What is the ph of a soft drink in which the major buffer ingredients are 6.6 g of nah2po4 and 8.0 g of na2hpo4 per 355 ml of solution?

First, we need to get the concentration of  [NaH2PO4]:
[NaH2PO4] =( mass / molar mass ) * volume L
when we have mass NaH2PO4 = 6.6 g & molar mass = 120g/mol & V = 0.355 L
So by substitution:
[NaH2PO4] = (6.6g / 120g/mol) * 0.355 L = 0.0195 M

then, we need to get the concentration of [Na2HPO4]:
[Na2HPO4]= (mass / molar mass ) * volume L
So by substitution:
[Na2HPO4] = (8g/ 142g/mol) * 0.355 L = 0.02 M

and when Pka of the 2nd ionization of phosphoric acid = 7.21 
So by substitution in the following formula, we can get the PH:

PH = Pka + ㏒[A]/[AH]

∴PH = 7.21 + ㏒[0.02]/[0.0195]
∴ PH = 7.2