Titration of Weak Acid with Strong Base

Titration Curve Overview

• Titration of acetic acid (CH₃COOH) with sodium hydroxide (NaOH).
• Various points on titration curve discussed.

Part A: Initial pH

• pH before any base is added.
• Location on titration curve established.

Part B: pH After 100 mL of Base

• Determination of pH after adding 100 mL of NaOH.
• Established location on titration curve.

Part C: pH After 200 mL of Base

• Objective: Find pH after adding 200 mL of 0.05 M NaOH.
  • Calculations:
    • Moles of hydroxide ions = 0.05 M * 0.2 L = 0.01 moles.
    • Initial moles of acetic acid = 0.2 M * 0.05 L = 0.01 moles.
    • Equal moles of acid and base (0.01 moles each).
  • Reaction:
    • Acetic acid neutralized by NaOH, forming acetate (CH₃COO⁻).
    • Acetate concentration = 0.01 moles / 0.25 L = 0.04 M.
  • Equilibrium Reaction:
    • Acetate reacts with water:
      • CH₃COO⁻ + H₂O ↔ CH₃COOH + OH⁻
    • Initial concentration of acetate = 0.04 M.
  • Equilibrium Expression:
    • Kb = [OH⁻][CH₃COOH] / [CH₃COO⁻]
    • Kb found using Ka for acetic acid (1.8 × 10⁻⁵); Kb = 5.6 × 10⁻¹⁰.
  • Calculating pH:
    • x = [OH⁻] = 4.7 × 10⁻⁶ M.
    • pOH = -log([OH⁻]) = 5.33.
    • pH = 14 - 5.33 = 8.67.
  • Result: pH at equivalence = 8.67 (basic solution).

Part D: pH After 300 mL of Base

• Objective: Determine pH after 300 mL of 0.05 M NaOH.
  • Calculations:
    • Moles of hydroxide ions added = 0.05 M * 0.3 L = 0.015 moles.
    • Neutralized acid requires 0.01 moles, excess hydroxide = 0.005 moles.
    • Concentration of remaining hydroxide = 0.005 moles / 0.35 L = 0.014 M.
  • Calculating pH:
    • pOH = -log([OH⁻]) = 1.85.
    • pH = 14 - 1.85 = 12.15.
  • Result: pH after 300 mL of base is 12.15.

Conclusion

• pH at different stages of titration calculated.
• Equivalence point and beyond shows basic pH.
• Titration curve visualized with pH values at added volumes.