Student X titrated 25 cm ^ 3 of Na 2 CO ^ 3 with 0.1 mol dm ^ -3 HCI, using methyl orange as indicator…

Analyze the question

Na₂CO₃ is a dibasic salt of a weak acid. It reacts with HCl in two stages, and each indicator detects a different stage.

Explain the concept

• Stage 1: Na₂CO₃ + HCl → NaHCO₃ + NaCl. The pH at the end of this stage is about 8.3, and phenolphthalein changes colour here (range 8.3–10).
• Stage 2: NaHCO₃ + HCl → NaCl + H₂O + CO₂. The pH at the end of this stage is about 3.5–4, and methyl orange changes colour here (range 3.1–4.4).

Solution

• Student Y (phenolphthalein) only reaches the end of stage 1. This needs 1 mole of HCl per mole of Na₂CO₃, so the titre is V.
• Student X (methyl orange) reaches the end of stage 2. This needs 2 moles of HCl per mole of Na₂CO₃, so the titre is 2V.

So X's titre is twice Y's titre.

Checking the other options

• A is wrong because HCl gas is not released. CO₂ is released, and only in the second stage.
• B is wrong because the titres are different (2V and V).
• D is wrong because it reverses the relationship. Y's titre is half of X's, not double.

Answer: C

Tip: Phenolphthalein "sees" only half of the carbonate neutralisation, while methyl orange sees all of it. So methyl orange titre = 2 × phenolphthalein titre. Remember "Methyl orange = Most acid = Maximum volume."