Finding our how much acid there is in a solution.
Results table
Titration
Rough
2
3
Initial burette reading
0.00 cm3
1.00 cm3
0.00 cm3
1.10 cm3
Final burette reading
1.00 cm3
21.90 cm3
1.10 cm3
22.10 cm3
Titre
1.00 cm3
0.90 cm3
1.10 cm3
1.00 cm3
Average titre = 11.0 cm3
Equation of reaction: Na2CO3 (aq) + H2SO4 (aq) H2O (l) + CO2 (g) + Na2SO4 (aq)
So, one mole of sulphuric acid reacts with one mole of sodium carbonate to form water, carbon dioxide and sodium sulphide.
In the sodium carbonate solution there was 1.06g of sodium carbonate made up to volume of 100cm3 with distilled water.
Na2CO3
Na = 23* 2 = 46g
C = 12 * 1 = 12g
O = 16* 3 = 48g
+106g
mole of Na2CO3 = 106g
We used 1.06 g in our solution, 1.06/106 = 0.01 moles
so, 0.01 moles of Na2CO3 in 100cm3
In each experiment we use 10cm3, so this contains 0.001 moles of Na2CO3
Since we know that the volumes of the two solutions need to be approximately the same for the two solutions to react we know that the sulphuric acid must have approximately 0.001moles in 11cm3.
Sulphuric acid = 0.001 moles in 11.00cm3
From this we can work the concentration of the sulphuric acid to be:
1.00/1000= 0.011 (changing cm3 to dm3)
0.001/0.011 = 0.09 mol/dm3 (number of moles/volume)
The results and calculations clearly show what I believe to be an accurate concentration of sulphuric acid. I carried out the experiment as stated in my method 3 times after doing a rough test. This rough test was not included when I worked out my average titre. All my results were within 0.2cm3 of each other, which I accept to be a reasonable degree of accuracy.
Results table
Titration
Rough
2
3
Initial burette reading
0.00 cm3
1.00 cm3
0.00 cm3
1.10 cm3
Final burette reading
1.00 cm3
21.90 cm3
1.10 cm3
22.10 cm3
Titre
1.00 cm3
0.90 cm3
1.10 cm3
1.00 cm3
Average titre = 11.0 cm3
Equation of reaction: Na2CO3 (aq) + H2SO4 (aq) H2O (l) + CO2 (g) + Na2SO4 (aq)
So, one mole of sulphuric acid reacts with one mole of sodium carbonate to form water, carbon dioxide and sodium sulphide.
In the sodium carbonate solution there was 1.06g of sodium carbonate made up to volume of 100cm3 with distilled water.
Na2CO3
Na = 23* 2 = 46g
C = 12 * 1 = 12g
O = 16* 3 = 48g
+106g
mole of Na2CO3 = 106g
We used 1.06 g in our solution, 1.06/106 = 0.01 moles
so, 0.01 moles of Na2CO3 in 100cm3
In each experiment we use 10cm3, so this contains 0.001 moles of Na2CO3
Since we know that the volumes of the two solutions need to be approximately the same for the two solutions to react we know that the sulphuric acid must have approximately 0.001moles in 11cm3.
Sulphuric acid = 0.001 moles in 11.00cm3
From this we can work the concentration of the sulphuric acid to be:
1.00/1000= 0.011 (changing cm3 to dm3)
0.001/0.011 = 0.09 mol/dm3 (number of moles/volume)
The results and calculations clearly show what I believe to be an accurate concentration of sulphuric acid. I carried out the experiment as stated in my method 3 times after doing a rough test. This rough test was not included when I worked out my average titre. All my results were within 0.2cm3 of each other, which I accept to be a reasonable degree of accuracy.