pH = 11.8 => uso pKa3 = 12.15 => coppia HPO4-- / PO4---

H3PO4 + NaOH -----------> NaH2PO4 + H2O
0.15_______X_______________0________/_
_0_______X-0.15____________0.15______/_

NaH2PO4 + NaOH ------------> Na2HPO4 + H2O
0.15______X-0.15______________0_______/_
__0_______X-2*0.15___________0.15_____/_

Na2HPO4 + NaOH ------------> Na3PO4 + H2O
0.15______X-0.30______________0______/_
0.15-X+0.30__0______________X-0.30____/_
=
0.45-X______________________X-0.30.

[H+] = 10^-11.8 = 7.1*10^-13 * (0.45-X) / (X-0.30)

=> X = 0.346mol/L

=> 0.346mol/L * 1.2L = 0.416mol

=> 0.416mol * 40g/mol (PM) = 16.63g di Na