Chem Help!! titration?

a)
Initial number of moles of NH₃ = (0.050 mol/L) × (30.0/1000 L) = 0.0015 mol
HCl completely dissociates to give H⁺ ions.
Initial number of moles of H⁺ ions = (0.025 mol/L) × (24.5 L/1000 L) = 0.0006125 mol
Volume of reaction mixture = (30.0 + 24.5) mL = 54.5 mL = 0.0545 L

The reaction between NH₃ and HCl :
NH₃(aq) + H⁺(aq) → NH₄⁺(aq)
Obviously, NH₃ is in excess. After reaction :
[NH₃] = (0.0015 - 0.0006125) / 0.0545 M = 0.0163 M
[NH₄⁺] = 0.0006125/0.0545 M = 0.0112 M

Consider the dissociation of NH₃ :
NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq) …… Kb = 1.8 × 10⁻⁵

pOH = pKa + log([NH₄⁺]/[NH₃]) = -log(1.8 × 10⁻⁵) + log(0.0112/0.0163) = 4.6
pH = pKw - pOH = 14.0 - 4.6 = 9.4


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b)
Initial number of moles of NH₃ = (0.050 mol/L) × (30.0/1000 L) = 0.0015 mol
Initial number of moles of H⁺ ions = (0.025 mol/L) × (64.0 L/1000 L) = 0.0016 mol
Volume of reaction mixture = (30.0 + 64.0) mL = 94.0 mL = 0.0940 L

The reaction between NH₃ and HCl :
NH₃(aq) + H⁺(aq) → NH₄⁺(aq)
Obviously, H⁺ ions are in excess. After reaction :
[H⁺] = (0.0016 - 0.0015) / 0.0940 M = 0.00106 M
[NH₄⁺] = 0.0015/0.094 M = 0.0160 M

Consider the dissociation of NH₄⁺ :
NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq) …… Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
As Ka is very small and due to the common ion effect in the presence of H⁺ ions, the dissociation of NH₄⁺ is negligible.

Hence, [H⁺] = 0.00106 M
pH = -log[H⁺] = -log(0.00106) = 3.0


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c)
Initial number of moles of NH₃ = (0.050 mol/L) × (30.0/1000 L) = 0.0015 mol
Initial number of moles of H⁺ ions = (0.025 mol/L) × (69.5 L/1000 L) = 0.00174 mol
Volume of reaction mixture = (30.0 + 69.5) mL = 99.5 mL = 0.0995 L

The reaction between NH₃ and HCl :
NH₃(aq) + H⁺(aq) → NH₄⁺(aq)
Obviously, H⁺ ions are in excess. After reaction :
[H⁺] = (0.00174 - 0.0015) / 0.0995 M = 0.00241 M
[NH₄⁺] = 0.0015/0.0995 M = 0.0151 M

Consider the dissociation of NH₄⁺ :
NH₄⁺(aq) ⇌ NH₃(aq) + H⁺(aq) …… Ka = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵)
As Ka is very small and due to the common ion effect in the presence of H⁺ ions, the dissociation of NH₄⁺ is negligible.

Hence, [H⁺] = 0.00241 M
pH = -log[H⁺] = -log(0.00241) = 2.6