Triprotic citric acid requires 3 moles NaOH per mole for complete neutralisation

Steps:

• Moles of citric acid = 0.050 mol dm^{-3} × 0.025 dm³ = 0.00125 mol (25 cm³ solution volume).
• Moles of NaOH needed = 3 × 0.00125 mol = 0.00375 mol (3:1 stoichiometry).
• Volume of NaOH solution = 0.00375 mol / 0.10 mol dm^{-3} = 0.0375 dm³ ≈ 37 cm³.

Why C is correct:

• 37 cm³ matches the volume for 3 equivalents of base per mole of triprotic acid in titration stoichiometry.

Why the others are wrong:

• A: 12 cm³ assumes monoprotic (1:1) behaviour, neutralising only the first proton.
• B: 25 cm³ assumes diprotic (2:1) behaviour, to the second equivalence point.
• D: 50 cm³ assumes lower NaOH concentration or incorrect 4:1 ratio.

Final answer: C