Acid-base Titration

Question

A standard solution was prepared by dissolving 0.63 g of pure oxalic acid dihydrate in water and making up to 250 mL. What volume of 0.10 M sodium hydroxide is needed to neutralise 25 mL of this solution?

RankGuru · Accepted Answer

Oxalic acid dihydrate, H_2C_2O_4·2H_2O, has a molar mass of 126 g mol^{-1}, so 0.63 g corresponds to 0.63/126 = 0.005 mol in 250 mL, giving a molarity of 0.005/0.250 = 0.02 M. Oxalic acid is diprotic, providing two ionisable protons per molecule, so 25 mL of the 0.02 M solution contains 0.02 × 25 × 2 = 1.0 milliequivalent of acid. At the equivalence point the milliequivalents of base equal those of acid, so the sodium hydroxide must supply 1.0 milliequivalent. Since sodium hydroxide is monoacidic, its normality equals its molarity at 0.10 N, and the required volume is 1.0/0.10 = 10 mL. Option 5 mL ignores the diprotic nature of oxalic acid and treats it as supplying only one proton. Option 20 mL doubles the acid equivalents incorrectly. Option 25 mL assumes equal concentrations and a one-to-one mole ratio, which is not the case here. This stoichiometric reasoning underlies the NCERT standard titration exercise. Plausibility check: the more dilute, diprotic acid needs a modest volume of the more concentrated base, and 10 mL is consistent in order of magnitude.

Acid-base Titration

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