Volumetric analysis (5) A Back Titration.

Here is the problem:

How can we determine the percentage of calcium carbonate (CaCO₃) in a sample of chalk from the cliffs around Flamborough Head in the UK?

Calcium carbonate is an insoluble base so it will combine with a strong acid such as hydrochloric acid forming soluble calcium chloride solution and water and carbon dioxide.

Here is the equation:

CaCO₃ (s)  +   2HCl (aq)   =     CaCl₂ (aq)  +   H₂O(l)    +   CO₂ (g)

The approach is known as a back titration.

A measured mass of chalk from the cliff is weighed out and then dissolved completely in an excess of hydrochloric acid.

The resultant solution is made up to the mark in a 250ml volumetric flask.

A titration is then carried to determine the concentration of the excess hydrochloric acid in the flask.

Knowing the excess and the original amount of acid we can work back (hence back titration) to find the amount of acid reacting with the calcium carbonate in the chalk and hence the percentage of calcium carbonate in the chalk. 

Here is how to carry out this experiment and a typical set of results and calculation:

Determination of the percentage purity of natural chalk.

Chalk occurs throughout the UK.

It is mainly composed of calcium carbonate CaCO₃ a white insoluble solid. 

The percentage purity of this material can be determined by a back titration.

Step 1: Weigh out accurately 1.30g of a sample of naturally occurring chalk. 

 Step 2: Dissolve this mass of chalk in 50ml of 1M hydrochloric acid solution in a 250ml beaker. 

Step 3: Transfer the contents of the beaker and all washings to a 250ml volumetric flask and make up to the mark with distilled water. 

Step 4: By pipette extract 25ml of the acid solution from the volumetric flask and transfer it to a 250ml conical flask. 

Step 5: Titrate this aliquot of the acid solution with 0.1M sodium hydroxide solution using phenolphthalein indicator: the end point occurs with the first faint pink tinge to the solution in the flask. 

Step 6: Repeat the titration until you have concordant results i.e. titres to within 0.1ml of each other.

Typical set of results:

Weighings:

Mass of weighing bottle and chalk:     11.36g

Mass of empty weighing bottle:            10.02g

Mass of chalk transferred to flask:      1.34g

Titration results:

Pipette solution:    Hydrochloric acid            x  mol/dm³

Burette solution:    Sodium hydroxide             0.1mol/dm³

Indicator:                 Phenolphthalein

Burette readings:               Rangefinder            1                     2                     3

Final reading (ml):           25.30                        25.10              25.20            25.20

First reading (ml):              0.00                           0.00               0.00               0.10

Volume NaOH (ml):           25.30                        25.10             25.20             25.10

Average titre (ml):                        25.1(3)

Calculation:

In this calculation I am going to use the millimole method which is where the concentration of solutions is just multiplied by the volume in millilitres.

A millimole (mmol) is a thousandth of a mole. 

First, calculate the amount (mmol) of sodium hydroxide used in the titration from the average titre and the concentration of sodium hydroxide.

n  =  cV   so amount(NaOH) =  0.1× 25.1  =  2.51 mmol

Second, from the equation for the reaction of sodium hydroxide with hydrochloric acid determine the number of millimols acid this alkali reacts with.

So if                NaOH   +   HCl   = NaCl   +   H₂O 

Then amount(HCl)  =   2.51 mmol

Third, work out amount (mmol) excess acid remained in the volumetric flask after all the chalk had reacted. 

This is           2.51 ×  10  =  25.1 mmol excess hydrochloric acid

because a tenth of the solution in the volumetric flask was pipetted into the conical flask. 

Fourth, we can now work out the amount (mmol) of acid that reacted with the calcium carbonate in the chalk sample.

Total amount acid used =  50  ×  1  = 50 mmol (HCl)

Excess amount acid  =  25.1 mmol 

So amount(HCl) reacting with chalk =  50 – 25.1 =  24.9mmol

Fifth, from the equation for the reaction of calcium carbonate with hydrochloric acid 2 moles acid react with 1 mole calcium carbonate.

So amount of calcium carbonate in the chalk sample is 24.9/2 =12.45mmol

Sixth, work out what this amount of calcium carbonate weighs like this:

M_r (CaCO₃)  = 100.1 g/mol

Mass(CaCO₃)  =  100.1  ×  12.45/1000   =  1.25g

Seventh, work out the percentage purity of the chalk sample like this:

% purity(chalk)  =  mass(CaCO₃)  ×   100/ mass of chalk

% purity(chalk)  =  1.25   ×   100/1.30   =   95.9%

Extension challenge:

All the solutions in this experiment are colourless so wouldn’t this experiment look so much more exciting if instead of using chalk we used a sample of verdigris scraped off a copper roof somewhere!!

Verdigris is basic copper carbonate (CuCO₃) and it is green in colour. 

Dissolve verdigris in sulphuric acid makes blue copper sulphate solution. 

All blue and green and then the indicator colour change would be interesting to see as the faint tinge of not pink but probably a brown colour in the pale blue solution.