[j4vier]

Hi everyone,

I am trying to prepare a 50 mM ammonium acetate buffer from acetic acid and ammonium hydroxide.

Here's what I find confusing. From the balanced reaction of the two reagents ( 1:1 ) I would have assumed that to make a 50 mM ammonium acetate I would at least need 50 mM of acetic acid and 50 mM of ammonium hydroxide.

Then for pH adjustments, I would add an excess of one or the other to decrease or increase the pH respectively before making to final volume. The reagent which is not added in excess would be the limiting reagent (the reagent which determines the amount of product formed) while the reagent added in excess would react only up to 50 mM and the rest would be present in excess.

What I find confusing is reports online stating that they prepare a 10 mM ammonium acetate buffer taking into account the concentration of acetic acid only, not ammonium hydroxide.

For example here is an example of a procedure for making a 10 mM ammonium acetate solution using the method described.

3.   Ammonium acetate in water (10 mM; pH~4.3):
a.   Stock glacial acetic acid solution is 100%
i.   MW = 60 grams
b.   Stock ammonium hydroxide solution is 33%
i.   MW = 35 grams
c.   Add 600 uL glacial acetic acid and 400 uL ammonium hydroxide to 1 L water
i.   10 mM is based on acetic acid only – ammonia is used to pH

In this example I calculated that they are reacting 10.5 mM acetic acid with 3.3 mM ammonium hydroxide. How is it possible then to obtain 10 mM ammonium acetate? I would have thought that the ammonium acetate concentration in this case would have been 3.3 mM due to ammonium hydroxide being the limiting reagent.

My question is, which is the right method to do this if I want to prepare a certain concentration of salt like 10 mM ammonium acetate from the method described?

Is it right to have 10 mM acetic acid and 10 mM ammonium hydroxide and then add 1 in excess to pH, or have a 10 mM acetic acid solution only and add just enough ammonium hydroxide to reach the desired pH?

I hope it is not too confusing

thanks

[Borek]

50 mM ammonium acetate buffer

Impossible, which is a reason why you have problems. You can have a 50 mM acetate buffer (then [CH₃COOH]+[CH₃COO^-] is 0.05 M), you can have 50 mM ammonium buffer (then [NH₃]+[NH₄⁺] is 0.05 M), you can't have 50 mM ammonium acetate  buffer, as it would require sums of concentrations of both things to equal 0.05 M. Nice, neutral solution of a salt, hardly a buffer.

[j4vier]

50 mM ammonium acetate buffer

Impossible, which is a reason why you have problems. You can have a 50 mM acetate buffer (then [CH₃COOH]+[CH₃COO^-] is 0.05 M), you can have 50 mM ammonium buffer (then [NH₃]+[NH₄⁺] is 0.05 M), you can't have 50 mM ammonium acetate  buffer, as it would require sums of concentrations of both things to equal 0.05 M. Nice, neutral solution of a salt, hardly a buffer.

You are right, apologies if phrased wrong. At pH 7, Ammonium Acetate is just a salt. However I'd like to focus on concentration only of the Ammonium Acetate for now to obtain a 50 mM Ammonium Acetate salt.

I found a solution to a problem similar to mine which instead of using Ammonium Hydroxide, uses Sodium Hydroxide to form Sodium Acetate.

http://www.chemteam.info/AcidBase/titration-problems-1.html

In particular, solution to problem 1, part c-2.

Here is clearly shows that if I react 2.52 x 10^-3 moles of acetic acid with 7.8 x 10^-4 moles of sodium hydroxide, I only obtain 7.8 x 10^-4 moles of sodium acetate due to the fact that sodium hydroxide is the limiting reagent.

My question is then, why for the example which I showed above in the first post, I am seeing reported a 10 mM Ammonium Acetate buffer based on Acetic Acid only which is obtained by reacting 10 mM Acetic Acid with 3.3 mM Ammonium Hydroxide.

Is the ammonium hydroxide not the limiting reagent in this case and should I not end up with only 3.3 mM Ammonium Acetate?

Thank you

[magician4]

you can't have your cake and eat it...

either you prepare a c₀=50 mM (NH₄⁺)(CH₃COO^-) solution, with whatever outcome with respect to the pH may result ( by chance, in this very case it  IS very near to pH = 7,00... )

... or you want (within reasonable limits) a specific pH , and that will force an asymmetric composition onto your system, by either increasing or reducing ONE of the components (however, you're free to decide which one)
  it's up to you which component becomes "limiting" under those conditions

regards

Ingo

[j4vier]

you can't have your cake and eat it...

either you prepare a c₀=50 mM (NH₄⁺)(CH₃COO^-) solution, with whatever outcome with respect to the pH may result ( by chance, in this very case it  IS very near to pH = 7,00... )

... or you want (within reasonable limits) a specific pH , and that will force an asymmetric composition onto your system, by either increasing or reducing ONE of the components (however, you're free to decide which one)
  it's up to you which component becomes "limiting" under those conditions

regards

Ingo

Hi Ingo,

I would like to focus on your first point which you suggested. I am aiming to prepared a 50 mM ammonium acetate solution. What concentration would you use of both reagents to achieve this? I Would say that I would need at least 50 millimoles acetic acid and 50 millimoles ammonium hydroxide in order to obtained the desired concentration of ammonium acetate.

Of course after I established that I have 50 millimoles ammonium acetate, I can add either more acid or base to change the pH before making it up to volume. But if I add only one of the 2 reagents, the amount of ammonium hydroxide formed should still be 50 millimoles as dictated by the limiting reagent which I haven't increased.

Regards

[AWK]

Hi everyone,

I am trying to prepare a 50 mM ammonium acetate buffer from acetic acid and ammonium hydroxide.

You started from a confusing statement. Ammonium acetate shows buffering properties, but usually is called simply - salt.
Term 50mM ammonium acetate buffer commonly means, as Borek said, a mixture of ammonium acetate and ammonia with combined concentration of both reagents 50 mM and
recommended pH.
From your concentrated solutions you can prepare solutions of salt or buffer solutions concentrated up to about 7 M.
These are very basic stoichiometry calculations concerning neutralization reaction with exact molar ratio of reagents or limited reagent problem.
Mixing both reagents you can obtain buffer solutions with best buffer capacity around pH ~5 or pH ~9.

[magician4]

I would like to focus on your first point which you suggested. I am aiming to prepared a 50 mM ammonium acetate solution.

if c₀(ammoniumacetate) = 50 mM  is your target, why not buy ammoniumacetate and prepare the respective solution thereof ? (should save you some trouble, and should be cheaper, if memory serves, too)

What concentration would you use of both reagents to achieve this?

i have a feeling, that the difference between concentration c ( or c₀ , which is very much identical in this case) and amount n is somewhat unclear to you?

I Would say that I would need at least 50 millimoles acetic acid and 50 millimoles ammonium hydroxide in order to obtained the desired concentration of ammonium acetate.

see above: this depends on the volume of solution you wish to prepare!

Of course after I established that I have 50 millimoles ammonium acetate, I can add either more acid or base to change the pH before making it up to volume.

you could do a lot of things to modify the pH after the fact ...
... only that by then we'd be talking a different system
again: what is your target?

But if I add only one of the 2 reagents, the amount of ammonium hydroxide formed should still be 50 millimoles

pls. note, that in an ammoniumacetate-solution, "unmodified", the amount of "ammoniumhydroxide" ( that should be NH₃, aq., free , I take it ?) is next to nill (i.e. negligible)

... which , by adding a strong additional base, for example, will change drastically

regards

Ingo

[j4vier]

I would like to focus on your first point which you suggested. I am aiming to prepared a 50 mM ammonium acetate solution.

if c₀(ammoniumacetate) = 50 mM  is your target, why not buy ammoniumacetate and prepare the respective solution thereof ? (should save you some trouble, and should be cheaper, if memory serves, too)

What concentration would you use of both reagents to achieve this?

i have a feeling, that the difference between concentration c ( or c₀ , which is very much identical in this case) and amount n is somewhat unclear to you?

I Would say that I would need at least 50 millimoles acetic acid and 50 millimoles ammonium hydroxide in order to obtained the desired concentration of ammonium acetate.

see above: this depends on the volume of solution you wish to prepare!

Of course after I established that I have 50 millimoles ammonium acetate, I can add either more acid or base to change the pH before making it up to volume.

you could do a lot of things to modify the pH after the fact ...
... only that by then we'd be talking a different system
again: what is your target?

But if I add only one of the 2 reagents, the amount of ammonium hydroxide formed should still be 50 millimoles

pls. note, that in an ammoniumacetate-solution, "unmodified", the amount of "ammoniumhydroxide" ( that should be NH₃, aq., free , I take it ?) is next to nill (i.e. negligible)

... which , by adding a strong additional base, for example, will change drastically

regards

Ingo

Hi,

I will split my answer into two parts, one with my calculations, one with an example found online which I'm hoping someone can explain.

Part 1:

The preparation of ammonium acetate from ammonium hydroxide and acetic acid is recommended for LC-MS buffer prep especially when running acetonitrile gradients, hence I will be sticking to this procedure and won't be using the readily available ammonium acetate salt.

Second,  I understand the difference between molarity and moles.

Third, volume is 1L so I meant 50 millimoles in 1L as I want to prepare 50 mM.

So here are my calculations and please note I am discussing concentration only (not pH).

Starting reagents:

Glacial acetic acid- 17.4 M 

Ammonium Hydroxide-13.2 M

Objective:

Prepare a solution containing 50 millimoles of ammonium acetate in 1 L volume (50mM concentration).

Mix: 2.87 mL of Acetic Acid  with 3.8 mL of Ammonium hydroxide and make up to 1 L with H₂O.

2.87 mL of Glacial Acetic acid contains 0.05046 moles

3.8 mL of Ammonium Hydroxide contains 0.05016 moles

Moles of Ammonium acetate expected: 0.05016 and since I will make it to 1L with water, I would say a final 50 mM concentration.

Is this correct or incorrect? If the latter, how many mL of both reagents would you mix to achieve a 50 mM ammonium acetate concentration?


Second part:

If you download the document at this link: emerald.tufts.edu/~mcourt01/Documents/HPLC-mass%20spec%20buffers.doc

You will see a similar procedure described for preparing a 10 mM ammonium acetate buffer.

They mention that the 10 mM is based on acidic acid only so I am wondering if the terminology is correct, to say that you prepared a 10 mM ammonium acetate buffer if in fact that is only the concentration of one of the reagents.

Based on my calculation

100% GAA: 17.5 M
33% Ammonium Hydroxide: 17.5 M

They mix 600 μL of GAA with 400 μL of Ammonium hydroxide so 0.0105 moles of GAA with 0.007 moles of ammonium hydoxide and they make the final volume to 1L.

In this case I would say that the ammonium acetate concentration is 0.007 M due to ammonium hydroxide being the limiting reagent.

What is your opinion?

Thanks

[magician4]

The preparation of ammonium acetate from ammonium hydroxide and acetic acid is recommended for LC-MS buffer prep especially when running acetonitrile gradients, (...)

just for the record: this sounds like complete rubbish to me, esp. the "especially when running acetonitrile gradients" part
... and, as already mentioned, we don't talk "buffer" here.

however, this is up to you: the procedure described will also lead to the respective solution, too, yes

pls. note: knowing the hygroscopy of glacial acid and the volatility of NH₃ from concentrated solution thereof, my feeling is that without prior determination of the real content of you given ingredients your result might be far worse (with respect to the precision of your desired content) than when starting from (NH₄⁺)(CH₃COO^-), and using scales

(...) Is this correct or incorrect?

neglecting that for example even "glacial acetic acid" more often than not is less than 100% ( ballpark of 98% would be my best guess), and neglecting the problems of precisely measuring " 2,87 mL " , and being generous about rounding , and then some...
... yes , this looks good to me

They mention that the 10 mM is based on acidic acid only so I am wondering if the terminology is correct, to say that you prepared a 10 mM ammonium acetate buffer if in fact that is only the concentration of one of the reagents.

if you mixed 10 mmoles of acetic acid, and 10 mmoles of NH₃, and solved/diluted the  result ad 1 L  aq. solution, your concentration would be 10 mmoles per liter with respect to c₀( ammoniumacetate), c₀ of NH₄⁺ and c₀ of CH₃COO^- , i.e. with respect to all components we're talking.
( the concentrations c of those would be a little smaller, due to the equilibria involved here: usually , this is negligible, however)

In this case (...)

my feeling is, that they simply were lazy / were unable to use better dosing techniques / simply didn't give a damn (as to be precise here doesn't matter anyway ) , that's all


regards

Ingo

[j4vier]

The preparation of ammonium acetate from ammonium hydroxide and acetic acid is recommended for LC-MS buffer prep especially when running acetonitrile gradients, (...)

just for the record: this sounds like complete rubbish to me, esp. the "especially when running acetonitrile gradients" part
... and, as already mentioned, we don't talk "buffer" here.

however, this is up to you: the procedure described will also lead to the respective solution, too, yes

pls. note: knowing the hygroscopy of glacial acid and the volatility of NH₃ from concentrated solution thereof, my feeling is that without prior determination of the real content of you given ingredients your result might be far worse (with respect to the precision of your desired content) than when starting from (NH₄⁺)(CH₃COO^-), and using scales

(...) Is this correct or incorrect?

neglecting that for example even "glacial acetic acid" more often than not is less than 100% ( ballpark of 98% would be my best guess), and neglecting the problems of precisely measuring " 2,87 mL " , and being generous about rounding , and then some...
... yes , this looks good to me

They mention that the 10 mM is based on acidic acid only so I am wondering if the terminology is correct, to say that you prepared a 10 mM ammonium acetate buffer if in fact that is only the concentration of one of the reagents.

if you mixed 10 mmoles of acetic acid, and 10 mmoles of NH₃, and solved/diluted the  result ad 1 L  aq. solution, your concentration would be 10 mmoles per liter with respect to c₀( ammoniumacetate), c₀ of NH₄⁺ and c₀ of CH₃COO^- , i.e. with respect to all components we're talking.
( the concentrations c of those would be a little smaller, due to the equilibria involved here: usually , this is negligible, however)

In this case (...)

my feeling is, that they simply were lazy / were unable to use better dosing techniques / simply didn't give a damn (as to be precise here doesn't matter anyway ) , that's all


regards

Ingo

Hi Ingo,

Another reason for using liquid solutions of acetic acid and ammonium hydroxide rather than salts, is due to the fact that you can get anyway without filtering the buffer if you use LC-MS grade liquid reagents. If I had to weight out salts I would have to filter the solution which in the past gave me trouble on the MS side, such as seeing a peak for nylon after filtering the buffer through a nylon filter!

By the way, the molarity that I quoted where calculated taking into account the % and the density which were 99.5% GAA and 25% NH4OH.

Now let's say I want to this ammonium acetate solution to be acidic  so rather than adding 2.87 ml of GAA, I add 30 ml of GAA and keep the ammonium hydroxyde volume the same as before 3.8 ml, making up to 1L.

Now the moles of GAA are 17.4 * 30/1000= 0.522 moles in 30 ml which react again with 0.05016 moles of ammonium hydroxide. Am I right in saying that the moles expected of ammonium acetate are still 0.05016 which would be 50 mM when made in 1L? and are the moles left of unreacted acetic acid 0.522-0.05016=0.466?

Also is the pH calculation below for this solution correct?

pH= 4.76+ log [Ac-]/[[HAc]
    = 4.76+ log [0.05016]/[0.466]
    = 4.76- 0.97=3.8
   
Thanks for your help

[magician4]

my first feeling is, that with both (a small amount of) ammoniumhydroxide and ( a somewhat higher excess) of GAA around, HH and most other approximated approaches will not satisfy: I'd hence prefer to use a simplified*⁾ charge balance

[H⁺] + [NH₄⁺] = [AcO^-]

side conditions:

(a) K_a= [AcO^-] * [H⁺]/[AcOH] =  [AcO^-] * [H⁺]/(c₀(AcOH) - [AcO^-])     [AcO^-] = K_a * c₀(AcOH)/([H⁺] + K_a)

(b) K_b = [NH₄⁺] * [OH^-] / [NH₃] = [NH₄⁺] * (K_w / [H⁺]) / (c₀(NH₃) - [NH₄⁺])     [NH₄⁺] = K_b * c₀(NH₃) /  (K_b + (K_w/[H⁺]) )

introducing side conditions to the main term:

[H⁺] + K_b * c₀(NH₃) /  (K_b + (K_w/[H⁺]) ) = K_a * c₀(AcOH)/([H⁺] + K_a)

this , however, is an expression  third order in [H⁺] with the (plausible of those three) solution**⁾

[H⁺] = 1,666.. * 10^-4 M : pH ~ 3.78


... and as your result with HH seems to be pretty close to this, looks like I overdid it, and your simplyified HH-approach will do just fine here, too


regards

Ingo

*⁾ with [OH^-] being neglected, as we're talking acidic conditions here

**⁾
considering c₀(AcOH) = 0.522 M and c₀(NH₃) = 0.05016 M

[magician4]

additional remark:

Another reason (...)

use one of those:


... and, btw., the solubility of (NH₄⁺)(AcO^-) is like 1480 g/L.
were talking 0.05 M here:  filtering ? ? ? ?


regards

Ingo