Choosing an Indicator for a Weak Acid–Strong Base Titration
In a Class 12 practical examination, a student must titrate 25.0 mL of 0.100 M acetic acid (Ka = 1.8 × 10−5 at 25 °C) with 0.100 M sodium hydroxide. Two indicators are on the bench: methyl orange (colour change pH 3.1–4.4) and phenolphthalein (pH 8.3–10.0). The student first argues that methyl orange is correct because acetic acid is a weak acid and the conical flask is acidic at the start. A classmate replies that the equivalence point is not at pH 7. The laboratory has only narrow-range pH paper, not a pH meter. In a trial run, 25.0 mL of NaOH was added and the phenolphthalein turned pink; after two minutes of swirling the pink faded. The teacher asks the student to calculate the pH at half-equivalence and at equivalence, justify an indicator, and explain the fading. Assume the solutions are at 25 °C and activity effects are negligible.
1. Calculate the pH at half-equivalence and at the equivalence point. Which indicator should be used, and why is the other choice wrong?
Read the worked answer
Half-equivalence: 12.5 mL of NaOH has been added, so half of the acid has been converted to acetate. Here [CH3COOH] ≈ [CH3COO−], and the Henderson–Hasselbalch equation gives
pH = pKa + log([A−]/[HA]) = 4.74 + log(1) = 4.74.
Equivalence: total volume = 25.0 + 25.0 = 50.0 mL. Moles of acetate formed = 0.100 × 0.0250 = 2.50 × 10−3 mol, so [CH3COO−] = 2.50 × 10−3 / 0.0500 = 0.0500 M.
Kb = Kw/Ka = 1.0 × 10−14 / 1.8 × 10−5 = 5.56 × 10−10.
[OH−] = √(Kb × C) = √(5.56 × 10−10 × 0.0500) = 5.27 × 10−6 M; pOH = 5.28; pH = 8.72.
The equivalence pH is therefore basic, not neutral. Phenolphthalein changes over pH 8.3–10.0 and brackets 8.72, so it is suitable. Methyl orange changes at pH 3.1–4.4, which lies in the buffer region well before equivalence; it would change colour after only a small fraction of the acid has been neutralised, giving a very low titre.
2. Account for the fading pink colour after 25.0 mL of NaOH had been added. Does it mean the equivalence point has not been reached? Give two procedural precautions.
Read the worked answer
At 25.0 mL the solution is at the equivalence point and pH ≈ 8.7, so phenolphthalein is pink. The fading is not evidence that the acid is still unneutralised. Two effects can lower the pH at the surface or in the bulk solution: absorption of atmospheric CO2, which produces carbonic acid and hydrogen carbonate and buffers the solution to a lower pH; and slight overshoot past the true endpoint, so that the excess alkali is slowly consumed by CO2. If the pink disappears within a minute or two and does not return on swirling, the endpoint was probably taken slightly late or the water contained dissolved CO2.
Precautions: (1) use freshly boiled and cooled distilled water, or boil the titration flask briefly to expel CO2; (2) titrate to the first permanent pale-pink colour that persists for about 30 s, adding NaOH dropwise near the endpoint, and keep the flask stoppered or covered as far as practicable. The appearance of a persistent pink after 25.0 mL is consistent with the calculated equivalence volume.
3. Estimate the titre that would be recorded if methyl orange were used, and state the sign and approximate size of the error.
Read the worked answer
Methyl orange changes over pH 3.1–4.4, so take the endpoint as pH ≈ 4.0–4.4. Rearranging the buffer equation:
[A−]/[HA] = 10^(pH − pKa).
At pH 4.0: ratio = 10^(−0.74) = 0.18, so the fraction neutralised is 0.18/(1 + 0.18) = 0.15. Titre ≈ 0.15 × 25.0 = 3.8 mL.
At pH 4.4: ratio = 10^(−0.34) = 0.46, fraction = 0.46/1.46 = 0.31. Titre ≈ 0.31 × 25.0 = 7.8 mL.
Thus methyl orange would give an endpoint somewhere around 4–8 mL instead of 25.0 mL. The recorded titre is far too low, so the calculated concentration of the acetic acid would be too low by roughly 70–85%. The error is systematic and large because the indicator changes in the buffer region, not at the steep part of the titration curve.