From: megha goyal
Date: 4 February 2012 02:42
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From: William G. Scott
http://en.wikipedia.org/wiki/Henderson–Hasselbalch_equation
or any intro chemistry text
William G. Scott
Professor
----------
From: Edward A. Berry
Well, by convention the concentration of a buffer refers to the concentration of the buffering species, OAc- in this case. So the sum of NaOAc and HOAc should be 10 mM.
If the MW of NaOAc(H2O)3 is 136, then .1123 mg is 0.90 umol.
Density of HOAc is 1.049, and MW is 60.05, so assuming 100%
it is 1049/60.05=17.47 M,
so 0.476 ul HOAc is 8.35 umol
8.35 umol + 0.90 umol = 9.25 umol, which in 925 ul will be .01 M = 10 mM
So if you finally dilute to 0.925 ml, the concentration will be right.
If you only add NaOH to get the right pH, composition will be right.
But volume and weight can be measured much more accurately and
reproducibly than pH, and pKa is affected by ionic strength,
so if you use the equilibrium expression for dissociation
of a weak acid (AKA Hendersen Hasselbach) to calulate the
proportions of NaOAc and acetic acid that will give the right pH
at your final dilution, you will probably get more accccurate pH
and definitely more reproducible.
x=[OAc]/[HOAc] = 1/H+ = 10^(pH-pK)
OAc/Conc = OAc/(Oac+HOAc) = 1/(1+1/x)
[OAc] = Conc/(1+1/x)
HOAc/Conc = HOAc/(OAc+HOAc) = 1/(1+x)
[HOAc] = Conc/(1+x)
All you need to know is
desired concentration (Conc),
desired pH
pKa of the buffer at final ionic strength and temperature.
(Or just google for a good buffer calculator)
Date: 4 February 2012 02:42
Our recombinant product is formulated in 10 mM sodium acetate buffer at pH 4.0 and the std composition mentions sodium 0.035 mg and acetate 0.59 mg per ml of sample. It mentions Sodium acetate is formed by titrating glacial acetic acid with sodium hydroxide.
Can anyone guide me on how this 10 mM buffer is prepared or on its calculation.
What currently I do is the old method that we have been following i.,e Use 0.123 mg Na Acetate trihydrate + 0.476 µl Gl. Acetic acid, adjust pH to 4.0 using 5 N NaOH. But we do not know how these values have been arrived at and if this calculation is correct.
Please guide me with this regards. Any help on this will be highly appreciated.
thanks,
Megha
----------
From: William G. Scott
http://en.wikipedia.org/wiki/Henderson–Hasselbalch_equation
or any intro chemistry text
William G. Scott
Professor
----------
From: Edward A. Berry
Well, by convention the concentration of a buffer refers to the concentration of the buffering species, OAc- in this case. So the sum of NaOAc and HOAc should be 10 mM.
If the MW of NaOAc(H2O)3 is 136, then .1123 mg is 0.90 umol.
Density of HOAc is 1.049, and MW is 60.05, so assuming 100%
it is 1049/60.05=17.47 M,
so 0.476 ul HOAc is 8.35 umol
8.35 umol + 0.90 umol = 9.25 umol, which in 925 ul will be .01 M = 10 mM
So if you finally dilute to 0.925 ml, the concentration will be right.
If you only add NaOH to get the right pH, composition will be right.
But volume and weight can be measured much more accurately and
reproducibly than pH, and pKa is affected by ionic strength,
so if you use the equilibrium expression for dissociation
of a weak acid (AKA Hendersen Hasselbach) to calulate the
proportions of NaOAc and acetic acid that will give the right pH
at your final dilution, you will probably get more accccurate pH
and definitely more reproducible.
x=[OAc]/[HOAc] = 1/H+ = 10^(pH-pK)
OAc/Conc = OAc/(Oac+HOAc) = 1/(1+1/x)
[OAc] = Conc/(1+1/x)
HOAc/Conc = HOAc/(OAc+HOAc) = 1/(1+x)
[HOAc] = Conc/(1+x)
All you need to know is
desired concentration (Conc),
desired pH
pKa of the buffer at final ionic strength and temperature.
(Or just google for a good buffer calculator)
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