Experiment 6:
Chemical Equilibrium—The Hydrolysis of Ethyl Acetate

 

Objectives:

ü       Determine the value of the equilibrium constant for a reaction

ü       Use acid-base titrations and solution stoichiometry in determining the equilibrium constant

 

Definitions:

ü       Equilibrium – a state of balance in a chemical reaction in which the forward and backward rates are equal.

 

Background Material:

For the reaction below, when A and B were mixed, the reaction proceeds in the forward direction to produce C and D.  However, as time progresses, the concentration of C and D increases causing an increase in the rate of the reverse reaction.  Concurrent with this increased rate of the reverse reaction is a reduction of the forward rate due to the decrease in the concentration of A and B.  At some point, the rate of the forward and reverse reactions will become the same and we will reach a state of dynamic equilibrium:

 

aA  +  bB    cC  +  dD

 

Equilibrium does not mean that the forward and reverse reactions have stopped.  Molecules of A and B are still reacting to form C and D and molecules of C and D are reacting to form A and B.  However, since the rate of the forward and reverse reactions is the same, it will appear that nothing is happening.  As such, all quantifiable physical and chemical properties such has pH, color, and concentration will remain constant.

 

For a general equilibrium equation we can specify an equilibrium constant, Kequil, that relates the concentrations of all product and reactant species,

where [A], [B], [C], and [D] are the molar concentration of all species present at equilibrium.  The exponents, a, b, c, and d represent the stoichiometry coefficients from the balance chemical reaction.  Kequil is really the ratio of the rate of the reverse reaction divided by the rate of the forward reaction and so is a dimensionless constant at a given temperature.

 

We will be studying the acid catalyzed (HCl) hydrolysis of an ester (ethyl acetate, EtAc), to form an alcohol (ethanol, EtOH) and an acid (acetic acid, HAc):

 

CH3CH2CO2CH +  H2O    CH3CH2OH  +  CH3CO2H

EtAc  +  H2O    EtOH  +  Hac

 

This hydrolysis reaction will occur spontaneously at room temperature and the addition of acid only affects the amount of time it takes the reaction to reach equilibrium.

 

In practice it is often difficult to determine the equilibrium concentration of all these species.  However, we can calculate the equilibrium constant by determining the concentration of a single species at equilibrium, if we know the initial concentration of all species.  We do not have to be concerned about the concentration of the hydrochloric acid since it only serves as a catalyst.  For our specific reaction, the equilibrium constant will equal:

 

Since we will know the initial concentration of each component, we only need to determine the equilibrium concentration of one component to calculate the equilibrium constant.  Although we have several chemical species to choose from, we will monitor the acetic acid concentration since it can be very accurately determined via a simple titration with standardized base.

 

In the above discussion, we decided to monitor the concentration of the HAc to determine the equilibrium constant.  However, the presence of the strong acid HCl complicates our task.  A simple titration with sodium hydroxide will neutralize all the acids that are present and we will not be able to differentiate the amount of base needed to neutralize the HCl versus the amount that was needed to neutralize the HAc.  This is a common problem in chemistry and is solved by titrating a blank, which only contains the HCl (Solution 1).  Since it contains a known amount of HCl and water, a simple titration will determine the amount of base needed to neutralize the HCl contained in the other solutions.  This amount of HCl will be constant in the other solutions because the HCl is a catalyst that is not consumed by the reaction.  Now that we know the amount of base needed to account for the HCl, any additional base used in the titration of Solution #3/4, Solution #5/6, and Solution #7/8, must represent the amount of HAc present.

 

Sample calculation:

The volume of 1.090 M NaOH used for titrations was:

 

Vials 1 & 2

Vials 3 & 4

Vials 5 & 6

Vials 7 & 8

13.48 ml

34.74 ml

20.25 ml

29.60 ml

 

Vials 1 and 2:

Remember that molarity x volume = moles so:

1.090 M x .01348 L = .0147moles HCl

 

To get grams of HCl present:

.0147 moles HCl x 36.45 g/mole = 0.536 g HCl in 5 mL

 

What is the total mass of the 5 mL of HCl?

1.0436 g/mL x 5 mL = 5.218 g HCl

 

What is the mass of water in the 5 mL of HCl?

5.218 g – 0.536 g = 4.682 g H2O

 

How many moles of water present in the 5 mL of HCl?

4.682 g H2O ÷ 18 g/mole = 0.260 mole water in 5 mL

 

Vials 3 and 4:

It may help you to set up a small table looking at the initial and final conditions of the reaction

 

   

EtAc 

H2O

EtOH 

HAc 

Initial: 

3 mL 

2 mL + 0.260 mol

0 mL

0 mL

Change: 

- x

- x

+ x

+ x

Calculate x and you’re set.  You get x from the titration information:

 

Moles base used in titration:

1.090 M NaOH x .03474 L = 0.03787 moles base

 

Since the stoichiometry of the reaction is 1 to 1, moles base = moles acid:

0.03787 moles base = 0.03787 moles acid

 

This is NOT x because the titration only gives you moles of total acid, don’t forget about HCl:

0.03787 moles acid – 0.0147 moles HCl = 0.02317 moles acetic acid = x

 

Final moles ethanol is x to so moles ethanol = 0.02317 moles = x

 

How many moles of ethyl acetate present initially:

3 mL x 0.894 g/ml ÷ 88g/mol = 0.03056 moles ethyl acetate initially

 

How many moles of water initially:

2 mL x 1 g/mL ÷ 18 g/mol = 0.1111 moles water initially

 

Let’s revisit the table:

 

 

EtAc(aq) 

H2O(soln) 

EtOH(aq) 

HAc(aq) 

Initial: 

.03056 mole

0.1111 + .026 mole

0 mole

0 mole

Change: 

-.02317 mole

-.02317 mole

+.02317 mole

+.02317 mole

Equilibrium: 

.00739

.34793 mole

.02317 mole

.02317 mole

 

Now we can calculate the equilibrium constant for this reaction as:

Kequil =

Kequil = 0.209

 

Vials 5 & 6:

Repeat the above.  Be careful with x!!!  Remember that the titration only tells you the moles of total acid present.  Vials 5 and 6 already have acid present initially.  Watch out!!!

 

Vials 7 & 8:

Repeat the above.


Items for lab report:

ü       Create a table like the one on page 59.

ü       Be sure to include in your table the volume of NaOH used in the titrations (you will have duplicates for each set of mixtures, average the volume of NaOH used and do 1 calculation for each set of mixtures).

ü       Show all calculations for sets 1/2, 3/4, 5/6, and 7/8

ü       Be very clear in your presentation—I WILL NOT search for your answers!!!