Chemistry Acid-Base pH Question?

a)
pH of 0.15 mol dm⁻³ of HX solution = 2.69
[H⁺] of 0.15 mol dm⁻³ of HX = 10⁻²˙⁶⁹ mol dm⁻³

Consider the dissociation of 0.15 mol dm⁻³ of HX:
____________________ HX(aq) ___ ⇌ ___ H⁺(aq) ___ + ___ X⁻(aq) _____ Ka
Initial (mol dm⁻³): _____ 0.15 ____________ 0 ____________ 0
Change (mol dm⁻³): __ -10⁻²˙⁶⁹ ________ +10⁻²˙⁶⁹ ______ +10⁻²˙⁶⁹
Eqm (mol dm⁻³) ___ 0.15 - 10⁻²˙⁶⁹ ______ 10⁻²˙⁶⁹ _______ 10⁻²˙⁶⁹

At equilibrium:
Ka = [H⁺] [X⁻] / [HX] = (10⁻²˙⁶⁹)² / (0.15 - 10⁻²˙⁶⁹) = 2.8 × 10⁻⁵ mol dm⁻³
pKa = -log(2.8 × 10⁻⁵) = 4.6

When HX is exactly half-neutralised and [HX] = [X-], [X⁻]/[HX] = 1
Henderson-Hasselbalch equation:
pH = pKa + log([X⁻]/[HX]) = pKa + log(1) = pKa = 4.6


b)
Equation for the solution:
NaOH + HX → NaX + H₂O

No. of moles of NaOH to completely neutralize the HX = (0.25 mol dm⁻³) × (15/1000 dm³) = 0.00375 mol
No. of moles of HX used in titration = 0.00375 mol
Volume of HX used in titration = (0.00375 mol) / (0.15 mol dm⁻³) = 0.025 dm³

No. of moles of excess NaOH = (0.25 mol dm⁻³) × [(25 - 15)/1000 dm³] = 0.0025 mol
Volume of the final solution = [0.025 + (25/1000) dm³] = 0.05 dm³

[OH⁻] = [excess NaOH] = (0.0025 mol) / (0.05 dm³) = 0.05 mol dm⁻³
pOH = -log[OH⁻] = -log(0.05) = 1.3
pH = pKw - pH = 14.0 - 1.3 = 12.7