If I want a 2 L solution with a pH of 5, how much acetic acid and how much H2O do I need?

Let y M CH₃COOH is needed.

pH = 5
Hence, [H₃O⁺] = 10⁻⁵ M

CH₃COOH(aq) + H₂O(l) ⇌ CH₃COO⁻(aq) + H₃O⁺(aq) ...... Kₐ = 1.8 × 10⁻⁵ (M)
At equilibrium :
[CH₃COOH] = (y - 10⁻⁵) M
[CH₃COO⁻] = [H₃O⁺] = 10⁻⁵ M

Kₐ = [CH₃COO⁻] [H₃O⁺] / [CH₃COOH]
1.8 × 10⁻⁵ = (10⁻⁵)² / (y - 10⁻⁵)
y - 10⁻⁵ = (10⁻⁵)² / (1.8 × 10⁻⁵)
y = 1.56 × 10⁻⁵

1.56 × 10⁻⁵ M CH₃COOH is needed.

When volume = 2 L :
No. of moles of CH₃COOH needed = (1.56 × 10⁻⁵ mol/L) × (2 L) = 3.12 × 10⁻⁵ mol

Molar mass of CH₃COOH = 12.0×2 + 1.0×4 + 16.0×2 = 60.0 g/mol
Mass of CH₃COOH needed = 0.00187 g
Volume of water needed ≈ 2 L