How do I make up a 0.01M Na2HP04/NaH2P04 buffer solution, pH adjusted to 7? Both chemicals that I have on hand are anhydrous. The molecular weight of Na2HP04 is 141.96 g/mol and the molecular weight of NaH2P04 is 119.98 g/mol. Do I just dissolve 1.4196 g of Na2HP04 and 1.1998 g of NaH2P04 into 950 ml distilled water, then adjust the pH to 7 and make up to 1000 ml in a volumetric flask? Most of the methods I find use Na2HP04.2H2O and NaH2P04.2H2O. Thanks! (Please note that chemistry is not my field but I am helping out in the chemistry lab because we are short staffed currently)
There are several online calculators, hope they do calculations right. Here is one:
http://www.biochempages.com/2017/06/online-phosphate-buffer-calculator.html
Dear Nirvana.
Yuo are almost right.
THe only change is that if you would like to obtain 0,01M in phospate mixing the 2 form, you need to ass 0,005M of Na2HP04 and 0,005M of NaH2P04 and not 0,01 + 0,01 otherwise you will obtain 0,02M phospate, therefore you have to divide your numbers by 2.
Regarding the salt hydratation (presence of water molecules), it will only affect the amout to be weighted due to the different MW of the two form.
The molecular weight of Na2HP04 is 141.96 g/mol and the molecular weight of NaH2P04 is 119.98 g/mol.
considering that MW of H20 is 18
it means that
Na2HP04*2H20 is 141.96 + 18X2 = 177,96g/mol
NaH2P04*2H20 is 119.98 + 18X2= 155,98g/mol
and therefore in case you have this salts in your hands, you have to consider this new MW instead the old to calculate the amout for produce the 0,005M + 0,005M
good luck
ciao
Manuele
Hello
I hope this could help you
https://www.flinnsci.com/api/library/Download/18ce587821c24fb3b0ad7d878bd6a3d9
Good Luck
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