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Commercial Vinegar Test

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  • 1. Ravi Prapharsavat IB chemistry SL2 September 16, 2010 Determine the acetic acid in commercial vinegar Introduction: The research being conducted in the lab is to prove that commercial vinegar contains five percent of acetic acid, by titration method. As acetic acid is an organic compound, it reacts with bases such as NaOH to form salt and water. CH3COOH+ NaOH  NaCH3COO+ H2O The balanced chemical equation for neutralization of acetic acid with NaOH, and the equivalence point of this neutralization reaction can be determine using a chemical indicator to signal the end point such as Phenolphthalein. The titration method is that a drop of a solution (substance, in this case NaOH and vinegar) to another solution (substance, in this case Vinegar and NaOH with indicator) to determined the concentration. The drops continue until the titration meets the End Point, which is the sudden change in a physical property, such as the indicator color (in this caseit turns pink). The standardization is then use, whereby the concentration of a reagent is determine by reaction with a known quantity of a second reagent. Therefore, according to the theory of titration and acetic acid reacting with NaOH, the standardization should give a percentage of how much acetic acid is in the commercial vinegar. Design: Research Question: Prove that commercial vinegar contained 5% of acetic acid. Control Variables: NaOH (constant amount), Vinegar (constant amount), and phenolphthalein (constant drops) Dependent Variables: the amount used by the NaOH and vinegar solution in the burret Independent Variable: the constant amount and concentration of the solutions (NaOH, Phenolphthalein, and vinegar) Equipment: 125 mL Erlenmeyer flasks, 50 mL beret, beret stand and clamp Chemical: o.5 M. of NaOH, Phenolphthalein (chemical indicator), and commercial vinegar claiming to contain 5% acetic acid Procedure: • Pour 20mL of vinegar into the 125 mL Erlenmeyer flask Equation: 1
  • 2. • Fill the burret with 50 mL of NaOH solution • Add 4 drops of the chemical indicator to the vinegar solution in the Erlenmeyer flask • Perform the titration until the vinegar and indicator solution turns and remains light pink • Record the final volume of NaOH solution inside of the burret • Perform this experimentation another two times. • When finish the trials, repeat it again three times but is time with vinegar in the burret • Using the volume and the molarities of NaOH, and the volume and density of vinegar, the percentage of acetic acid will be calculated Data Collection and Processing: NaOH Reading Trials Initial volume of NaOH in Burret (mL) ±0.05 Final volume of NaOH in Burret (mL) ±0.6 Volume of NaOH used (initial volume-Final volume) (mL) ± 0.6 1 50.0 13.5 36.5 2 50.0 14.5 35.5 3 50.0 15.8 34.2 Average 14.6 35.4 Vinegar Reading Trials Initial volume of Vinegar in Burret (mL) ±0.05 Final volume of Vinegar in Burret (mL) ±1 Volume of Vinegar used (initial volume- Final volume) (mL) ± 1 1 50.0 28.1 31.1 Table 1: indicate the NaOH solution as it comes to the end point, where the solution used in the Erlenmeyer flask.
  • 3. 2 50.0 29.0 33.2 3 50.0 30.2 32.1 Average 29.1 32.1 Density of Vinegar = 1.01 g/mL Molarity of Vinegar= 0.5 M True value of CH3COOH= 5.00 % Sample Calculation: Uncertainties: Uncertainties= range 2 Example: Uncertainties∈the final volume of vinegar in burret= {30.2-28.1} over {2} ¿1.0(¿1d.p) Calculation of the percentage of acetic acid in commercial vinegar: Molesof NaOH =Volume×Molarity ¿35.4×0.5 mol 1000 ¿0.0177moles Molesof acetic acid =0.0177moles of NaOH ×(1mol acetic acid ÷ 1mol NaOH ) Table 2: indicate the vinegar solution as it comes to the end point, where the solution used in the Erlenmeyer flask.
  • 4. ¿0.0177×(60.0g mol ) ¿1.062 grams Mass of Vinegar sample=Density×Volume ¿ 1.01g mL ×32.1mL ¿32.421 grams of aceticacid ∈Vinegar=(massof aceticacid massof vinegar )×100 ¿( 1.062g 32.412g )×100 ¿3.3 Error= true−observe true ×100 ¿ 5.00−3.3 5.00 ×100 ¿4.5 (¿1d.p) Conclusion and Evaluation: The commercial vinegar does not contained 5% of acetic acid as it is stated in the bottle, however it is possible. The percentage of CH3COOH in the sample vinegar is found to be 3.3%, however, the label indicate that its 5.00%, which gives the deviation of the experimental value from the true value to be 1.7. By using the true value, a percentage error was calculated to be 4.5%; the experimental percentage of CH3COOH in the vinegar sample is lower than the true value. There are two weaknesses that have been encounter in this experimentation. One possible error that might be taken to an account is how the titration could past the true equivalence point of the reaction. By being accurate in the color change will result in a less NaOH or vinegar value used in the
  • 5. burret. Another weakness that is encountered in the experimentation is: the solution is sometimes mixed with water, and therefore causes the acetic acid as well as NaOH to be less concentrated; resulting in reacting slower (turning pink slower). The solutions, such as the vinegar and NaOH, are left open on the table. Therefore the molecule evaporate, leacing it with mostly water, which the acetic acid and NaOH will become less concentrated. If the NaOH and the vinegar are new and the lids are close, the result might indicate a smaller volume of NaOH used, which increase the percentage of the acetic acid.
  • 6. burret. Another weakness that is encountered in the experimentation is: the solution is sometimes mixed with water, and therefore causes the acetic acid as well as NaOH to be less concentrated; resulting in reacting slower (turning pink slower). The solutions, such as the vinegar and NaOH, are left open on the table. Therefore the molecule evaporate, leacing it with mostly water, which the acetic acid and NaOH will become less concentrated. If the NaOH and the vinegar are new and the lids are close, the result might indicate a smaller volume of NaOH used, which increase the percentage of the acetic acid.

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