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Outline whether the acidity of a bottle of vinegar (ethanoic acid) is labelled correctly.
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| Submitted: Mon Jun 19 2006
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Aim: My aim in this investigation is to outline whether the acidity of a bottle of vinegar (ethanoic acid) is labelled correctly. Prediction: If the vinegar bottle is labelled correctly at 5% acidity then I predict that the amount of NaOH needed for the titration will be: Ethanoic acid + Sodium hydroxide --> Sodium ethonate + Water CH3COOH(aq) + NaOH(aq) --> NaCH3COOH(aq) + H2O(l) 5% acidity of vinegar (ethanoic acid) with a volume of 25cm3 will be used. 0.1moldm-3 of NaOH will be used to neutralise the ethanoic acid. Therefore: 1. We can work out the cm3 of the ethanoic acid in 25cm3 of vinegar 5% 100 X 25cm3 =1.25cm3 2. We can then work out the grams of ethanoic acid followed by the moles of ethanoic acid. Moles = Mass Molar mass Density of ethanoic acid is equal to 1.049g/cm3 (reference bibliography 1) Therefore 1.25cm3 X 1.049 = 1.31125 this is the mass...
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