Question

A classic experiment in equilibrium studies dating from 1862 involved the reaction in aqueous solution of ethanol and acetic acid to produce ethyl acetate and water.

C2H5OH(aq) + CH3COOH(aq) <=> CH3COOC2H5(aq) + H2O(l)

The reaction can be followed by analyzing the equilibrium mixture for its acetic acid content using a titration with Ba(OH)2 as shown below:

2CH3COOH(aq) + Ba(OH)2(aq) -> Ba(CH3COO)2(aq) + 2H2O(l)

In one experiment, a mixture of 1.000moles acetic acid and 0.5000 moles ethanol is brought to equilibrium in a 1.000L flask. A 20.00 mL sample of the equilibrium mixture requires 56.74 mL of 0.1000 M Ba(OH)2 for its titration.

A) What is the concentation of the acetic acid found in the titration?

B) What is the mass action expression for the reaction of ethanol and acetic acid?

C) What is the value of the equilibrium constant for the reaction of ethanol and acetic acid?

Answer 1

From the second reaction, 2 moles of acetic acid requires 1 mole of Ba(OH)2

moles of Ba(OH)2 in 0.1M and 56.74ml =0.1*56.74/1000=0.005674moles

0.005674 moles of Ba(OH)2   requires 0.005674*2=0.011348 moles of acetic acid

this moles are present in 20ml Concentration of acetic acid = 0.011348*1000/20=0.5674 M

b) According to law of mass action, Keq= [CH3COOC2H5] {[ C2H5OH] [CH3COOH]}

Concentration of acetic acid at equilibrium= 0.5674

20ml sample contains 0.011348 moles of acetic acid

1000ml sample contains 1000*0.011348/20 =0.5674 moles

Moles of acetic acid consumed = 1-0.5674=0.4326 moles

Moles of Ethanol consumed=0.4326 moles of ethanol at equilibrium =0.5-0.4326=0.0674 moles

Moles of water and CH3COOC2H5 formed= 0.4326 and moles of water fomed =0.4326

Kc= 0.4326*0.4326/ (0.0674*0.4326)=6.42