A student performs a titration of a 25.0 mL sample of 0.100 M acetic acid (CH₃COOH) with 0.100 M sodium hydroxide (NaOH). What is the pH at the equivalence point of this titration? (K_a for acetic acid = 1.8 × 10⁻⁵)
- A2.87
- B8.72
- C11.13
- D7.00
Show answer & explanationAnswer & explanation
Correct answer: B. 8.72
At the equivalence point of a weak acid-strong base titration, all the weak acid has reacted to form its conjugate base. The solution will contain the conjugate base (acetate, CH₃COO⁻), which will hydrolyze water to produce OH⁻ ions, making the solution basic. 1. Moles of CH₃COOH = 0.025 L × 0.100 M = 0.0025 mol. 2. Moles of NaOH needed = 0.0025 mol. Volume of NaOH needed = 0.0025 mol / 0.100 M = 0.025 L = 25.0 mL. 3. Total volume at equivalence point = 25.0 mL + 25.0 mL = 50.0 mL = 0.050 L. 4. Concentration of CH₃COO⁻ at equivalence point = 0.0025 mol / 0.050 L = 0.050 M. 5. The acetate ion hydrolyzes: CH₃COO⁻(aq) + H₂O(l) ⇌ CH₃COOH(aq) + OH⁻(aq). 6. K_b for CH₃COO⁻ = K_w / K_a = (1.0 × 10⁻¹⁴) / (1.8 × 10⁻⁵) = 5.56 × 10⁻¹⁰. 7. Set up ICE table: [CH₃COO⁻] = 0.050 - x, [CH₃COOH] = x, [OH⁻] = x. 8. K_b = x² / (0.050 - x) ≈ x² / 0.050 (assuming x is small). 9. x² = 5.56 × 10⁻¹⁰ × 0.050 = 2.78 × 10⁻¹¹. 10. x = [OH⁻] = √(2.78 × 10⁻¹¹) = 5.27 × 10⁻⁶ M. 11. pOH = -log(5.27 × 10⁻⁶) = 5.28. 12. pH = 14.00 - pOH = 14.00 - 5.28 = 8.72.
Why the other options are wrong
- A. This pH would be characteristic of a weak acid solution before the equivalence point.
- C. This pH would be characteristic of a much stronger base or a solution well past the equivalence point.
- D. pH 7.00 occurs at the equivalence point only when titrating a strong acid with a strong base.
Weak Acid-Strong Base Titration Equivalence Point
At the equivalence point of a weak acid-strong base titration, the solution contains the conjugate base of the weak acid, which hydrolyzes water to produce hydroxide ions, resulting in a basic pH.
- Moles of acid = moles of base at equivalence point.
- The pH at the equivalence point is > 7 due to conjugate base hydrolysis.
- Need to calculate K_b from K_a (K_a * K_b = K_w) for the conjugate base.
- Use an ICE table to find [OH⁻] from the conjugate base hydrolysis, then calculate pOH and pH.
Memory trick: Conjugate base makes it basic; remember K_w/K_a!