an Ammonium Chloride Solution. How?
« Reply #22 on: July 15, 2007, 07:54:34 PM »
how could I use the trial version of the program to setup this formula so that I can print off the nessecary information?

I gotta be honest, I will probably only ever use it once, but it would be nice to have this all well documented for future reference.

you never know though, I could end up using the program alot more, if I knew how lol.
I dont know even half of the information that I would need to enter to have everything filled in properly.