Sulphuric Acid investigation evaluation

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Analysis:

Sodium Carbonate Solution:

250 cm3 of sodium carbonate (Na2CO3)

Number of moles = mass

                                      Ar

Mass: 2.65g of sodium carbonate

Mr:  Na =23g   C=12g   O=16g

 Ar of Na2CO3 = (23x2) + (12) + (16x3)

                          = 106g

Number of moles = 2.65g

                                106g

                             = 0.025 moles in 250 cm3 of sodium carbonate 

Volume = 250 = 0.25 dm3

               1000

Concentration (mol/dm3) = Amount in moles of solute

                                                       Volume (dm3)

Concentration of Na2CO3 = 0.025 = 0.1 mol/dm3

                                       0.25

Titration Results:

Na2CO3 + H2SO4         Na2SO4 + CO2 + H2O

As my results were all the same, the Average Titre = 25.50 cm3

This means that 25.50 cm3 of sulphuric acid neutralises 25 cm3 of my sodium carbonate solution.

25.50 cm3 = 0.0255 dm3

Na2CO3 + H2SO4           Na2SO4 + CO2 + H2O

        1     :     1

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As the stoichiometry of the reaction is one to one, therefore the number of moles of sodium carbonate is equal to the number of moles of sulphuric acid.

Previously in my coursework I worked out that I had 0.025 moles of sodium carbonate in my 250cm3 solution.  

Number of moles of sodium carbonate in 25 cm3 = 0.025 = 0.0025 moles

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