Preparing potassium phosphate buffer(pH 7.4) - (May/12/2006 )
Hi all,
According to http://molbiol.ru/eng/protocol/01_02b.html#a48a, to make 1L of 1 M potassium phosphate buffer (pH 7.4), I would need 802 mL of 1M K2HPO4 and 198 mL of 1M KH2PO4.
But if I want to make 1L of 0.1 M potassium phosphate buffer (pH 7.4), do I:
1) Add 802 mL of 0.1M K2HPO4 and 198 mL of 0.1M KH2PO4 OR
2) Dilute 10x from the 1 M potassium phosphate buffer (i.e. 80.2 mL of 1M K2HPO4 and 19.8 mL of 1M KH2PO4 + 900mL of water)?
Please help! Because I getting different microsomal t1/2 values due to inconsistencies in preparing this buffer... 
which one is easier for you?
If your assay is highly sensitive to pH, then you should check and adjust the pH of the final solution in either case. Relying on mixing buffers to achieve precise pH is unlikely to work. Also, the pH changes when buffers are diluted. You should check and adjust pH at the temperature you intend to use for the assay.
hi
well, preparing two 0.1M solutions is better than diluting 10x your 1M solution, as your preparation is done under pH monitoring. So your final solution is at required pH.
Due to atmospheric inconveniences, pH of a ddH2O decreases, and can alterate the pH too much if you dilute your stock solution.
I agree with phage434 and Fred -- make two 0.1 M solutions, one of the K2HPO4 (dibasic) and one of the KH2PO4 (monobasic). The pH of the dibasic will be alkaline (around pH 8), and the monobasic will be weakly acidic (around 6.5, maybe 6.8).
What I usually do is take a volume of monobasic close to what I want my final volume to be, and add dibasic to it dropwise with stirring until the pH is correct. I usually wind up with slightly more than I need (which is much better than winding up with slightly less than I need 
)
Hi, I have some question regarding the preparation of Potassium phosphate buffer too..hope to get some help here
The experiment req me to have 5mM of potassium phosphate buffer (pH7.4)
Does it means that I have to prepare two 5mM solutions; one of K2HPO4 and the other of KH2PO4? And then slowly adding the monobasic stock to the dibasic stock untill the req pH is reach?
Use Henderson Hasselbach's equation: pH = pka + log [salt]/[acid]
7.4 = 7.1 + log [salt]/[acid]
7.4 - 7.1 = log [salt]/[acid]
0.3 = log [salt]/[acid]
Taking antilogs: 1.99 = [salt]/[acid]
Now, you need 0.005M so
0.005M = [salt] + [acid]
so [salt] = 0.005 - [acid]
substituting into HH equation:
7.4 = 7.1 + log 0.005 - [acid]/[acid]
Take antilogs: 1.99 = 0.005 - [acid]/[acid]
Rearrange terms: 1.99[acid]+[acid] = 0.005M
2.99[acid] = 0.005M
[acid] = 0.005/2.99
[acid] = 0.00167M (not yet grams, matey. Still need to convert)
moles = mass (g)/ Mr thus 0.00167 x 174 = 0.2906g
Easy to work out the [salt]
0.005M = [acid] + [salt], we know [acid] now so
0.005 - 0.00167 = 0.0033
moles to grams: 0.0033 x 136 = 0.4488g
Hey presto! Job done.
Easier way is to google for the web-based algorithms that do the sums for you. (This does seem a very low capacity system. I don't think it will buffer cells for more than a few minutes to an hour).
you can do it this way but you may want to prepare a more concentrated stock and dilute to your final concentration, especially if you have other components in your final buffer.
you can make the concentrated stock by mixing equimolar mono and dibasic until you reach the desired pH.
have a look here how to prepare PBS with different pH
http://molbiol.ru/eng/protocol/01_02b.html#a48a
I can't believe it!! All of this quarrel about preparing PBS??? ok, let's take the grandma's bill out:
For 1L PBS 0,1M:
- 8 gr NaCl
- 0,22 gr KCl
-1,44 gr Na2HPO4*2H2O
-0,2 gr KH2PO4
...adjust to pH 7.4 by mixing with necessary amounts of HCl or NaOH..
take 100 ml of the buffer and add it to 900 ml ddH20...that's your buffer!!!
Easy, isn'it??