UPSC Combined Geo-Scientist, Mains · 2025 · Mains Paper I
Q1(d). Phenolphthalein in dilute and in concentrated alkali
Question
Phenolphthalein shows red colour in presence of dilute alkali. However, if treated with excess of concentrated alcoholic alkali, the red colour first produced disappears. Explain how.
Answer in brief
Dilute alkali converts the colourless lactone into the red quinonoid dianion; excess concentrated alkali adds OH⁻ to the central carbon, giving the colourless carbinol trianion with broken conjugation.
Key points
- pH < 8.2: lactone; central C sp³ isolates the rings; absorbs only in UV → colourless.
- pH 8.2–12: loss of 2H⁺ opens the lactone; central C becomes sp² and one ring quinonoid; the long chromophore absorbs near 550 nm → red.
- Excess conc. OH⁻ attacks the electrophilic central C (nucleophilic addition); sp³ again → colourless. Change is slow and reversible on dilution.
- Phenolphthalein is therefore not used above pH 12; thymolphthalein (9.3–10.5) serves at higher pH.
The full model answer
As written for the answer booklet: the diagram carries the working, the lines stay short. Read here only; it is not offered as a download.
Coversphenolphthaleinquinonoid formcarbinolacid-base indicatorcolour change
All parts of Question 1
- 1(a) Defect in silver chloride5 marks
- 1(b) Isomerism in [(NH₃)₅CoONO]Cl₂ and [(NH₃)₅CoNO₂]Cl₂5 marks
- 1(c) Trans effect: Niᴵᴵ, Ptᴵᴵ, Pdᴵᴵ5 marks
- 1(d) Phenolphthalein in dilute and in concentrated alkali5 marks
- 1(e) Solubility product of silver chromate5 marks
- 1(f) Colour change interval of diphenylamine5 marks
- 1(g) Eriochrome Black T and calcium5 marks
- 1(h) Tendency to form metal clusters5 marks
- 1(i) Complete the nuclear equations5 marks
- 1(j) First ionisation enthalpy of 5d elements5 marks