Determining the amount of CaCO3 in eggshell of hen's egg
| Design | DCP | CE |
Aspect 1 | | | |
Aspect 2 | | | |
Aspect 3 | | | |
Introduction: The back titration is a method used in determining the amount of excess of the reagent. The calcium carbonate is a substance which gives the eggshell stiffness.
Research question: What is the amount of calcium carbonate in the eggshell measured by back titration?
Table 1. Variables.
Type of variable | Variable | Unit |
Dependent | Amount of calcium carbonate in eggshell | % by mass | Independent | Volume of titrated excess of hydrochloric aced | cm3 | Controlled | Volume of hydrochloric acid Weight of eggshell Temperature Amount of phenolophateine | cm3 g oC drop | Uncontrolled | Purity of solutions Biological diversity of eggs Pressure | - - hPa |
5 beakers 50 cm3
1 plastic pipette
2,5 g of eggshell
100 cm3 of 1moldm-3 hydrochloric acid
ap. 70 cm3 of 1moldm-3 sodium hydroxide
1 20 cm3 pipette
Risk assessment: you have to remember to wear gloves, goggles and apron. Solutions may be irritating.
Crush to dust eggshell in the mortar.
Fill each of the 5 beakers with 20 cm3 of hydrochloric acid measured by glass pipette. Add 0.5 g of eggshell dust to each beaker, measured by balance. While the reaction of eggshell with acid occurs, prepare the buret and clamp for titration. Make sure they are clean. Pour NaOH solution into the buret to the '0' level.
Make sure all of the eggshell reacted with the HCl. If not, you can help the reaction by using the baguette. Put two drops of phenolophateine into each beaker using the plastic pipette. Take the first beaker and titrate the excess of hydrochloric acid. When the solution starts to be pinkish, record the volume of titrated NaOH. Refill the buret to the '0' level and repeat the procedure for each beaker. Remember to record the results. Remember to be careful and to leave your workplace clean!
Table 2. Raw data. The weight of eggshell reacting with HCl and titrated NaOH. Number of trial | Weight of eggshell [g±0,01g] | Volume of HCl [cm3±0,05cm3] | Volume of titrated NaOH [cm3±0,05cm3] | 1 | 0.49 | 20.00 | 9.60 |
2 | 0.50 | 20.00 | 11.50 |
3 | 0.51 | 20.00 | 11.60 |
4 | 0.50 | 20.00 | 9.90 |
5 | 0.50 | 20.00 | 10.30 |
Mean | 0.50±0,01 | 20.00±0,05 | 9.93±0,05 | Standard deviation | 0.00047 | 0.00 | 0.29 | Uncertanties were taken as in measurments, not calculated by formula, to avoid too large and unreliable uncertainties in further calculations in which they're calculated according to formulas:
in case of division and multiplication:
d=dA/A+dB/B, where d is overall uncertainty, dA is uncertainty of A and dB is uncertainty of B
in case of addition and subtraction:
d=dA+dB, where where d is overall uncertainty, dA is uncertainty of A and dB is uncertainty of B
Trials 2 and 3 were rejected because of too large differentiation of results.
Two reactions occured in the experiment. Firstly, the HCl reacted with CaCO3 and secondly, the excess of HCl was neutralized by NaOH.
2HCl + CaCO3 → CaCl2 + CO2 + H2O
HCl + NaOH → NaCl + H2O
1. Calculating the amount of HCl at the beginning of reaction
CmHCl = 1.0 moldm-3
VHCl = 20.00 cm3 = 0.020 dm3 ± 0.00005
n = Cm * V
n = 1.0* 0.020 = 0.020 mol ± 0.00025
2. Calculating mean amount of NaOH which neutralized the excess of HCl
CmNaOH = 1.0 moldm-3
VNaOH = 9.93 cm3 = 0.0099 dm3 ± 0.00005
n = Cm * V
n = 1.0 * 0.0099 = 0.0099 mol ± 0.0005...