- Room temperature at which the experiment is performed
There can be a difference in the concentration of the solution if the solution prepared and the experiment performed is at different temperatures .Hence an ideal way to control the variable would be to prepare the solution and perform the experiment at the same temperature.
The mass of the egg shells that is taken is controlled by the person performing the experiment. This is the major variable , as according ti this variable all other variable are dependent of this.If we take a higher mass of egg shell then a greater concentration of acid will be required containing greater number of moles and greater concentration of NaOH will be needed. The issue with this will be that there might not be lab equipment that will suffice the required amount to perform the experiment. So we would then have to repeat the experiment multiple times to adjust the increase , hence increasing the uncertainity error. Similarly if the mass of egg shells taken is too less , then there will be increased chances of errors due to greater inaccuracy.
Relevant data
Calculation
CaCO3(s) + 2 HCl(aq) -> CO2(g) + H20 + CaCl2(aq)
Volume of Hcl = 100cm3
Concentration of Hcl = 2 M
Moles = C x V = 2x0.1 = 0.2 mol
Mass of egg shell = 4.931g
Mols = 0.04931 mol
The solution was made up to 250cm3 by adding distilled water.
Hcl+NaOH -> NaCl+ H20
Volume of alkali = 250cm3
Mols of alkali = 0.125 mol
Concentration = n/v = 0.125/0.250dm3 = 0.5M
Materials Required
- Egg shells
- Mortar and pestle
- Conc. HCl with a concentration of 2M
- Conc. NaOH with a concentration of 0.5M
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3 Beaker 100cm3
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50cm3 pipette (±0.10 cm3)
- Phenolphthalein indicator
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Burette 50cm3 (±0.02 cm3)
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Volumetric flask 250cm3 (±0.12 cm3)
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7 conical flask 250cm3
- Dropping pipette
- Stirrer
- Stirring bar
- Stirring rod
- Glass funnel
- Plastic funnel
- Distilled water
Procedure
- Take the egg shell and make sure it’s completely dry
- Weigh the egg shell and take 5 grams of the shell
- Obtain mortar and pestle and grind the shell until it is fine powder.
- Put the egg shell powder in the beaker
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Take another beaker pour 50cm3 of acid.
- Take the pipette and rinse the pipette make sure that while rinsing the acid does not go inside the balloon. Pour some more acid into the beaker
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Take 50cm3 Hcl acid.
- Release the acid in the beaker which contains the egg shell.
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Repeat this step once again so as to obtain a solution of volume of 100cm3 .
- Wait for some time for the acid to react with egg shell.
- Now stir the solution until the foam disappears.
- Take a volumetric flask , glass funnel and filter paper.
- Fold the filter paper onto the glass funnel and attach it on the top of the volumetric flask.
- Take distilled water.
- Slowly rinse the water on the filter paper so as to make the filter paper stick onto the funnel.
- Now pour the solution of egg shell and acid from the beaker in to the glass funnel carefully.
- The solution is being filtered , it will take time.
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The solution will not however add up to 250cm3 volume, so we need to add distilled water into the volumetric flask. Make sure that the water that is added stays under the filter paper level so as that it is being filtered.
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After the solution has reached the 250cm3 , take a stirrer bar drop it in the volumetric flask , put the flask on the stirrer and let the solution stir for 2 hours.
- Now take a pipette, make sure that the pipette is clean.
- Rinse the pipette
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Take a conical flask of 250cm3 and rinse it.
- Using the pipette transfer the solution into the conical flask.
- Take a burette and rinse it with NaOH , the desired alkali.
- Put the plastic funnel on top of the beaker and carefully pour the alkali into the beaker.
- Make sure that the meniscus is just above the zero mark.
- Add two drops of phenolphthalein into the solution
- Hold the tap with your left hand and the conical flask with your
- Slowly open the tap of the burette and let the alkali flow into the conical flask.
- When the alkali is being poured into the flask, keep swirling the flask with your right hand.
- When you see the pink color coming on the solution , slow down the flow of acid.
- Close the tap of the burette when the solution obtains a light pink color and doesn’t go when the flask is swirled.
- Subtract the reading of the burette from 0 and the value obtained is the volume of alkali used to titrate/neutralize the excess acid.
Qualitative Data
- Sodium hydroxide solution is colorless before titration
- HCL is colorless before titration
- The end result of the solution is slight pink
- When the solution is off-shot the solution has a deep purple color
- before reaching the endpoint the solution turns pink and then fades away as we swirl the conical flask
- the color of phenolphthalein is colorless before titration
- the color of phenolphthalein is slight pink after titration
- a drop of the solution is left behind in the pipette
- ideal color of the solution is fade pink
Quantitative Observation
Table below shows the various trials of titration that were taken in order to neutralize the acid and the base and find the acid left over and hence complete the back titration process
Table below shows the various apparatus and the readings taken
Table shows the chemicals that were used and there respective volumes
Error calculation
Uncertainty of Weighing balance = (Uncertainty/the mass measured) x 100 = (0.001/4.931)x100 = 0.02%
Uncertainty of pipette of 25cm3 = (Uncertainty / the volume taken)x100 = (0.06/25.00) x 100 = 0.24%
Uncertainty of the volumetric flask of 250cm3= (uncertainty /the volume taken) x100 = (0.12/250)x100 = 0.048%
Uncertainty of the burette of 50cm3 = (uncertainty/the total volume)x100 = (0.02/50)x100 = 0.04%
Total uncertainty = sum of all the uncertainties = 0.02%+0.24%+0.048+0.04%= 0.348%
Calculation
Table below shows the various trials of titration that were taken in order to neutralize the acid and the base and find the acid left over and hence complete the back titration process
Mols of HCl initially = c x v = 2.0 x 0.1 = 0.2 mol
NaOH+Hcl -> NaCl+ H20
# moles of NaOH needed for back titration = c x v = 0.021 = 0.0105 mol
# moles Hcl present in 25.00cm3 = 0.0105 mol
# moles Hcl present in 250.00 = 0.0105 x 10 = 0.105 mol
CaCO3(s) + 2 HCl(aq) -> CO2(g) + H20 + CaCl2(aq)
Moles of Hcl reacting with CaCO3 = total moles - moles left over = 0.200 - 0.105 = 0.095 mol
2 moles of Hcl is reacting with one mole of CaCO3
# moles CaCO3 is ½ x 0.0950 = 0.04750 mol
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mass CaCO3 = 0.0475 x 100 = 4.750 g
uncertainty = 0.348% = (0.348/100) x 4.750 = 0.1653g
Mass of CaCO3 = 4.750 g ± 0.1653 g
Percentage purity = (Mass of CaCO3/Mass of egg shell) x 100 = (4.750/4.931)x100 = 96.32%
Conclusion
The main aim of this experiment was to find the percentage purity of CaCO3 in an egg shell. This was found by using the procedure of back titration. Back titration is a titration done in reverse; instead of titrating the original sample, a known excess of standard reagent is added to the solution, and the excess is titrated. A back titration is useful if the endpoint of the reverse titration is easier to identify than the endpoint of the normal titration, as with reactions. Back titrations are also useful if the reaction between the analyte and the titrate is very slow, or when the analyte is in a non- solid.
First measured mass of egg shell was taken and it was reacted with excess of HCl acid in a beaker. The solution was then taken and the volume was increased to 250cm3.Then using a pipette 25cm3 volume of the solution was taken and was poured into a volumetric flask and it was titrated against a known concentration of NaOH.This was repeated 7 times and the mean titre was found to be 21.11cm3. After finding the titre calculations were done to find the mass of CaCO3 present in the egg shell.
The egg shell weighed 4.931 g ± 0.001 and was found to contain 4.750 g ± 0.1653 g CaCO3 .
The % purity of the egg shell containing CaCO3 was found by using the give formula :- (Mass of CaCO3/Mass of egg shell) x 100 which gave 96.32% purity.