pH of Weak Acids
Practice Questions
Kₐ Expression:
Kₐ = [CH₃COO⁻][H⁺] / [CH₃COOH]
Calculation:
Assume [H⁺] ≈ √ (Kₐ × [HA])
[H⁺] = √ (1.74×10⁻⁵ × 0.5) = 2.95×10⁻³ mol dm⁻³
pH = -log(2.95×10⁻³) = 2.53
Kₐ Expression:
Kₐ = [CH₃CH₂COO⁻][H⁺] / [CH₃CH₂COOH]
Calculation:
Kₐ = 10⁻⁴·⁸⁷ = 1.35×10⁻⁵ mol dm⁻³
[H⁺] = √ (1.35×10⁻⁵ × 1.0) = 3.67×10⁻³ mol dm⁻³
pH = -log(3.67×10⁻³) = 2.44
[H⁺] = √ (6.6×10⁻⁴ × 2) = 0.0363 mol dm⁻³
pH = -log(0.0363) = 1.44
[H⁺] = 10⁻³·²² = 6.03×10⁻⁴ mol dm⁻³
Rearranging approximation: [HA] = [H⁺]² / Kₐ
[HA] = (6.03×10⁻⁴)² / 1.74×10⁻⁵
[HA] = 0.0209 mol dm⁻³
Kₐ = 10⁻⁴·⁸² = 1.51×10⁻⁵ mol dm⁻³
[H⁺] = √ (1.51×10⁻⁵ × 0.1) = 1.23×10⁻³ mol dm⁻³
pH = -log(1.23×10⁻³) = 2.91
Kₐ = 10⁻⁴·⁸⁷ = 1.35×10⁻⁵ mol dm⁻³
[H⁺] = 10⁻⁴·¹⁰ = 7.94×10⁻⁵ mol dm⁻³
[HA] = [H⁺]² / Kₐ
[HA] = (7.94×10⁻⁵)² / 1.35×10⁻⁵
[HA] = 4.67×10⁻⁴ mol dm⁻³
Step 1: New Concentration
Total Volume = 150cm³
Dilution Factor = 100 / 150 = 2/3
[HA] = 0.2 × (2/3) = 0.133 mol dm⁻³
Step 2: Calculate pH
[H⁺] = √ (1.74×10⁻⁵ × 0.133) = 1.52×10⁻³ mol dm⁻³
pH = -log(1.52×10⁻³) = 2.82
Step 1: Find Diluted Concentration
Final pH = 4.0 → [H⁺] = 10⁻⁴ mol dm⁻³
[HA]final = [H⁺]² / Kₐ = (10⁻⁴)² / 1.78×10⁻⁴
[HA]final = 5.62×10⁻⁵ mol dm⁻³
Step 2: Find Original Concentration
Total Volume = 700cm³; Initial Volume = 200cm³
Dilution Factor = 200 / 700 = 2/7
[HA]original = 5.62×10⁻⁵ / (2/7) = 1.97×10⁻⁴ mol dm⁻³
[H⁺] = √ (1.78×10⁻⁴ × 1.97×10⁻⁴)
[H⁺] = 1.87×10⁻⁴ mol dm⁻³
pH = -log(1.87×10⁻⁴) = 3.73