Choosing an Indicator for a Weak Acid–Strong Base Titration
A student titrates 25.00 mL of 0.100 M ethanoic acid (Ka = 1.8 × 10⁻⁵) with 0.100 M sodium hydroxide. They have four indicators available: methyl orange (pH range 3.1–4.4), bromothymol blue (6.0–7.6), phenolphthalein (8.3–10.0) and alizarin yellow (10.1–12.0). The student predicts the equivalence point will be at pH 7 because 'acid plus base makes a neutral salt', and chooses bromothymol blue. A trial run gives an endpoint that appears at 24.6 mL, noticeably before the true equivalence volume of 25.00 mL. The teacher asks the student to justify the indicator choice quantitatively and to explain why the observed endpoint volume is short.
1. Calculate the pH at the equivalence point and use it to decide which indicator is appropriate. Show your working.
Read the worked answer
At equivalence, all ethanoic acid has been converted to ethanoate ion, so the solution contains the conjugate base of a weak acid. Its concentration is diluted by the added volume: total volume ≈ 50.00 mL, so [CH₃COO⁻] = (0.100 × 25.00)/50.00 = 0.0500 M.
For the conjugate base, Kb = Kw/Ka = 1.0 × 10⁻¹⁴ / 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰.
Using Kb = x²/c: x² = 5.6 × 10⁻¹⁰ × 0.0500 = 2.8 × 10⁻¹¹, so x = [OH⁻] ≈ 5.3 × 10⁻⁶ M. Then pOH ≈ 5.28 and pH ≈ 8.72.
The equivalence point is basic, not neutral, because ethanoate hydrolyses: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Phenolphthalein (8.3–10.0) brackets pH 8.72; bromothymol blue (6.0–7.6) does not. The 'neutral salt' prediction is wrong because it ignores hydrolysis of the conjugate base.
2. Explain why the trial endpoint appeared at 24.6 mL and predict the direction of the systematic error in the calculated acid concentration.
Read the worked answer
Bromothymol blue changes colour near pH 7, which is reached before the equivalence point on the titration curve. On the steep region the pH rises rapidly, so a small volume of NaOH produces a large pH change; the indicator signals at 24.6 mL while the true equivalence volume is 25.00 mL.
Because the student would use V(NaOH) = 24.6 mL in c(acid) = c(base)V(base)/V(acid), the calculated acid concentration is lower than the true value by a factor of 24.6/25.00 ≈ 0.984, i.e. about 1.6 % low. This is a systematic error, not random: repeating the titration would give the same short volume. Switching to phenolphthalein, whose range brackets pH 8.72, removes the bias.