Re: Ammonium Formate
Derek Stein
Hi,
How would I go about making a 10mM ammonium formate,pH3, 25% ACN buffer and then a 500mM ammonium formate,pH6.8, 25% ACN buffer. I want to just weigh out the crystal form ammonium formate and then adjust the pH. But this may be wrong. Someone has told me to start with formic acid and adjust the pH with ammonium-OH. This is all well an good for the first solution but the second solution would (500mM) have way to much formic acid in it for me to get the pH up to 6.8. Not sure how to properly make each of these buffers.
Derek
Derek Stein Masters Student 507-745 Bannatyne Avenue Department of Medical Microbiology University of Manitoba Winnipeg, Manitoba R3E 0J9Ph: (204) 789-3202 Cell: (204) 290-5996
Web Site: http://www.mrsi.ca
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Dr Engelbert Buxbaum
<derek_ri...@hotmail.com>:
> How would I go about making a 10mM ammonium formate,pH3, 25% ACN buffer
> and then a 500mM ammonium formate,pH6.8, 25% ACN buffer. I want to just
> weigh out the crystal form ammonium formate and then adjust the pH. But
> this may be wrong. Someone has told me to start with formic acid and
> adjust the pH with ammonium-OH. This is all well an good for the first
> solution but the second solution would (500mM) have way to much formic
> acid in it for me to get the pH up to 6.8. Not sure how to properly
> make each of these buffers.
For a purely aqueous solution you need a mixture of formic acid and
ammonium formate, the ratio calculated from the
Henderson-Hasselbalch-equation,
log([A]/[HA]) + pKa = pH
Example:
Your pH is 3.0, the pKa of formic acid is 3.75. Hence log([A]/[HA]) = 3 -
3.75 = -0.75 and [A]/[HA] = antilog(-0.75) = 0.178. At the same time your
total concentration [A]+[HA] = 10 mM. Hence 10 mM = 1.178 [HA] and [HA] =
8.5 mM.
So you need to prepare a solution of 8.5 mM formic acid and 10 mM - 8.5 mM
= 1.5 mM ammonium formate. It is probably easiest to prepare that as a
concentrate, say, 10-fold. The molecular mass of ammonium formate is
63.06, hence you need 946 mg/l. The formic acid is best added from a
solution of known (from titration) concentration. If that solution were 1
M, you'd need 85 ml/l.
Note however that adding organic solvent changes the activity of the ions
and hence the pH. Whether or not the original author has corrected for
that effect is a different question. If the paper was written properly
(giving all the info necessary to repeat the experiment) it should state
this.
The other problem with your question is that formate has essentially no
buffering capacity at pH 6.8. Buffers work only within +/- 1 pH-unit of
their pKa, hence you would use formate only between about 3 and 5. Neither
would the ammonium ion (pKa = 9.2) have much buffering capacity at this pH.
One further aspect: Ammonium formate is often used as volatile buffer,
that vanishes upon lyophilisation. In my experience however in this
situation the water and ammonia evaporate first, leaving the sample in
concentrated formic acid, which may well melt due to its relatively low
vapor pressure. Whether or not the sample survives that depends on its
chemistry. Ammonium bicarbonate is truly volatile and buffers well in the
pH 7-9 range.