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You are provided with a sample of vinegar which contains approximately 5g of ethanoic acid (CH3COOH) per 100cm. The aim of this experiment is to determine the exact concentration of this chemical in vinegar.

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Introduction

Georgina Garfield EXPERIMENT 4: TO DETERMINE THE CONCENTRATION OF ETHANOIC ACID IN A SAMPLE OF VINEGAR You are provided with a sample of vinegar which contains approximately 5g of ethanoic acid (CH3COOH) per 100cm�. The aim of this experiment is to determine the exact concentration of this chemical in vinegar. To do this you must carry out an acid-base titration because it is a very accurate procedure if done correctly. The solution of vinegar is the acid, and the base you will titrate it against is sodium hydroxide (NaOH). This solution has a concentration of exactly 0.100 mol dm��. The first thing you must do is convert the approximate concentration of Ethanoic Acid in vinegar from grams per 100cm�, to moles per dm�. To do this you must carry out the following calculations: *Number of moles of CH3COOH in 5g. You know that the number of moles can be calculated by dividing the mass by the M.R. (Relative Molecular Mass). You can find the M.R by adding the relative atomic mass of each of the elements in the molecule in the proportions given in the molecular formula. There are 2 Carbons, 4 Hydrogens and 2 Oxygens. Atomic Masses: Carbon � 12.0 Hydrogen � 1.0 Oxygen � 16.0 Therefore: M.R CH3COOH = 2 � 12 + 4 � 1 + 2 � ...read more.

Middle

PROCEDURE: * Rinse the 250cm� pipette with vinegar * Pipette 25cm� of vinegar into the volumetric flask. (note: always make sure the bottom of the meniscus is on the graduation mark) * Carefully fill the volumetric flask with distilled water to the graduation mark. * Stopper the flask and shake until the solution is homogeneous. The vinegar has now been diluted by a factor of �10 and will be used in the titration. TITRATION: Apparatus * Conical flask * 25 cm� pipette * Pipette filler * Burette * Filter funnel Chemicals: * Diluted vinegar solution * Aqueous NaOH solution * Phenolphthalein * Distilled water (*The end point in this weak acid - strong base titration is approximately 9.3. Methyl orange has a transition pH range of 3.1-4.4 and is therefore completely useless for this experiment. Phenolphthalein on the other hand has a pH range of 8.3-10.0, so it will change colour right on the end point. For this reason, the indicator which will be used in this titration is Phenolphthalein.) HEALTH AND SAFETY CONSIDERATIONS: * Sodium Hydroxide in its pure form is very corrosive and can cause severe burns. In its diluted form it is less harmful, but still an irritant. It can also cause severe permanent eye damage. ...read more.

Conclusion

CH3COOH + NaOH � CH3COONa + H2O v v v v 1 mole 1 mole 1 mole 1 mole 25 cm� x cm� Now, you have to calculate the number of moles of NaOH that reacted in this experiment. This is done as follows: Number of moles (n) = Volume (v) � concentration (c) 1000 The exact concentration of NaOH has been given: 0.100mol/dm�; and the volume you will use is your average titre. .?. n = x cm� � 0.100 mol dm�� 1000 = y moles According to the mole ratio, the number of moles of CH3COOH is the same as that of NaOH. Therefore we can rearrange the former equation for c, to work out the concentration of Ethanoic Acid in diluted water. If n = v � c 1000 .? . c = n �1000 v Now you can substitute into the equation. C = y moles � 1000 25 cm� C = Z mol dm� Since the vinegar was diluted by a factor of �10, if you multiply this concentration by 10 you will have the concentration of CH?COOH in undiluted vinegar. C = Z � 10 C = 10z mol dm�� This now has to be converted to g/dm�. To do this you multiply the number of moles by the M.R, which we already know is 60. Mass = n � M. ...read more.

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