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 N-NaOH. When the mixture reaches the equilibrium, 10

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 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? ​

TestHub · Accepted Answer

Acid initially in 10 mL = 10 meq Acid at equilibrium in 10 mL = 5 meq Ester formed = 10 - 5 = 5 meq, water formed = 5 meq Alcohol left = 10 - 5 = 5 meq K c ​ = [ester][H 2 ​ O]/([acid][alcohol]) = (5)(5)/[(5)(5)] The volume cancels as Δ n = 0 K c ​ = 1 ​

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