Determining the percentage purity of calcium carbonate in a sample of limestone.
In my experiment, I hoped to find the amount of calcium carbonate in some mineral limestone using the back titration method
The equation of the reaction is as follows:
2HCl + CaCO3 ï¿½ CaCl2 + CO2 + H2O
As not all the acid will be used up in the above reaction, I plan to obtain the amount of acid not used up and consequently the amount of calcium carbonate in the limestone, by titrating it with known sodium hydroxide solution. The equation of the reaction is as follows:
HCl + NaOH ï¿½ NaCl + H2O
The equipment and reagents that I used are as follows:
250 cm3 beaker
Electronic balance (ï¿½ 0.01 g)
250 cm3 volumetric flask (ï¿½ 0.2 cm3)
50.0 cm3 burette (ï¿½ 0.05 cm3)
25. 0 cm3 pipette (ï¿½ 0.1 cm3)
3 x conical flasks
Clamp and retort stand
1.0 mol dm-3 hydrochloric acid, HCL
1.5 g of limestone
0.1 mol dm-3 Sodium hydroxide, NaOH
When I had all this, I started my investigation by weighing 1.5 g of limestone on the electric balance. After obtaining the weighed crystals I dissolved them with 1.0 mol dm-3 hydrochloric acid, in the beaker provided and stirred the mixture till bubbling stopped. I then added the solution into the volumetric flask and topped it with distilled water till I reached the 250 cm3 mark.
I than mounted the burette on the retort stand, and filled it with the 0.10 mol dm-3 Sodium Hydroxide solution up to the 50 cm3 mark.
After doing all this, I titrated the solution of the limestone against the base from the burette, until the colour of the solution in the conical flask just turned pink. I repeated the procedure to obtain 3 sets of readings and will take the average value as my result.
Data processing and presentation:
With all the above processes being done, I wrote down the values which I obtained in the raw data table below.
Raw Data Measure
Volume of acid used / cm3 (ï¿½ 0.05 cm3)
I then processed my data to find the average volume of acid used.
Vavg = V1+V2+V3
Vavg = 41.6 + 41.4 +
Vavg = 41.5 (ï¿½ 0.05 cm3)
Now I used the various formulas related to the mole concept to find the amount of calcium carbonate in the sample of limestone.
The equations of the reactions are as follows:
1) 2HCl + CaCO3 ï¿½ CaCl2 + CO2 + H2O
2) HCl + NaOH ï¿½ NaCl + H2O
Moles of acid in first reaction = Molarity x Average volume
= 1 x (50 ï¿½ 0.02 cm3)
= 0.05 moles
Moles of base used = Molarity x Average volume
= 0.1 x (41.5 ï¿½ 0.05 cm3)
=4.15 x 10-3 moles of Sodium Hydroxide.
Moles of acid = 4.15 x 10-3 moles
Molarity of HCl = Moles x 1000
= 4.15 x 10-3 x 1000
25 ï¿½ 0.1
= 0.166 mol dm-3
Moles of HCl in 250 cm3 solution = Morality x Average volume
= 0.166 x (250 x ï¿½ 0.2 cm3)
= 0.0415 moles of HCl in 250 cm3 solution
Moles of acid used in reaction = 0.05 – 0.0415
= 8.15 x 10-3 moles
Ratio of moles of CaCO3 : HCl
1 : 2
x : 8.15 x 10-3
x = 8.15 x 10-3 x 1
= 4.075 x 10-3 moles
Mass of CaCO3 in 1.5g of limestone = Moles x RMM (RMM of CaCO3 = 100)
= 4.075 x 10-3 x 100
Percentage purity of calcium carbonate = 0.41 x 100
= 27.3 ï¿½ 0.35 %
A possible source of error in this experiment is the determination of the end-point, which is characterised by the solution just turning orange. This is because a slightly greater volume of acid may have been used than required to produce the pink colour. To try to reduce the effects of this error I would like to carry out a large number of titrations and their average used in the calculation.
Another possible way to reduce error is by using more accurate measuring instruments like a more precise burette so as to reduce the uncertainty of the measurements.