Acid Base titration to determine the percentage by mass of calcium carbonate in an egg shell .

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Gerald Veliz

11°C - IB

Acid-base Titration


  • To determine the percentage by mass of calcium carbonate in an egg shell .


The first product is composed by the reactants: 0.6 grams of a ¨impure¨ sample of calcium carbonate and 20cm3 of 1.00 mol/dm3 hydrochloric acid solution, and it forms the following equation:

CaCO3 + 2HCl → CaCl2 + CO2 + H2O

, which will be later use to complete the objective of the experiment.

As having the acidic solution, sodium nitrate of 0.100 mol/dm3 was used to titrate the hydrochloric acid, and with the use of Vernier instruments, the exact volume of sodium nitrate needed to reach the equivalence point, would be obtained. The second equation for this process is the following:

Hydrochloric acid being titrated by sodium hydroxide.

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HCl + NaOH → NaCl + H2O


Moles of HCl= n=0.02dm3 x 1.00moldm3=0.02 moles of HCl

Moles of NaOH n=0.0107dm3 x 0.10moldm3= … as 1:1 relation with HCl then n=0.00107 moles of HCl

Total number of moles used=0.02-(0.00107*10)=0.00930 moles

As 2:1 relation with CaCO3 then 0.00932=0.00465 moles of CaCO3 

Conversion to grams= 0.00465 moles x 100.08 grams1.00 moles=0.46537 grams of CaCO3 

Percentage by mass= 0.465g0.60 g x 100=77.50%


 From the graph obtained, the point of equivalence can´t be easily identify by just observing the graph because in this case we titrate a weak acid with a strong base, so ...

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