How to Determine the % of Ethanoic Acid in Vinegar

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  • Topic: Titration, Base, Erlenmeyer flask
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  • Published : April 12, 2011
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Chemistry Internal Assessment
Title: Determination of the percentage (%) of ethanoic acid in vinegar Date Experiment was performed: 1st April 2011
Criteria Assessed: DCP, CE
Apparatus:
2 conical flasks
1 250cm3 volumetric flask ±0.30cm3
Electronic Balance ±0.10g
50cm3 Burette±0.1cm3
25cm3 Pipette±0.06cm3
Materials:
0.20M HCl (±0.02moldm-3)
1.1g solid NaOH (±0.1g)
A solution of vinegar of unknown concentration (density= 1.05gcm-3) Phenolphthalein
Method:
* A solution of NaOH was prepared by dissolving 1.1g in 250cm3 of water * 1.1g of solid NaOH was weighed and then dissolved in the stirred until dissolved in 250cm3 of water * 25cm3 of this solution was placed into a conical flask and 3 drops of phenolphthalein indicator were added * A burette was filled with 0.2±0.02moldm-3 HCl

* The HCl was titrated against the Na2CO3 and the data recorded in a table * The same procedure was repeated but this time using CH3COOH instead of HCl * The % of ethanoic acid was determined by titrating the standardized NaOH with it * Given that the % of ethanoic acid in this sample was 4%, the % error was calculated and a conclusion drawn.

Results:
Table showing titration of 0.2M (±0.02moldm-3) HCl against NaOH Burette Reading/cm³±0.1cm³| Trial| | 1| 2| 3|
Final Volume/cm³ (±0.1cm³)| 27.5| 27.3| 27.4|
Initial Volume/cm³ (±0.1cm³) | 0.0| 0.0| 0.0|
Volume of acid required/cm³ (±0.2cm³)| 27.5| 27.3| 27.4|
Colours of solutions: acid, base, and phenolphthalein were all initially colourless. The base turned dark pink when phenolphthalein was added to it. Add the end point the trials were colourless.

Table showing titration of CH3COOH unknown concentration against NaOH Burette Reading/cm³±0.1cm³| Trial|
| 1| 2| 3|
Final Volume/cm³ (±0.1cm³)| 10.9| 20.6| 39.6|
Initial Volume/cm³ (±0.1cm³) | 0| 10.9| 30.6|
Volume of acid required/cm³ (±0.2cm³)| 10.9| 9.7| 9.4|

Colours of solutions: acid, base, and phenolphthalein were all initially colourless. The base turned dark pink when phenolphthalein was added to it. Add the end point the trials were colourless. From the quantitative results it was observed that the titration of CH3COOH took a less amount of acid and occurred within a shorter period of time.

Calculations:
Average volume of HCl used = 27.3+27.42
=27.35cm3±0.73%
Percentage uncertainty = 0.227.35 × 100
=0.73%
Concentration of HCl = 0.20moldm-3(±0.02moldm-3)
Percentage uncertainty = 0.020.2 × 100
=10%
Volume of HCl = 27.351000 = 0.02735dm3±0.73%
Number of Moles of HCl = 0.20moldm-3×0.02735dm3
= 0.005470mols± (10%+0.73%) = 5.47×10-3± 10.73% = 5.47×10-3± 5.87×10-4 The balanced equation for the reaction is as follows

NaOH + HCl NaCl + H2O
From the equation above 1mol NaOH: 1mol HCl
∴ Moles of NaOH = 5.47×10-3±10.73%
Volume of NaOH = 25cm3± 0.06cm3 = 25cm3±0.24%
= 0.025dm3± 0.24%
Concentration of NaOH = 5.47×10¯30.025dm¯3
= 0.2188moldm-3
Percentage uncertainty = 10.73%± 0.24%
= 10.97%
∴ Concentration of NaOH = 0.2188moldm-3±10.97%
= 0.2188moldm-3±0.0240moldm-3
Mass of NaOH = Number of Moles × relative molecular mass
= (5.47×10-3) mols ×40g
= 0.2188g±10.97%
In gdm-3 = 0.2188g0.025dm¯3
= 8.752gdm-3±10.97%

For CH3COOH
Average volume of CH3COOH used = 9.7+9.62
=9.65cm3±2.07%...
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