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Determine the concentration or molarity of Ethanoic acid (CH3COOH) in two types of commercial vinegar.

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Introduction/Plan: The aim of this investigation is to determine the concentration or molarity of Ethanoic acid (CH3COOH) in two types of commercial vinegar. To set about this, values of percent by mass have been noted from the internet and the modal value for this was 5%. The concentration can be calculated from percentage by mass by doing this: 5% ethanoic acid in vinegar can be interpreted by saying that it is the same as 5 grams of acid per every 100 grams of solution so: The number of moles of ethanoic acid must first be calculated: No. Mole = Mass/Mass of one Mole. The molecular mass of ethanoic acid must then be calculated which is 60. No. mole = 5/60 = 0.083 mole (to 2 d.p.) Therefore the number of mole in 1000 grams of solution can be found which will be useful as this is approximately equal to 1 dm3 and means that the final result can be displayed in mol/dm-3, and so to do this it will have to be multiplied by ten: No. mole of CH3COOH in 1000 grams of solution = 0.083 . 10 =0.83 mole. Therefore the predicted concentration is: 0.83 mol/dm3. To find out the concentration accurately a titration will have to be done to basically find out how much alkali will ...read more.


The Values collected in cm3 for the amount of NaOH required to neutralise the diluted vinegar Clear Vinegar Coloured Vinegar (Brown) Rough: 19.7 Rough: 15.4 18.8 14.5 18.8 14.4 18.7 14.5 Average Value: 18.77 Average Value: 14.47 The results were taken three times but had the three I had taken seemed completely random then I would have kept taking them until I found values which match up as being close to each other. The average values have been taken and used to find the concentration of both vinegars: Analysis of Results: Transparent Vinegar: In this titration 18.77 cm3 of 0.1 M sodium hydroxide (NaOH) solution reacted with 20 cm3 of a solution of ethanoic acid of unknown concentration. 1 CH3COOH + 1 NaOH � 1 CH3COONa + 1 H2O 20 cm3 18.77 cm3 (sodium ? M 0.1 M ethanoate) The first stage is to find the No. mole of the sodium hydroxide: No. mole NaOH in 18.77 cm3 of 0.1 M solution = Molarity � Volume / 1000 = 0.1 � 18.77 / 1000 = 1.877 � 10-3 mole This now makes it possible to find out the number of moles of ethanoic acid because we know that one mole of the acid (CH3COOH) ...read more.


This error boundary is likely to be added to by human error when stopping the sodium hydroxide being added once the acid has been neutralised (when the indicator changes colour). The awareness of the colour change of the indicator is perhaps slightly impaired when the brown vinegar is being neutralised because it is more difficult to see past the brown colour when the actual indicator begins to change colour itself. This problem could have been overcome by filtering the brown vinegar through carbon to remove its brown colour and to make it transparent. A similar type of error boundary as described previously can be applied to the measuring of the diluted vinegar in the volumetric flask and in the pipettes. When the various artefacts (burette, conical flask and pipettes) were rinsed out ready for the titration, they had to be rinsed out with whatever they were going to contain before they could be used so to counteract whatever had been in there before hand. Even though they might have been rinsed out some of whatever had been there previously may still remain and adversely affect the results of the titration. Finally, the results collected clearly reflect the prediction in the plan. Chemistry Coursework: Vinegar Concentration Investigation. ...read more.

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