The Indicator That Should Not Have Been Chosen: Weak Acid, Strong Base
In a school laboratory a student titrates 25.0 mL of 0.100 mol L⁻¹ ethanoic acid (CH₃COOH, Ka = 1.8 × 10⁻⁵) with 0.100 mol L⁻¹ sodium hydroxide solution, using a pH meter and phenolphthalein. She calculates that 25.0 mL of alkali is needed to reach the equivalence point and writes in her report: “At the equivalence point the acid and the base have reacted completely, so the solution is neutral and pH = 7. The correct indicator is therefore bromothymol blue.” A classmate objects that a weak acid should leave a basic solution at the end point, but cannot say how basic or why. Indicator ranges supplied: methyl orange 3.1–4.4, bromothymol blue 6.0–7.6, phenolphthalein 8.2–10.0. The teacher asks for a calculation rather than an opinion, and draws attention to the pH meter reading the student recorded just before the end point: 7.4 at 24.95 mL.
1. Calculate the pH at the equivalence point and decide whether the claim “pH = 7” survives.
Read the worked answer
Amount of acid: n = 0.0250 × 0.100 = 2.50 × 10⁻³ mol. This is exactly neutralised by 25.0 mL of 0.100 mol L⁻¹ NaOH, so the total volume is 50.0 mL and the only solute of consequence is ethanoate ion:
[CH₃COO⁻] = 2.50 × 10⁻³ / 0.0500 = 0.0500 mol L⁻¹
Ethanoate is the conjugate base of a weak acid, so it hydrolyses:
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 mol L⁻¹ pOH = 5.28 → pH = 8.72
The approximation is safe because C/Kb ≈ 9 × 10⁷, far above the usual threshold. The claim pH = 7 does not survive: neutralisation is complete, but “complete reaction” is not the same as “neutral solution”. The product is a weak base, and the equivalence point must be basic.
2. Choose the indicator and justify the choice using the calculated pH and the ranges given.
Read the worked answer
The indicator must change colour over a range that brackets the equivalence pH of 8.72.
- Methyl orange (3.1–4.4) changes long before the end point, while excess acid is still present - it would signal at a fraction of the required volume and give a grossly low titre.
- Bromothymol blue (6.0–7.6) changes while the solution is still acidic; the first permanent colour change would be recorded early and the calculated concentration would be too low.
- Phenolphthalein (8.2–10.0) brackets 8.72 almost centrally and is therefore correct.
The general rule is that the indicator is chosen to match the pH at the equivalence point, and that pH is decided by the salt formed, not by the strength of the acid or base on its own. Choosing an indicator whose range ends at 7 is the characteristic error of treating every titration as strong acid–strong base.
3. The student recorded pH 7.4 at 24.95 mL. Explain why the titration is still reliable with phenolphthalein, using that reading and the pH at the end point.
Read the worked answer
At 24.95 mL, 5.0 × 10⁻⁶ mol of acid remains unneutralised in 49.95 mL, while 2.495 × 10⁻³ mol of ethanoate is present. This is a buffer:
pH = pKa + log(salt/acid) = 4.74 + log(2.495 × 10⁻³ / 5.0 × 10⁻⁶) = 4.74 + 2.70 = 7.44
which matches the meter. So at 0.05 mL before equivalence the solution is still only faintly basic, and phenolphthalein is colourless. One drop later the remaining acid is consumed and the solution jumps to about pH 10, where phenolphthalein is pink. The curve is nearly vertical between 24.95 mL and 25.05 mL, so the end point is passed one drop after equivalence: an error of roughly 0.05 mL in 25 mL, i.e. 0.2 %, well inside ordinary volumetric precision. The indicator is reliable precisely because a very small volume change produces a very large pH change.