How much 1.5 m sodium hydroxide is necessary to exactly neutralize 20.0 ml of 2.5 m phosphoric acid?

Answer is: 100 mL of sodium hydroxide.
Chemical reaction: 3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O.
V(H₃PO₄) = 20.0 mL ÷ 1000 mL/L = 0.02 L.
n(H₃PO₄) = V(H₃PO₄) · c(H₃PO₄).
n(H₃PO₄) = 0.02 L · 2.5 mol/L.
n(H₃PO₄) = 0.05 mol.
From chemical reaction: n(H₃PO₄) : n(NaOH) = 1 : 3.
n(NaOH) = 0.15 mol.
V(NaOH) = n(NaOH) ÷ c(NaOH).
V(NaOH) = 0.15 mol ÷ 1.5 mol/L.
V(NaOH) = 0.1 L · 1000 mL/L = 100 mL.

The closer the atoms are to becoming ionic the more highly dense the negatively charged side gets.