Calculate the pH of the solution after adding 25,00 mL of 0,200M NaOH to 35,00 mL of 0,100 M acetic acid solution(HC2H3O2).?

Initial number of moles of NaOH = (0,200 mol/L) × (25,00/1000) = 0,005000 mol
Initial number of moles of HC₂H₃O₂ = (0,100 mol/L) × (35,00/1000) = 0,003500 mol

The equation for the reaction between NaOH and HC₂H₃O₂ :
NaOH + HC₂H₃O₂ → NaC₂H₃O₂ + H₂O
OR: Mole ratio NaOH : HC₂H₃O₂ = 1 : 1

Obviously, HC₂H₃O₂ is the limiting reactant (completely reacts).

Number of moles of HC₂H₃O₂ reacted = 0,003500 mol
Number of moles of NaOH reacted = 0,003500 mol
Number of moles of NaOH left unreacted = (0,005000 - 0,003500) mol = 0,001500 mol
Volume of the final solution = (25,00 + 35,00) mL = 60,00mL = 0,06000 L
Molarity of NaOH of the final solution = (0,001500 mol) / (0,06000 L) = 0.02500 M

NaOH completely dissociates in water to give Na⁺ and OH⁻ ions.
In the final solution, [OH⁻] = 0.025 M
pOH = -log[OH⁻] = -log(0.02500) = 1.6
pH = 14 - pOH = 14 - 1.6 = 12.4