Volumetric Analysis (9) Yet another back titration

Measuring the amount of Nitrogen in a Fertiliser

Introduction

Ammonium salts are used as fertlisers e.g. Nitram or ammonium nitrate NH₄NO₃

Ammonium salts release ammonia gas on treatment with sodium hydroxide solution.  If the ammonia released is caught in a solution of a strong acid then it is possible to measure the proportion of nitrogen in the fertiliser.

The residual unreacted acid can be determined using a titration and you can then use the results of the titration to work back to the proportion of nitrogen in the fertiliser.  

This approach can be used with fertilisers that might contain trace elements to be added to the soil such as zinc, as these trace elements will not affect the release of ammonia.  

Procedure

1. Place 50ml of Mhydrochloric acid (HCl) in a 250ml conical flask.  This is the acid that will be used to trap the ammonia.
2. Weigh accurately about 1.70g of ammonium sulphate ((NH₄)₂SO₄ ) into a 50ml distillation flask.  Add an excess of 25ml 2M sodium hydroxide solution (NaOH) to the distillation flask and a few anti-bumping granules. 
3. Quickly put the distillation apparatus together and gently warm the distillation flask so that it boils VERY slowly.
4. Watch out for rapid sucking back of the solution from the conical flask.  This can happen because the ammoina gas is so soluble in the acid. 
5. Opening the tap of the funnel (see diagram) equalises the pressures in and out of the apparatus and prevents suck back.  (This is a typical apparatus to use but you would need to change the receiving vessel from the 50ml round bottomed flask in the diagram a 250ml conical flask with the same fitting.)

6. You will need to ensure you distill over 10-15ml of the solution to ensure all the ammonia has been removed.
7. Then wash the apparatus and make sure all the washings enter the conical flask.
8. Now add the contents of the receiving flask to a 250ml volumetric falsk and make up to the mark with distilled water.  
9. Titrate 25ml samples of the distillate with 0.1M sodium hydroxide solution using methyl orange indicator to determine the residual acid in the distillate.

Typical analysis and calculation

In this analysis, we begin from the final titration results and work back to the percentage of nitrogen in the original fertiliser.  

The titration results show that 25.0ml of 0.1M sodium hydroxide neutralised the excess acid in the distillate.  This is

25.0  ×  0.1    =    0.0025 moles NaOH     

  1000

This number of moles of sodium hydroxide was contained in 25ml of the distillate from the volumetric flask.  This is a tenth of the volumetric flask so the actual number of moles of excess acid is 

10  ×   0.0025    =    0.025moles  HCl          

This now allows us to determine the number of moles of hydrochloric acid that reacted with ammonia from the fertiliser.

First, let’s calculate the total number of moles of hydrochloric acid in the receiver.

50  ×   1    =    0.050moles     

  1000 

Therefore, the number of moles of hydrochloric acid reacting with the ammonia from the fertiliser was    

0.050   —   0.025    =    0.025moles            

And as 1 mole of ammonia reacts with one mole of hydrochloric acid this is also the number of moles of ammonia from the fertiliser.

And as 1 mole of ammonia contains 1 mole of nitrogen atoms this is the number of moles of nitrogen atoms in the sample of fertiliser.

Therefore the mass of nitrogen in the fertiliser is 

0.025    ×    14    =   0.35g            

And therefore the percentage of nitrogen in the fertiliser is

0.35   ×   100    =    21%      

1.70

This is the percentage of available nitrogen in ammonium sulphate.

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?

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.