Ammonia solutions

[PolyEster]

Can someone help me?
I have a bottle of ammonia 25% But I need to know how to make up to the different molarities. Also have a couple of other bottles of Ammonia but they are all different percentages.
sorry if I sound dumb. But I just cant figure out how to work it out.
Gayle

[Robb]

Hi Gayle,

To begin, we need to know what the molarity is from the 25% Ammonia Solution. And we know that this is 13.3 Molar. Let’s use 13M. This way our calculations will be easier.

All you need to do then is this,

List what you know;

C₁ = Initial; Concentration

V₁ = Initial Volume BUT we will find this value out when we get the answer.

C₂ = Final Concentration, The concentration we need.

V₂ = Final Volume, The volume we need.

Equation to use

C₁ V₁ = C₂ V₂

To convert the equation because we need to know V₁, (It is as follows):

V₁ = C₂xV₂ Divided by C₁

Therefore:

Now we know this all we need to do is Multiply C₂ by V₂ then divide this by C₁ and the answer we get is how much we need of the 25% (13 Molar) Ammonia Solution we need to make up the volume we want. Maybe 2 Litres.

The Volume of the Volumetric Flask will be the volume you want. (Maybe you want to make up 2 Litres then V₂ will be 2000 or 2000mL)

Then we divide the answer to 13 x 2000 by the concentration we need Say 2M and this will give us (2 x 2000) divided by 13 this gives the answer of 307.69 or 308mL now all we need to do is take 308mLs of 25% Ammonia Solution add this to a 2L Volumetric Flask dilute it to the mark and you have 2L of 2M Ammonium Hydroxide Solution.

Just Remember you need to keep all volumes expressed in Millilitres because this will tell you in the end how much of the concentrated liquid you need to take in Millilitres. This makes it much easier because otherwise you have to convert back.

Once you have made up one concentration you can then substitute into the equation 2 Molar instead of 13 Molar because you have now made the 2M solution up and then dilute these again by 2000mL (2L) so the final thing to do is work out what your final concentration will be and calculate the answer again and you have the next diluted solution with the Molarity you wanted.

I hope I haven't confused the issue.

Robb....

[PolyEster]

Hi Robb
Thanks for the Reply on Ammonia. I think I can work out finally how to figure out the molarity now thanks to your explaination. Well I will give it a go anyway today.
Thanks Gayle

[fibreweb]

I wrote earlier today about my problems with Nitric acid solutions and my recipes, they pale into insignifigance compared to the numerous dilutions recipes I have for Ammonia. I havd recipes from several different sources, none seem to correlate or add up.

To make up a 1M solutions the figures I have are:
From a 10% solution I need 180 mls made up to 1 L
from an 18% solution I need 194 mls made up to 1 L
from a 25% solution I need 190 mls made up to 1 L and
from a 29% solution I need only 56 mls made up to 1 L

These don't seem to follow any logical pattern and I don't know which is correct. I have a bottle of 10% and one of 25% in my store cupboard so theye are the ones I really need to know.

If it was just molarity I would have no problem, but these percent solutions have me bamboozled.

Thanks
Wendy

[Robb]

Hi Wendy,

The Calculations listed above are consistent if you are using the same units. For example Liters and Molarity or Mililitres and Percantage.

It does not cover mixed units for example Molarity and Percentage.

What we need to do is convert the percentages into Molarity first, so this gives us something to work with, and to at least make things look more sensible for calculations.

Cheers,

Robb.....

[JudyM]

So does that mean !5% ammonia soln has a molarity of 7.98?

And does anyone know the concentration of "strong ammonia solution" - syn liquid ammonia forte?

Thanks in advance
Judy

[fibreweb]

A question for Robb to clarifyy if he would.

I have spent time this morning googling and punching figures into the calculator and have found these results, would you be able to tell me if they are correct before I go and write them up into " the Laboratory" as extra information as it only has 29% quantities.

From a variety of sources I have:

Ammonia 10% calculates about 5.62 M
Ammonia 25% calculates about 13.3 M
Ammonia 28% calculates about 14.8 M
Ammonia 35% calculates about 18.1 M

When I tried to work backwards the ratios of molarity/percentage was a bit different for each one but still all fairly close.

Are these figures accurate enough to put into writing?

Thanks in anticipation,

Wendy

[Robb]

Hi Wendy,

Firstly I would like to apologise for the time it has taken for a reply to you.. But no doubt you can see how busy I am with all things great and small...

I have perused through each of your Concentrations and I can say that you are on the ball as these are correct.. Just one thing though, that I have found that rounding off values should be left until the end.

That way you can get accurate results pretty close to the real value. When working with percentages and converting them, (Particularly here as per your results), the concentrations will be out by a small margin but not enough to cause any issues. (as probably you have found(and as you quoted in your previous post))..

Excellent!!
These results would not be an issue to have documented.

Cheers,

Robb.....

[fibreweb]

Hi Robb,

Thanks very much, I knew I could rely on you.

I have spent time this morning calculating the amounts of each concentration that is needed to make 1M solutions as that is what we use most often. (not that is used that often, essentially only with year 8 acids and bases). I will table them all and put them into my "bible" for future reference.

This is an invaluable resource!!!!!! Thanks once again.

Wendy

p.s I still don't totally understand how how 25% ammonia (which with other solutions as I understand is 25 gm/100ml therefor 250 gm/Litre) works out as 13.3 M. 250gms divided by the molecular weight of ammonia being 17 is 14.7 M. I accept that it is for some reason, but I don't know why. I have the same trouble with the acid concentrations.

Actually that is probably how someone else has worked it out and then changed all the figures in my home made recipe book.

[Robb]

Hi Wendy,

Keep in mind that concentration of Ammonia will be reduced at high concentrations as there is some loss due to fuming... We know that Ammoia is produced by the Haber Process.

But we must understand that Ammonia Solution is gas dissolved in water where the Ammonia is slightly soluble. This is why you get a kick when you open the lid.

Cheers,

Robb.....