0.1 M x 0.5L = 0.05 mols HCl. Adding 25 mL 2M NaOH is 2M x 0.025 L = 0.05 mols NaOH. What you want to do is to back off very slightly with the NaOH (you might try something like 24.95 mL which I calculate to give 0.00019 M or a pH of 3.7. Inching closer, 24.99 mL would leave H^+ of 4E-5 for pH 4.4. The problem here is two-fold. hope it helps
