Volumetric Analysis (7): A Back Titration and Equilibria

How to use a Back Titration to determine the Kc for the ethanol/ethanoic acid equilibrium.

We know that ethanol and ethanoic acid react reversibly to form ethyl ethanoate and water according to this equation:

CH₃COOH + CH₃CH₂OH  →  CH₃COOCH₂CH₃ + H₂O

A mixture of 8.00 × 10^–2 mol of ethanoic acid and 1.20 × 10^–1 mol of ethanol is allowed to reach equilibrium at 20 °C and the equilibrium mixture then placed in a 250 cm³ graduated flask.

The volume is then made up to 250 cm³ with distilled water.

A 10.0 cm³ sample of this equilibrium mixture is titrated with sodium hydroxide added from a burette.

The ethanoic acid in this sample reacts with 3.20 cm³ of 2.00 × 10^–1 mol dm³ (0.2M) sodium hydroxide solution.

Use the above results to calculate the value for Kc for the reaction of ethanoic acid and ethanol at 20 °C.

Give your answer to the appropriate number of significant figures.

How to calculate the Kc value

Let’s begin with the titration results.

First thing to think about is why the sodium hydroxide was used and what it was doing in the practical.

You need to realise that the sodium hydroxide, being a strong alkali, is going to react with an acid and the only acid available is the ethanoic acid in the equilibrium mixture.  

So using the sodium hydroxide data will get us to the number of moles of ethanoic acid left after the equilibrium was set up.  

Let’s then calculate the number of moles of sodium hydroxide solution that was used to neutralise the residual ethanoic acid in the equilibrium mixture.

Here we need to use n=cV

So           n (mol NaOH)  =     3.2ml   ×   0.2moldm^-3/1000  =  0.00064moles

The other thing we need now to refer to is the equation for the reaction between sodium hydroxide and ethanoic acid i.e.

CH₃COOH   +    NaOH     →       CH₃COONa      +   H₂O                       

The reacting ratio is 1:1 i.e.1 mole ethanoic acid to 1 mole sodium hydroxide.

So the titration calculation tells us the moles of residual ethanoic acid i.e.  0.00064moles, the same as the moles of sodium hydroxide.

But another thing you now need to bring into play in the calculation is that the titration involved a sample of the equilibrium mixture i.e. a 10ml sample.  

We need the total moles of residual ethanoic acid so we must multiply the titration result up by 25 since 10ml is 1/25th of 250ml.

Therefore, total residual ethanoic acid is  25  ×    0.00064   =    0.016moles  

It’s now possible to work out the equilibrium mixture composition using this method:

Draw up a table like this:

	CH3COOH	CH3CH2OH	CH3COOCH2CH3	H2O
Initial moles	0.08	0.12	0.00	0.00
Change in moles to reach equilibrium	–0.064	–0.064	0.064	0.064
Equilibrium moles	0.016	0.056	0.064	0.064
Equilibrium concentration  mol dm3	0.064	0.224	0.256	0.256

The equilibrium moles of ethanoic acid is bold and from this value we can fill in the table.

Note that the equilibrium concentration has been calculated from the fact that the equilibrium mixture was 250ml or 1/4 of a litre.  Multiplying the mole values by 4 gives the molar concentrations in moles per litre. 

Having calculated the equilibrium concentrations you can substitute these values into the expression for the equilibrium constant for the reaction between ethanoic acid and ethanol. 

Kc       =          [CH₃COOCH₂CH₃ ] [  H₂O  ]                        

                        [ CH₃COOH] [  CH₃CH₂OH]              

Kc       =           0.256 × 0.256                       

                         0.064 × 0.224                                                          

Kc       =           4.57        (no units)         

Kc has no units because that are the same number of units on the top of the expression and the bottom of the expression i.e.

Kc       =          moldm^-3.   moldm^-3        
                       moldm^-3.   moldm^-3
so no overall units as all cancel.