Chemistry help anyone? No Idea what to do?

NH₃(aq) + HCl(aq) → NH₄Cl(aq)
OR Mole ratio NH₃ : HCl : NH₄Cl = 1 : 1 : 1

No. of milli-moles of NH₃ = (0.440 mmol/mL) × (20.0 mL) = 8.80 mmol
No. of milli-moles of HCl used = 8.80 mmol
Volume of HCl used = (8.80 mmol) / (0.200 mmol/mL) = 44.0 mL
No. of mill-moles of NH₄Cl formed = 8.80 mmol
Molarity of NH₄Cl formed = (8.80 mmol) / [(20.0 + 44.0) mL] = 0.138 M

NH₄Cl completely dissociates in aqueous solution.
[NH₄⁺] formed in titration = 0.138 M

Consider the dissociation (hydrolysis) of NH₄⁺ ions :
NH₄⁺(aq) + H₂O(l) ⇌ NH₃(aq) + H₃O⁺(aq) ...... Ka = Kw/Kb(NH₃) = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵)

Initially concentrations :
[NH₄⁺]ₒ = 0.138 M
[NH₃]ₒ = [H₃O⁺]ₒ = 0 M

Equilibrium concentrations :
[NH₄⁺] = (0.138 + y) M ≈ 0.138 M, assuming that 0.138 ≫ y
[NH₃] = [H₃O⁺] = y M

Kw = [NH₃] [H₃O⁺] / [NH₄⁺]
y²/0.138 = (1.0 × 10⁻¹⁴)/(1.8 × 10⁻⁵)
y = 8.76 × 10⁻⁶ (The assumption that 0.138 ≫ y holds.)

pH = -log[H₃O⁺] = -log(8.76 × 10⁻⁶) = 5.06

Ok but first can you tell me what I should make for lunch.