Hi,
I am preparing a sodium phosphate buffer solution at 0.1 M pH 7. For the preparation I am following this instruction that requires to mix 1 M Na2HPO4 and 1 M NaH2PO4 like this:
57.7 ml 1 M Na2HPO4 and
42.3 ml 1 M NaH2PO4
For a final volume of 1 L.
I have a doubt because for the preparation of the stock solution 1 M Na2HPO4 I used the heptahydrated salt (Na2HPO4 x 7H2O) that has a mw of 268.07, while the recipe refers to the unhydrated salt that has a mw of 142.00. However, my stock solution of 1 M was prepared with this calculation
1M X 0.2L X 268.07 = 53.6 g of Na2HPO4 x 7H2O dissolved in 0.2 L water
Then, I took the volumes specified in the recipe for the preparation of the mixture..So:
5.77 ml 1 M Na2HPO4 and
4.23 ml 1 M NaH2PO4
For the preparation of 100ml of buffer (then I adjusted the pH). Is it correct that I used these volumes specified in the recipe even if the stock solution was made from the heptahydrated salt, while the recipe refers to the stock solution of unhydrated salt (different mw)?
Thank you very much in advance,
Silvia
It is correct, although I wonder why the pH is adjusted. If you are mixing monobasic ans dibasic phosphate it is only in the interest that the resulting solution has the desired pH.
Hi Silvia,
If everything would be correct you did not had to adjust pH.
Because, old days phosphate buffers were used to calibrate pH meters.
I guess the problem is in how much water you are adding to your (Na2HPO4 x 7H2O) salt. In 268.07 g of this compound you have 142 g of Na2HPO4 and 126 g of H2O.
If you are going to prepare 1M Na2HPO4 in 200 mL of water, you have to
53.6 g of (Na2HPO4 x 7H2O) dilute in 174.8 mL (200 - 25.2) mL. For water 1 g is equal to 1 mL, and you had 25.2 g (or mL) already in your salt.
It seems to me, that you added more water than needed. So, at the end your Na2HPO4 solution was not 1M. And, probably, to bring pH of your solution to 7.0 you added more of (Na2HPO4 x 7H2O) salt.
Cheers.
George.
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