chem help needed?

a)
[H₃O⁺] = 0.08991 M
pH = -log[H₃O⁺] = -log(0.08991) = 1.05

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b)
Initial moles of H₃O⁺ = (0.08991 mol/L) × (10.00/1000 L) = 0.0008991 M
Moles of OH⁻ = (0.07613 mol/L) × (5.91/1000) = 0.0004499 M
Volume of the final solution = (10.00 + 5.91) mL = 15.91 mL = 0.01591 L

H₃O⁺(aq) + OH⁻(aq) → 2H₂O(ℓ)
H₃O⁺ is in excess.
In the final solution, [H₃O⁺] = (0.0008991 - 0.0004499)/0.01591 M = 0.02823 M
pH = -log[H₃O⁺] = -log(0.02823) = 1.55

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c)
All the equivalence point, both H₃O⁺ and OH⁻ completely react. The final solution is the solution of NaCl which is neutral.
Hence, pH = 7.00

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d)
Initial moles of H₃O⁺ = 0.0008991 M
Moles of OH⁻ = (0.07613 mol/L) × (15.00/1000) = 0.001142 M
Volume of the final solution = (10.00 + 15.00) mL = 25.00 mL = .025 L

H₃O⁺(aq) + OH⁻(aq) → 2H₂O(ℓ)
OH⁻ is in excess.
In the final solution, [OH⁻] = (0.001142 - 0.0008991)/0.025 M = 0.09716 M
pOH = -log[OH⁻] = -log(0.09716) = 2.01
pH = pKw - pOH = 14.00 - 2.01 = 11.99