Posted by cat on Wednesday, May 30, 2012 at 10:27pm.
You want to use some of the NaH2PO4 as the acid and some of the Na2HPO4 as the base.
pH = pKa2 + log (250+x)*0.1/(250-x)*0.1
Solve for x and add to 250 to find mL of base to use and 250-x to find mL of acid to use. I would plug those numbers back into the Henderson-Hasselbalch equation and check that a pH of 7.60 is what you have prepared.
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