UPSC Combined Geo-Scientist, Mains · 2016 · Mains Paper III

Q1(f). Butyric-acid extraction

Question

20 mL of 0.10 M butyric acid is shaken with 10 mL ether. The paper says 0.5 mol remains in water. Find distribution ratio and percentage extracted.

The printed quantity is impossible. If 0.5 mmol was intended, D=6 and extraction is 75%.

Initial amount is 0.020(0.10)=0.0020 mol, so 0.5 mol cannot remain.
With the stated correction, n_(aq)=0.0005 mol and n_(org)=0.0015 mol.
D=C_(org)/C_(aq)=(0.0015/0.010)/(0.0005/0.020)=6. Extracted fraction =0.0015/0.0020=0.75.
This is a total analytical distribution ratio; ionisation in water can make it differ from the partition coefficient of neutral acid.

The full model answer

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CoversButyric-acid extractionSolvent extraction