Finding the Equilibrium Constant for the Ester Formation of Ethyl ethanoate

Authors Avatar
Finding the Equilibrium Constant for the Ester Formation of Ethyl ethanoate

Michael Zuber

Introduction:

In this experiment 6 different mixtures of known amount of acid, alcohol and/or ester were mixed with dilute hydrochloric acid in a stoppered bottle. The mixture is then set aside for a week to reach equilibrium at room temperature. The mixture is then titrated against a solution of 1 mol.dm-3 NaOH to determine the total amount of acid present in the equilibrium. From this the amount of ethanoic acid can be found and therefore the amount of the other reagents can be found. These values can then be used to find the equilibrium constant for the reaction.

The equation for the reaction is:

Raw Data:

Below is a table of the different volumes of the chemicals initially added in each solution:

Solution

HCl (cm3)

Water (cm3)

Ethyl ethanoate (cm3)

Ethanoic acid (cm3)

Ethanol (cm3)

5

0

5

0

0

2

5

4

0

0

3

5

3

2

0

0

4

5

0

4

0

5

5

0

0

4

6

5

0

0

2

3

Below is a table of the volume of NaOH needed to neutralize the different solutions at equilibrium:

Solution

NaOH (cm3) used to neutralise soltuion (1)

NaOH (cm3) used to neutralize soltuion (2)

NaOH (cm3) used to neutralise soltuion (3)

44.2

43.7

42.9

2

44.5

45.2

3

32.4

32.1

31.1

4

52.9

50.9

51.8

5

22.8

23

6

37.5

36.4

From these values we can find an average amount of 1 mol.dm-3 NaOH needed to neutralize each solution can be found. Below is a table with these values:

Solution

Average amount of NaOH (cm3)

Random Error (±cm3)

Reading Error (±cm3)

Absolute Error (±cm3)

43.6

0.7

0.05

0.7

2

44.9

0.4

0.05

0.4

3

31.9

0.6

0.05

0.7

4

51.9

.0

0.05

.1

5

22.9

0.1

0.05

0.2

6

37.0

0.6

0.05

0.6

By taking an average there will be an inherent random error. This random error is equal to the (max titer - min titer)/2.

Along with the random error is a reading error. This reading error is equal to 1/2 the smallest division of the instrument used to make the measurement. Since we used a biurette the smallest division was 0.1cm3 therefore the reading error is 0.05cm3.

Since both of these errors are of the same units they can be added together to find an absolute error. The random error, reading error and absolute error are all shown in the above table.
Join now!


From these averages we can determine the amount of moles NaOH needed to neutralize the solution and since both HCl and ethanoic acid react 1 : 1 with NaOH the amount of moles of NaOH reacted will equal the total amount of moles of acid reacted.

Since we know that the concentration of the HCl is 3 mol.dm3 and in each solution 5cm3 of HCl was used, we can find the moles of HCl in the left over acid.

Mole HCl = 0.005 x 3 = 0.015 mol

This value can be subtracted from each ...

This is a preview of the whole essay