A mixture of equimolar quantities of propan-1-ol and propanoic acid is prepared. Immediately after mixing, 10 mL of the mixture was neutralized by 30 mL of 0.5 N NaOH. When the mixture reaches equilib

Question

A mixture of equimolar quantities of propan-1-ol and propanoic acid is prepared. Immediately after mixing, 10 mL of the mixture was neutralized by 30 mL of 0.5 N NaOH. When the mixture reaches equilibrium, 10 mL requires 10 mL of 0.5 N NaOH for neutralization. What is the equilibrium constant for the formation of ester?

TestHub · Accepted Answer

Acid initially in 10 mL = (30)(0.5) = 15 meq Acid at equilibrium in 10 mL = (10)(0.5) = 5 meq

Ester formed = 15 - 5 = 10 meq, water formed = 10 meq

Alcohol left = 15 - 10 = 5 meq

K_c = [ester][H₂O]/([acid][alcohol]) = (10)(10)/[(5)(5)]

The volume cancels as Δn = 0

K_c = 4

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