A mixture of equimolar quantities of ethyl alcohol and acetic acid is prepared. Immediately after mixing, 10 ml of mixture was neutralized by 10 ml of N-NaOH. When the mixture reaches the equilibrium,

Question

A mixture of equimolar quantities of ethyl alcohol and acetic acid is prepared. Immediately after mixing, 10 ml of mixture was neutralized by 10 ml of N-NaOH. When the mixture reaches the equilibrium, 10 ml requires 5 ml of N-NaOH for neutralization. What is the equilibrium constant for the formation of ester?

Prashant Kotian · Accepted Answer

Correct answer: 1. The reaction is Alcohol + Acid ⇌ Ester + Water. Initially, 10 ml of the mixture needs 10 ml NaOH, which measures the initial moles of acid. Let initial moles of acid = a (proportional to 10 ml NaOH). At equilibrium, 10 ml of the mixture requires 5 ml NaOH. So, remaining acid = a - x (proportional to 5 ml NaOH). Thus, a - x = a/2, meaning x = a/2. Equilibrium moles: Acid = a/2, Alcohol = a/2 (since equimolar initially), Ester = a/2, Water = a/2. K_c = ([Ester][Water]) / ([Acid][Alcohol]) = (a/2 × a/2) / (a/2 × a/2) = 1. Therefore, correct answer is 1.

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