Weighting Data:
Mass of vial: 6.281g
Mass of vial and Na2CO3: 7.583g
Mass of Na2CO3: 1.302g
Na2CO3 solution made up to: 250.0cm3
Titration I: Standardization of 0.1M hydrochloric acid solution
Titrant (in burette): Hydrochloric acid
Titrate (in conical flask): 25.0cm3 of Na2CO3
Indicator used: Methyl Orange
Colour of indicator changed from: yellow to pink
Average volume: 23.18 cm3
Calculation:
Na2CO3 (s) + 2HCl → 2NaCl(s) + H2O(l) + CO2(g)
Number of mole of Na2CO3(aq) = 1.302g / 105.8089gmol-1
= 0.0123
∵ HCl : Na2CO3 = 2 : 1
∴ Number of mole of HCl = (0.0123)(2)(25/250)
= 0.00246
Molarity of HCl = 0.00246 / (23.183 / 1000)
= 0.1062
~0.11M
Titration II: Determination of the given sodium hydroxide solution
Titrant (in burette): Hydrochloric acid
Titrate (in Conical flask): 25.0cm3 of NaOH
Indicator used: Methyl Orange
Colour of indicator changed from: yellow to pink
Average volume: 23.83 cm3
Calculation:
HCl(aq) + NaOH(aq) → NaCl(aq) + H20(l)
Number of mole of HCl = (0.1062)(23.83 / 1000)
= 0.00253
∵ HCl : NaOH = 1 : 1
∴ number of mole of NaOH = number of mole of HCl = 0.00253
Molarity of NaOH = 0.00253 / (25/1000)
= 0.1012
~0.10M
Titration III: Estimation of acetic acid content in commercial vinegar
Titrant (in burette): dilute commercial vinegar solution
Titrate (in Conical flask): 25.0cm3 of NaOH
Indicator used: Phenolphthalein
Colour of indicator changed from: purple to colourless
Average volume: 16.98 cm3
Calculation:
CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l)
Number of mole of NaOH = (0.1012)(25/1000)
= 0.00253
∵ CH3COOH : NaOH = 1 : 1
∴ number of mole of CH3COOH(dil) = number of mole of NaOH = 0.00253
Molarity of CH3COOH(dil) = 0.00253 / (16.98 / 1000)
= 0.1489999
Molarity of CH3COOH = (0.1489999)(250 / 25)
= 1.489999
Number of mole of CH3COOH in commercial vinegar = (1.489999)(250/1000)
= 0.3725
Mass of CH3COOH in commercial vinegar = (0.3725)[(2)(12.0107)+(4)(1.00794)+(2)(15.9994)]
= 22.369
The percentage of CH3COOH in commercial vinegar = (22.369 / 250)(100%)
= 8.948
~ 8.95%
Conclusion:
We prepared standard solution 0.05M NaOH.
Through the titration I, we know that the molarity of HCl was 0.11M.
In the titration II, we determinate NaOH solution is 0.1012M.
At the titration III, we found out the molarity of diluted commercial vinegar is ~0.15M and also calculation out the original vinegar was ~1.49M. In calculation, we found out there are 8.95% ethanoic acid in commercial vinegar.