Volumetric Analysis (8) Another Back Titration

Another example of a back titration

Preamble

A “back” titration is used when a particular reagent (acid, base or redox reagent) is treated with a definite amount of another reagent and then the excess is determined by titration.

Take the determination of the purity of a base such as a carbonate, this can be determined if the base is treated with a given amount of acid and then the excess acid is titrated with a standard alkali.

In the example I am going to discuss below, the weight of aspirin in an aspirin tablet is determined by back titration.

Background

Aspirin is an ester, a compound of two acids—ethanoic acid and hydroxybenzoic acid. 

Aspirin

Salicylic acid or 2–hydroxybenzoic acid:

Ethanoic Acid:

Aspirin can be hydrolysed using a strong alkali such as sodium hydroxide and at the same time the two acids formed are neutralised.  

equation

You can see from the equation above that 1 mole of aspirin reacts with 2 moles of sodium hydroxide.

The titration can be carried out using phenolphthalein as the indicator, it going slightly pink/magenta at the end point.

Apparatus

To carry out such an experiment you will need the following apparatus:

75mg aspirin tablets

The usual titration apparatus

250ml volumetric flask

Sodium hydroxide solution standardised at 0.5M

Phenolphthalein indicator

Phenol red indicator

Sulphuric acid solution 0.02M

Procedure

Put 5 small 75mg aspirin tablets into a large conical flask.

Add 25ml of 0.5M sodium hydroxide (NaOH) to the flask together with 25ml of pure water.

This mixture needs to be warmed gently for about 15minutes to ensure all the aspirin has been hydrolysed. 

After 15 minutes allow the reaction mixture to cool down and transfer the entire contents to a 250ml volumetric flask and make up to the mark with distilled water.

Titrate 25ml of this solution with 0.02M sulphuric acid (H₂SO₄) using phenol red indicator.

Note the volume of sulphuric acid required to neutralise the excess sodium hydroxide solution from the reaction mixture.  This is the back titration technique at this point.  

Analysis

1.  Calculate the number of moles of sulphuric acid required to neutralised the excess sodium hydroxide.

Suppose the excess sodium hydroxide requires 20ml of 0.02M sulphuric acid.  

Using n=cV. Then n(H₂SO₄) = 20×0.02   =    0.0004moles 

                                               1000

2.  Now sodium hydroxide reacts in a 2:1 ratio with sulphuric acid so the number of moles excess sodium hydroxide is   0.0004  ×  2  =  0.0008moles

3.  This number of moles of sodium hydroxide is contained in 25ml of the reaction mixture, that is a tenth of the mixture so the total number of moles of excess sodium hydroxide is 

0.0008  ×  10    =  0.008 moles

4.  From this we can determine the amount of sodium hydroxide that reacted with the aspirin in the 5 tablets.

First we need to know the total amount of sodium hydroxide that was added to the tablets.  This amount was  25ml of 0.5M sodium hydroxide 

Total       n(NaOH) =   25 × 0.5     =    0.0125moles

                                1000

So the amount of sodium hydroxide reacting with the aspirin is 0.0125. – 0.008 = 0.0045moles

5.  We now need to know how many moles of aspirin this amount pf sodium hydroxide hydrolyses.  

The hydrolysis reaction shows that 1 mole aspirin requires 2 moles sodium hydroxide for complete hydrolysis.  

Therefore, the amount of aspirin involved is 0.0045/2   =   0.00225moles.

6.  But what does this amount of aspirin weigh?

Use   m=  n×M. 

Therefore 

mass of aspirin m  =  0.00225   ×    180  =  0.405g

7.  So if 5 tablets were used then each tablet contains 

0.405/5  =  0.081g. or 81mg.

The value we have obtained is higher than that specified on the label.  

Can you think of reasons why it is the case that the tablet contains just slightly more than the specified amount of aspirin?