NEED HELP! Two solutions are mixed: 100.0 mL of HCl(aq) with pH 2.50 and 100.0 mL of NaOH(aq) with pH 11.20.?

pOH of the NaOH(aq) = 14.00 - 11.20 = 2.80

Initial no. of moles of HCl = (10⁻²˙⁵ mol/L) × (100.0/1000 L) = 0.0003162 mol
Initial no. of moles of NaOH = (10⁻²˙⁸ mol/L) × (100.0/1000 L) = 0.0001585 mol
Volume of the final solution = (100.0 + 100.0) mL = 200.0 mL = 0.2000 L

1 mole of HCl reacts with 1 mole of NaOH. Hence, HCl is in excess.
In the final solution, [H⁺] = [HCl] = {(0.0003162 - 0.0001585) mol} / (0.2000 L) = 0.0007885 M

pH of the resulting solution = -log[H⁺] = -log(0.0007885) = 3.10