0.1 mole of each C 2 H 5 OH and CH 3 COOH when allowed to react in 100 ml of non aqueous solution, it is seen 10 ml of the equilibrium mixture require 80 ml of 0.1 (N) NaOH for complete neutralisation. The equilibrium constant, For the reaction CH 3 COOH + C 2 H 5 OH 
CH 3 COO C 2 H 5 + H 2 O is expressed as K C and the value of 32 K C is –
Text Solution
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2
Sol. 
In 10 ml solution at equilibrium moles of
CH 3 COOH present = 0.1 (0.1 – x) ≡ ≡ 0.1 (0.1 – x)
moles of NaOH
∴ 0.1 (0.1 – x) = 80 × 10 –4 = 8 × 10 –3 or 0.01 – 0.1 x = 0.008
or x = 0.02
K C = 
= 32 K C = 32 × 
= 2
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