Assessed Practical Titration Write-up

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Thom O’Dell

Assessed Practical Titration Write-up

Equation:

Na2CO3 + H2SO4  Na2SO4 + CO2 + H2O

One mol of Na2CO3 reacts with one mol of H2SO4.

Results:

The weight of my sodium carbonate crystals was 2.67g and the results of the titrations are as follows:

So the average of the closest three titration results are is:

25.95 + 25.90 + 25.90 / 3 = 25.92

The mass of Na2CO3 I used is 2.67g and the relative molecular mass of Na2CO3 is 106. So the number of mols of Na2CO3 I used was:

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2.67 / 106 = 0.0251 mols in 250 cm3

So the concentration of the Na2CO3 solution was: 0.1004 mol dm-3.

So in the 25cm3 of Na2CO3 solution I used, there were:

0.0251 / 10 = 0.00251 mols.

The equation shows how 1 mol of Na2CO3 reacts with 1 mol of H2SO4, so 0.00251 mols of Na2CO3 reacts with 0.00251 mol of H2SO4. So in the 25.92 cm3 of acid I reacted with the Na2CO3 solution, there are 0.00251 mols. So the concentration of the acid is:

( 0.00251 / 25.92 ) x 1000 = 0.0968 mol dm-3

So the concentration of the acid is 0.0968 mol dm-3.

Evaluation:

There ...

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