Choosing an Indicator for a Weak-Acid Titration
A student must titrate 25.0 mL of 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵) with 0.100 M NaOH. Three indicators are available: methyl orange (colour change pH 3.1–4.4), bromothymol blue (pH 6.0–7.6) and phenolphthalein (pH 8.2–10.0). The student argues that because the acid and base react in a 1:1 ratio, any indicator will do, and picks methyl orange because 'the solution starts acidic'. The equivalence point is reached after 25.0 mL of NaOH is added.
1. Calculate the pH at the equivalence point and use it to decide which indicator is appropriate.
Read the worked answer
At equivalence: moles of acid = 0.100 × 0.0250 = 2.50 × 10⁻³ mol. Total volume = 25.0 + 25.0 = 50.0 mL. The solution contains acetate ion at concentration 2.50 × 10⁻³ / 0.0500 = 0.0500 M.
Acetate is a weak base: 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 ≈ 8.7.
The equivalence point is basic, so phenolphthalein (8.2–10.0) is correct. Methyl orange changes at pH 3.1–4.4, far too early - it would signal the end point long before equivalence, giving a large negative error.
2. Estimate the pH after only 12.5 mL of NaOH has been added and explain why this rules out methyl orange even more strongly.
Read the worked answer
Half-equivalence point: 12.5 mL of NaOH neutralises half the acid, so [CH₃COOH] = [CH₃COO⁻].
By the Henderson–Hasselbalch equation, pH = pKa + log([A⁻]/[HA]) = pKa + log(1) = pKa.
pKa = −log(1.8 × 10⁻⁵) ≈ 4.74.
At this point the solution is still acidic (pH 4.74), which is exactly where methyl orange is finishing its colour change. So methyl orange would appear to 'end' near half-neutralisation - the student would stop the titration after adding only about half the required base, a ~50% error.
Judgement: the 1:1 stoichiometry fixes how much base is needed, not what pH the mixture has at that point. Indicator choice depends on the pH at equivalence, which depends on the salt formed.