buffers and salt pair buffers

[MikeT]

Hi everyone,
I was wondering, an hplc method for creatine calls for ammonium acetate and another for the sulphate salt in the mobile phase. It does not say that the pH is brought to a certain point with an acid. Is it possible to have these salts alone in the mobile phase without adding an acid (say acetic acid) to it? When a method calls for a buffer, like ammonium acetate, is it assumed that you use the weak acid to bring it to the need pH?
Also, if you want to make a pH 6.0 phosphate buffer and you add a salt like KH2PO4*H2O which makes lets say a pH 6 buffer, why then do you need an acid? And why can you make a buffer with the previously mentioned salt and K2HPO4*7H20, without having to add an acid or base? Thanks guys

-Mike

[tom jupille]

Is it possible to have these salts alone in the mobile phase without adding an acid (say acetic acid) to it?

Yes, but then you will have a salt solution, not a buffer. In order to be a buffer, the weak ion (in this case, acetate) must be *partially* dissociated so that it can both donate and accept protons. In other words, an acetate buffer must contain significant levels of both the acetate (ionized) and acetic acid (unionized) form.

When a method calls for a buffer, like ammonium acetate, is it assumed that you use the weak acid to bring it to the need pH?

Yes, but there are a couple of "gotchas". If you are trying to buffer near the pKa of the acid (4.8 for acetate), you could do it by adding acetic acid, but that would increase the concentration of acetate. It would be better to start with acetic acid at a known concentration and add ammonium hydroxide to adjust the pH upward. If you were trying to buffer near the pKa of the base (9.4 for ammonium) it would be the other way around.

add a salt like KH2PO4*H2O which makes lets say a pH 6 buffer, why then do you need an acid?

Because it won't be a buffer and it won't be at pH 6 (probably more like 4-5). For pH 6 you would either use a appropriate mixture of the mono basic (KH2PO4) and dibasic (K2HPO4) salt or else start with the monbasic and add KOH as necessary to bring the pH up to 6 (by the way, pH 6 would be on the ragged edge of usability for phosphate (the pKa for the second ionization is listed as 7.2).

And why can you make a buffer with the previously mentioned salt and K2HPO4*7H20, without having to add an acid or base?

You can't. I suspect the issue here is that not everyone uses the word "buffer" correctly. I know people who use "buffer" to describe *any* mobile phase (e.g., "the buffer used for the separation was 45% acetonitrile / 55% water").

[ph/Amr Tarek]

you can reach to the desired pH by using the single salt only (ammonium acetate) or (KH2PO4) but you must use its weak acid or base (buffer form) because the buffer form suppress pH change .
Upon adjusting ammonium acetate pH 6 , if the pH become > 6 ; you can get it down by acetic acid , if pH become < 6 you can raise it by ammonia solution .

Upon adjusting KH2PO4 pH 6 , if pH become > 6 , you can get it down by phosphoric acid , if pH become < 6 you can raise it by K2HPO4 or KOH .

[MikeT]

But how do two salts make a buffer (example: NaH2PO4•H2O + Na2HPO4•7H2O)? Thanks guys

[MikeT]

Or NaHCO3 + Na2CO3

[tom jupille]

But how do two salts make a buffer

From my earlier post:

In order to be a buffer, the weak ion (in this case, acetate) must be *partially* dissociated so that it can both donate and accept protons.

The buffering ion must be partially dissociated, which means that you need a equilibrium mixture of two different charged versions of the same molecule.
- For acetate that would be CH3COOH <-> CH3COO-
- For phosphate, that could be H3PO4 <-> H2PO4- (at around pH 2), or it could be H2PO4- <-> HPO4= (around pH 7), or it could even be HPO4= <-> PO4-3 (around pH 12)
- For carbonate it could be H2CO3 <-> HCO3- (around pH 6) or it could be HCO3- <-> CO3= (around pH 10)
- for ammonium it would be NH4+ <-> NH3 (around pH 9) . . .

- . . . and so on

[ph/Amr Tarek]

Preparation of 0.1M sodium phosohate buffer pH 7.2
1) weigh 26.8 gm Na2HPO4.7H2O and dissolve in amount of water then complete to 1000 ml . (0.1M Na2HPO4.7H2O)
2) weigh 13.8 gm NaH2PO4.H2O and dissolve in amount of water then complete to 1000 ml . (0.1M NaH2PO4.H2O)
3) add from solution (2) to solution (1) till pH 7.2


ph/Amr Tarek
Instrumental Unit Head at Otsuka pharmaceuticals

[lmh]

To add to Tom's excelent explanation:
The point of a buffer is that it resists further change to pH. You can think of pH as just a measure of how many hydrogen ions are floating around in solution, so a buffer is just a mixture that can mop up or release hydrogen ions. Any reversible reaction that involves hydrogen ions can be a buffer:
A ---> H(+) + B

(I'm not worrying about charges of A and B; obviously charges have to balance). That's why you can make a buffer from Na2Hphosphate and NaH2phosphate: you can write a reaction converting one to the other and it involves H(+) ions.
Na + H2NaPhosphate <---> H + HNa2Phosphate

The reaction has to be reversible because it will release hydrogen ions going forwards, and mop them up going backwards. This is also why you have to have at least some of both sides of the reaction present in the mixture. If it's going to mop up hydrogen, it needs HNa2phosphate, and if it's going to release hydrogen, it needs H2Naphosphate.

(the more theoretical bit: I've over-simplified a lot. What's really happening in the buffer is that the equilibrium constant has to stay constant, and this is actually [H]*[B]/[A]. When the buffer mops up or releases hydrogen, [B] and [A] change. If they're both big then a small fractional decrease in [B] with a corresponding small fractional increase in [A] actually mops up an awful lot of hydrogen ions, while requiring only a small change in the concentration of [H] to bring the equilibrium constant back to its proper value, so you only get a small change of pH even if you make the buffer mop up a lot of added acid. Hope that helps.).