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Investigating the concentration of ethanoic acid in vinegar by titration method
 

Investigating the concentration of ethanoic acid in vinegar by titration method

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Investigating the concentration of ethanoic acid in vinegar by titration method. Please give proper reference to my IB student, Azam if you use his material.

Investigating the concentration of ethanoic acid in vinegar by titration method. Please give proper reference to my IB student, Azam if you use his material.

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    Investigating the concentration of ethanoic acid in vinegar by titration method Investigating the concentration of ethanoic acid in vinegar by titration method Document Transcript

    • Candidate Name: Muhammad Azam bin IsmailCandidate Number: 02206 - 007Date of experiment: 26th January 2009PRACTICAL 5: Determining the concentration of ethanoic acid in vinegarAimTo determine the concentration of ethanoic acid, CH3COOH in commercial vinegar by titrating itagainst a standard solution of sodium hydroxide of 0.0947M concentration.IntroductionVinegar is mainly composed of water and a small percentage of ethanoic acid. By law, vinegar shouldcontain no less than 4% of ethanoic acid. By titrating a diluted solution of vinegar against a standardsodium hydroxide solution of concentration 0.0947M, its concentration can be calculated usingstoichiometric theory. The acid-base titration is represented by the following equation: CH3COOH + NaOH CH3COONa + H2OIn this experiment, phenolphthalein indicator will be used to determine the end point of titration.The colour of the indicator should change from deep pink to pale pink at the end point.Apparatus and Materials o Pipette (25.00 ± 0.03)cm3 o Pipette filler o Volumetric flask (250.00 ± 0.12)cm3 o Stopper o Burette (50.00 ± 0.05)cm3 o 3 conical flasks (250cm3) o Retort stand and clamp o White tile o Filter funnel o Wash bottle o Commercial vinegar o 0.0947M sodium hydroxide solution o Distilled water o Phenolphthalein indicator Page | 1
    • Candidate Name: Muhammad Azam bin IsmailCandidate Number: 02206 - 007Date of experiment: 26th January 2009Procedure 1. 25.00cm3 of commercial vinegar is transferred to a 250.00cm3 volumetric flask using a pipette. Distilled water is added to the volumetric flask until the calibration mark. 2. The volumetric flask is capped with a stopper and inverted several times to form a homogenous solution. 3. 25.00cm3 of standard NaOH solution is transferred into 3 conical flasks each by using a fresh pipette. Three drops of phenolphthalein is added to the NaOH solution. The colour of the indicator should change from colourless to pink. 4. A burette is rinsed with distilled water to remove any impurities and is later rinsed with some vinegar solution. 5. The burette was placed securely on the retort stand and is filled with vinegar solution by using a filter funnel. 6. The initial reading on the burette is recorded. The titration is started by carefully adding the vinegar solution into the conical flask while swirling the conical flask. 7. When the colour of the indicator changes from deep pink to pale pink, the titration is stopped as the end point has been reached. 8. The final reading on the burette is recorded. 9. Steps 1 to 8 are repeated until the volume of titre falls within a 0.1cm3 difference in reading.Data CollectionQuantitative Data: Titration readings Titration number Rough 1st accurate 2nd accurate 3 3Initial burette reading/cm (±0.05 cm ) 1.50 0.50 1.60 3 3Final burette reading/cm (±0.05 cm ) 27.10 26.10 27.00Volume of titre, V/cm3 (±0.1 cm3) 25.6 25.6 25.4 Table 1: Collected data from titration - The result from the 2nd accurate titration is omitted as it does not fall within a 0.1cm3 difference from other readings.Qualitative Data: 1. The colour of the indicator changes from deep pink to light pink at the end point of titration. Page | 2
    • Candidate Name: Muhammad Azam bin IsmailCandidate Number: 02206 - 007Date of experiment: 26th January 2009Data Processing 25.6+25.6Average volume of titre = 2 = 25.6cm3The chemical equation of this reaction is: CH3COOH + NaOH CH3COONa + H2OBased on the equation, 1 mol of CH3COOH reacts with 1 mol of NaOH to produce 1 mol of CH3COOH.Thus, the mole ratio of ethanoic acid to sodium hydroxide is 1:1.Therefore, 1000 = 1000 (25.6) 0.0947 (25.00) 1000 = 1000 (25.6) = (2.37) 2.37 = 25.6 = 9.26 × 10−3 mol dm-3Since the vinegar was diluted 10 times, the actual concentration of ethanoic acid is = 0.0926 x 10 = 0.926 mol dm-3Hence, mass of ethanoic acid in 1000cm3 of vinegar is Mass = 0.926 x Mr of ethanoic acid = 0.926 x 60.06 = 55.6 gMass of ethanoic acid in 100cm3 of vinegar is 55.6 Mass = 10 = 5.56gTherefore, the concentration of ethanoic acid in the commercial vinegar is 5.56% Page | 3
    • Candidate Name: Muhammad Azam bin IsmailCandidate Number: 02206 - 007Date of experiment: 26th January 2009UncertaintiesUncertainty due to volumetric flask: 0.12Percentage uncertainty in volume of diluted vinegar solution = 250.00 × 100 = ±0.048%Uncertainty due to pipette: 0.03Percentage uncertainty in volume of sodium hydroxide solution = 25.00 × 100 = ±0.12%Uncertainty due to burette:Uncertainty from burette = Uncertainty from initial reading + Uncertainty from final reading = 0.05 + 0.05 = ±0.1cm3 0.1Percentage uncertainty for average titre = 25.6 × 100 = ±0.39%Total percentage uncertainty due to apparatus = 0.048% + 0.12% + 0.39% = ±0.56% 0.56Absolute uncertainty for percentage of ethanoic acid = 100 × 5.56 = ±0.03Therefore, percentage of ethanoic acid in commercial vinegar is (5.56 ± 0.03) % Page | 4
    • Candidate Name: Muhammad Azam bin IsmailCandidate Number: 02206 - 007Date of experiment: 26th January 2009ConclusionThe percentage of ethanoic acid in commercial vinegar is (5.56 ± 0.03) %. This means that there are(5.56 ± 0.03)g of ethanoic acid in every 100cm3 of vinegar. As such, this vinegar is a legal solution as itcontains more than 4% ethanoic acid as required by law.The literature value given for percentage of ethanoic acid in commercial vinegar is 5.6%Therefore, 5.56−5.6 Percentage errors = 5.6 × 100 = ±0.71%EvaluationThe total percentage error of the experiment is ±0.71% while the total percentage uncertainty fromapparatus is only ±0.56%. This means that the remaining 0.15% arises from systematic errors.One of the possible reasons why the result has deviated is due to calibration of apparatus. If theapparatus is inaccurate in reading, volume of the required solutions cannot be measured exactly.This can affect the final result of the experiment as a titration relies on accuracy of measurements.To rectify this problem, the apparatus should be sent for recalibration by the manufacturer beforecommencing the experiment.Another reason could be impurities which enter the bottle of commercial vinegar. One bottle ofvinegar was shared among several students in the class and impurities may have been accidentallyadded when transferring it to the volumetric flask. The best way to rectify this problem is to pour outthe vinegar into individual beakers and using a pipette to make the transfer.Also, vigorous swirling of the conical flask will allow atmospheric carbon dioxide to enter the conicalflask, thus increasing the acidity of the solution. Although this error may be small, it affects the finalresult of the experiment significantly. To avoid this from happening, the conical flask should beswirled gently.To improve the results obtained even further, a third accurate reading can be recorded to increaseaccuracy and to ensure consistency in readings. Page | 5