19.6 mL solution of 0.100 M HCl is titrated using 0.150 M  NaOH. What is the pH of the solution after adding 2.51 mL of the NaOH solution?

Initial moles of H⁺ (or H₃O⁺) = Initial moles of HCl = (0.100 mol/L) × (19.6/1000 L) = 0.00196 mol
Moles of OH⁻ added = Moles of NaOH added = (0.150 mol/L) × (2.51/1000 L) = 0.0003765 mol

Equation for the reaction: H⁺(aq) + OH⁻(aq) → H₂O(l)
Mole ratio H⁺ : OH⁻ = 1 : 1

Obviously, H⁺ is in excess.
Moles of H⁺ left after reaction = (0.00196 - 0.0003765) mol = 0.0015835 mol
Volume of the final solution = (19.6 + 2.51) mL = 22.11 mL = 0.02211 L
[H⁺] in the final solution = (0.0015835 mol) / (0.02211 L) = 0.0716 M

pH = -log[H⁺] = -log(0.0716) = 1.15