Explanation:Firstly you need to do the calculations via Handerson Hasselbach equation:
pKa's of phosphoric acid are 2.3, 7.21 and 12.35. If required pH is 6, then, 7.21 will be used. This means, monopotassium dihydrogen phosphate and dipotassium monohydrogen phosphate (diprotic (H2PO4-) and monoprotic (HPO4--) potassium salts) will be used.
pH = pKa + log ([A-]/[HA])
Now, for A- you put K2HPO4 concentration, and for HA you put KH2PO4 concentration:
6 = 7.21 + log(A/HA)
log(A/HA)=-1.2
A/HA = 0.063
So, now you know the fold difference between these salts, and the Molarity of the solution would be given to you, and you can calculate the required amount:
HA+A=0.05
A/HA=0.063
HA=0.047M
A=0.003M
This means, you need to put 0.047 moles of KH2PO4 and 0.003 moles of K2HPO4 salts into 1 liters of solution.
for molecular weights: KH2PO4= 39+97=136amu; K2HPO4= 39*2+96=174amu
Thus, 0.047*136=6.392g of KH2PO4 and 0.003*174= 0.522g of K2HPO4 should be added.
But, you should also note that, if |pH-pKa| >1 , then buffer capacity of solution decreases. In this case, it is equal to 1.2
