4g NaOH = 0.1 mol. 500 ml of 0.5 N HCl = 0.25 equivalents of acid. Moles of acid = 0.25. After neutralization, 0.15 mol of HCl remains in 500 ml. Molarity = 0.15/0.5 = 0.3 M. pH = -log(0.3) ≈ 0.52. The answer key points to 2.50, which implies a calculation error in the question or options.