3) After heating, the solution was then transferred into a volumetric flask via a funnel. More distilled water was then added to the flask to make a final volume of 250 a cm3.
4) 25 cm3 of the above solution was pipetted into a conical flask.
5) A titrating burette was then set up. First of all it was rinsed with the 0.1 M HCl, before the whole burette was filled with the acid.
- Three drops of phenolphthalein to the conical flask. The solution is now pink in colour.
- The base solution in the flask was the titrated with the HCl till the colour became colourless, which is the end point.
7) The above titration is repeated 3 times, and the result is recorded.
Result and Calculation:
Mass of aspirin = 1.504 gm
Concentration of NaOH = 1 mole/L
Concentration of HCl = 0.1 mole/L
Reaction after adding of NaOH is :
2H2O
1 mole of aspirin will react with 2 moles of NaOH
Reaction for the titration is :
NaOH + HCl ➔ NaCl + H2O
1 mole of NaOH will react with 1 mole of HCl
The result of the titration is tabulated as follows:
Calculation : Sample calculation using result of titration 2
# moles of HCl use = Molarity * Volume in litres = 0.1 * 0.009 = 0.0009 moles
So the number of moles of NaOH in the titrant is 0.0009 moles of NaOH.
We are only titrating 25 mls of NaOH from the bulk of the solution in the beaker, which was diluted 10 times. (25 ml topped up to 250 ml). So the number of moles of NaOH in the beaker was 0.0009 * 10 = 0.009 mole of NaOH.
We started out with 25 mls of 1 M solution of NaOH.
# moles of starting NaOH = 1 * 0.025 = 0.025 moles.
Of the above only 0.09 moles was left behind, so the aspirin has reacted with
0.025 – 0.009 = 0.016 moles of NaOH.
1 mole of aspirin will react with 2 moles of NaOH
So the number of moles of aspirin must be 0.016/2 = 0.008 moles
Molecular mass of aspirin (C9H8O4) = 180,
so mass of acetylsalicylic acid = 0.008 * 180 = 1.44 g
So % purity of aspirin = (1.44/1.504) * 100 = 95.74%
The other titration results was calculated the same way.
Average purity = 1/3 (94.55 + 95.74 + 95.74) = 95.34%
Conclusion and discussion:
The error in this lab is in the judgement of the end-point, in which the solution will completely becomes colourless.
The result was quite consistent, the average % purity is 95.34%