posted by cat .
you have 1.00 liter each of 0.10 M Na3PO4, Na2HPO4, NaH2po4 ka 1= 7.5*10^-3 ka2=6.2*10^-8 ka3=4.8*10^-13 a) prepare 500 ml of a buffer with PH= 7.6 b) how much 0.10 M HCL must be added to the 500ml of buffer to result in a PH=7.2 after the addition?
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.