Need help with figuring out how to solve this problem?

(a)
In water, 1 mol of H₂SO₄ dissociates to give 2 mol of H₃O⁺ ions, while 1 mol of NaOH dissociates to give 1 mol of OH⁻ ions.

[H₃O⁺] = (0.40 M) × 2 = 0.80 M
pH = -log[H₃O⁺] = -log(0.80) = 0.10

====
(b)
H₃O⁺(aq) + OH⁻(aq) → 2H₂O(ℓ)
Mole ratio H₃O⁺ : OH⁻ = 1 : 1

Initial moles of H₃O⁺ = (0.40 mol/L) × (50.00/1000 L) × 2 = 0.04 mol
Moles of OH⁻ added = (0.40 mol/L) × (25/1000 L) = 0.01 mol
Volume of the final solution = (50.00 + 25) mL = 75 mL = 0.075 L

After addition, [H₃O⁺] = [(0.04 - 0.01) mol] / (0.075 L) = 0.40 M
pH = -log[H₃O⁺] = -log(0.40) = 0.40

====
(c)
Initial moles of H₃O⁺ = 0.04 mol
Moles of OH⁻ added = (0.40 mol/L) × (50/1000 L) = 0.02 mol
Volume of the final solution = (50.00 + 50) mL = 100 mL = 0.1 L

After addition, [H₃O⁺] = [(0.04 - 0.02) mol] / (0.1 L) = 0.20 M
pH = -log[H₃O⁺] = -log(0.20) = 0.70

====
(d)
Initial moles of H₃O⁺ = 0.04 mol
Moles of OH⁻ added = (0.40 mol/L) × (75/1000 L) = 0.03 mol
Volume of the final solution = (50.00 + 75) mL = 125 mL = 0.125 L

After addition, [H₃O⁺] = [(0.04 - 0.03) mol] / (0.125 L) = 0.08 M
pH = -log[H₃O⁺] = -log(0.08) = 1.10

====
(e)
Initial moles of H₃O⁺ = 0.04 mol
Moles of OH⁻ added = (0.40 mol/L) × (100/1000 L) = 0.04 mol

H₃O⁺ and OH⁻ completely react to form H₂O, and the final solution becomes neutral.
pH = 7

====
(f)
Initial moles of H₃O⁺ = 0.04 mol
Moles of OH⁻ added = (0.40 mol/L) × (125/1000 L) = 0.05 mol
Volume of the final solution = (50.00 + 125) mL = 175 mL = 0.175 L

After addition, [OH⁻] = [(0.05 - 0.04) mol] / (0.175 L) = 0.057 M
p0H = -log[OH⁻] = -log(0.057) = 1.24
pH = pKw - pOH = 14.00 - 1.24 = 12.76