Choosing an Indicator and Buffering in a Weak-Acid Titration
A student titrates 25.00 mL of 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵) with 0.100 M NaOH. The lab has three indicators: methyl orange (pH 3.1–4.4), bromothymol blue (pH 6.0–7.6), and phenolphthalein (pH 8.2–10.0). The student plans to use bromothymol blue because 'neutralization makes a neutral solution, so a pH-7 indicator is correct.' A second student argues the equivalence point is basic because acetate hydrolyzes. The instructor asks them to settle it with calculations and to predict the pH at the half-equivalence point.
1. Calculate the pH at the equivalence point and decide which indicator is appropriate.
Read the worked answer
Moles acid: 0.02500 L × 0.100 M = 2.500 × 10⁻³ mol. At equivalence, an equal moles of NaOH is added (25.00 mL), so total volume = 50.00 mL and [acetate] = 2.500×10⁻³ / 0.05000 = 0.0500 M.
Hydrolysis: CH3COO⁻ + H2O ⇌ CH3COOH + OH⁻, Kb = Kw/Ka = 1.0×10⁻¹⁴ / 1.8×10⁻⁵ = 5.6×10⁻¹⁰. [OH⁻] = √(Kb·C) = √(5.6×10⁻¹⁰ × 0.0500) = √(2.8×10⁻¹¹) ≈ 5.3×10⁻⁶ M. pOH ≈ 5.28, so pH ≈ 8.72.
Indicator choice: The equivalence pH (~8.7) lies in phenolphthalein's range (8.2–10.0), not bromothymol blue's. The second student is right: the salt of a weak acid is basic, so the endpoint is basic. Bromothymol blue would change color before the true equivalence point, causing a systematic low bias in the calculated acid concentration.
2. Calculate the pH at the half-equivalence point and explain why this value is a useful check on Ka.
Read the worked answer
At half-equivalence, half the acid has been converted to acetate, so [CH3COOH] = [CH3COO⁻]. Using the Henderson–Hasselbalch form: pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa.
pKa = −log(1.8×10⁻⁵) = 4.74, so pH ≈ 4.74 at half-equivalence.
This is a powerful check: the pH at half-equivalence equals pKa regardless of concentration, so a measured pH there directly yields Ka. If the student's measured half-equivalence pH disagreed with 4.74, it would signal a concentration error, a bad electrode calibration, or contamination - not a changed Ka.
3. The student accidentally overshoots and adds 5.00 mL of NaOH past the equivalence point. Estimate the pH and explain why the curve is flat there.
Read the worked answer
Excess OH⁻: 0.00500 L × 0.100 M = 5.00 × 10⁻⁴ mol in total volume 25.00 + 25.00 + 5.00 = 55.00 mL. [OH⁻] = 5.00×10⁻⁴ / 0.05500 = 9.09×10⁻³ M → pOH ≈ 2.04 → pH ≈ 11.96.
Why flat: Past equivalence, added OH⁻ is essentially unneutralized (acetate is a very weak base and does not consume strong base appreciably), so pH is governed by simple dilution of excess strong base. Each added drop changes [OH⁻] only modestly, producing the long, nearly horizontal post-equivalence plateau. This is why phenolphthalein's broad range tolerates small overshoots - but the student should still stop at the first persistent pink, not 5 mL past.