Chemical Forums
Specialty Chemistry Forums => Biochemistry and Chemical Biology Forum => Topic started by: Isx on June 21, 2023, 01:37:39 AM
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I need to prepare a 1.0L 5X 90mM citrate buffer at pH 4.8. I did this problem 2 years ago for a prelab and remember receiving feedback that my previous thought process was incorrect but I no longer have the correct work (it was turned it) so I'm trying to relearn how to fix it.
These are the available reagents: CAPS free acid (3-cyclohexylamino-1-propanesulfonic acid); sodium phosphate monobasic; sodium phosphate dibasic; TRIS free base (tris(hydroxymethyl)aminomethane); glycine; glacial acetic acid; EDTA disodium salt dehydrate (Ethylenediaminetetraacetic acid); EDTA (Ethylenediaminetetraacetic acid); trisodium citrate; disodium citrate; citric acid; 1 M HCl; 1 M NaOH; 3 M HCl; 3 M NaOH.
Citrate has three acidic protons at pKa's 3.13, 4.76, and 6.4. I want to use the 4.76 value but that pKa2 would correspond to a monosodium citrate which I do not have access to. So would I use the disodium citrate as my acid?
With that thought,
4.8 = 6.4 + log(A-/HA)
-1.6 = log(A-/HA)
0.0251 = (A-/HA) this means there would be a lot more acid in my buffer ratio than base...
The X part is really confusing to me, but my procedure used this blurb as an example and I will continue my thinking based on this: "For example, you need 40 L of 0.001 M TAE buffer that you will use over the course of a week, and you only have a 1 liter storage bottle. Instead of actually making 40 liters, you make one liter of a 0.040 M TAE buffer by
one of the methods stored above, and you would call it “40X 0.001 M TAE”. Then, every time you need one liter
(V2) at a concentration of 0.001 M (M2), you take 25 mL (V1) of your 40X buffer and add 975 mL of deionized
water to it, and you have resulted in 1000 mL of a 1X 0.001 M (M1) TAE buffer."
If I need a 1L 5X 90mM (0.090M) buffer, then it's 0.2L of 0.45M citrate buffer? 0.2L * 0.45M = 0.09 mol of citrate total.
Back to this
0.0251 = (A-/HA)
0.0251 = ((0.09 - HA) / HA) from 0.09 = HA + A- then rearrange the equation
solving for HA I get 0.0878
0.09 - 0.0878 = 0.002204 of A-. That would check out because we have a lot more HA than A- like I said before.
So to prepare a buffer we need 0.0878 mol * 236.09 g/mol = 20.7 g of disodium citrate and 0.002204 mol * 258.06 g/mol = 0.569g of trisodium citrate. You would combine these dry reagents together in a beaker with a little less than 0.2L deionized water then check the pH, then adjust with 1M HCl or NaOH as needed to hit 4.8 and then dilute to 0.2L.
Thank you in advance, I hope my thought process makes sense, this wasn't the same incorrect version I turned in then- I tried to redo it, but I don't think I'm doing it correct again.
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4.8 is a perfect candidate for acetic acid, any particular reason you prefer to use citric?
You can't use pKa1 here, that would mean ignoring pKa2 completely. You need to take it into account.
Basically you need a buffer consisting of a mixture of mono- and disodium. That means starting with a citric acid and adding enough NaOH to convert it to monosodium citrate, then some more to convert monosodium to disodium. ALternatively you can just add disodium citrate to the freshly prepared monosodium citrate.
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This is a slight tangent. Citrate has a couple of possible disadvantages as a biochemical buffer. It is labile to microbial degradation, and it can chelate metal ions. Homopipes is a good choice for a buffer in this pH range. Another issue with citrate is that the practical pKa values depend upon ionic strength, making calculations of how much acid or base to add more difficult.
If you must use citrate buffers, then one way to do so is to prepare stock solutions of citric acid and trisodium citrate and mix them in carefully measured portions for reproducibility.