n acid : 0,005×0,04285 = 2,1425×10^-4 moli
n baza: 0,0045×0,09523 = 4,28535×10^-4 moli
H3PO4 + 2NaOH → Na2HPO4 + 2H2O
1 mol acid..................2 moli baza
0,00021425 moli.......x = 0,0004285 moli baza
1 mol acid...................2 moli baza
x moli acid..................0,000428535 moli baza
x = 0,0002142675 moli acid
n baza exces = 0,000428535-0,0004285 = 3,5×10⁻⁸ moli
Cf = n baza exces/Vsf = 3,5×10⁻⁸/9,5×10⁻³ = 3,684×10⁻⁶ M = [OH⁻]
pOH = -lg[OH⁻] = -lg 3,684×10⁻⁶ = 5,4336
pH = 14-5,433 = 8,56
