A mixture of 0.5 mole of ethanoic acid and 1 mole of ethanol was allowed to reach equilibrium at room temperature. The equilibrium mixture was quickly titrated and 40cm3 of 2moldm-3 sodium hydroxide was used in the titration.

Question

a) Why is the reaction titrated quickly?
b) Using the titration results, calculate the number of moles of
i) ethanoic acid left in the equilibrium mixture
ii) ethanoic acid reacted
iii) ethanol reacted
iv)ethanol left in the equilibrium mixture

c) Calculate the equilibrium constant for the reaction between ethanoic acid and ethanol at room temperature.

Answer 1

This is an equilibrium reaction which means that it proceeds in the forward direction BUT ALSO in the reverse direction. If you change the concn of acetic acid by neutralizing part of it with the NaOH titrant you change the point of equilibrium. If you do it quickly you don't make much error since the change is slow.

b) Using the titration results, calculate the number of moles of
i) ethanoic acid left in the equilibrium mixture
ii) ethanoic acid reacted
iii) ethanol reacted
iv)ethanol left in the equilibrium mixture

c) Calculate the equilibrium constant for the reaction between ethanoic acid and ethanol at room temperature.

mols NaOH used = M x L = 2 x 0.04 = 0.08
mols HAc (CH3COOH) used = 0.08
mols HAc left at equilibrium = 0.5-0.08 = 0.43
Now set up and ICE chart.
....................CH3COOH + HOC2H5 --> CH3COOC2H5 + H2O
I....................0.5....................1.0..................0.......................0
C.................-0.08.................-0.08..............+0.08...............+0.08
E....................?..........................?....................?.......................?

Plug the E line into Kc expression and calculate Kc.
Post your work if you get stuck. Check my work.