Monday, August 19, 2019
Vinegar :: essays research papers
Who’s been cheating on the vinegar? Aim: To investigate if the vinegars have been diluted by working out their concentrations. Scientific Background: Vinegar is an acid and there fore can be neutralised by an alkali. Acid + Alkali à salt + water Hydrochloric acid + Sodium Hydroxide à Sodium chloride + water Vinegar is a weak acid and therefore it only partially ionises in solution. CH3COOH ßà CH3COO- + H+ An alkali such as Sodium Hydroxide has a formula NaOH. If the alkali is strong then it ionises fully in solution. NaOH ßà Na+ + OH- When the acid H+ reacts with the alkali OH- the solution neutralises. Prediction: If the acid has been diluted there will be less H+ ions present in the volume used. Therefore to cancel the alkali more acid will be required. This is on the assumption that the alkali volume and concentration stay constant. Method:  ·Ã‚     Measure 10cm3 of 0.1m alkali into a measuring cylinder then place the solution into a 75cm3 conical flask.  ·Ã‚     Then add 2-3 drops of indicator to the alkali (it will go purple in colour)  ·Ã‚     Measure out 50 cm3 of the vinegar into a measuring cylinder and then place into a glass beaker.  ·Ã‚     Then using a dropper add the vinegar drop wise to the alkali until 1-drop changes the solution to colourless.  ·Ã‚     Replace the vinegar back into the measuring cylinder and record the volume of vinegar used.  ·Ã‚     Repeat this process twice more, recording results.  ·Ã‚     Then repeat the whole procedure for each different concentration of vinegar. Preliminary Work: Aim:- To find out which concentration of alkali will give accurate, measurable values. Concentration(M) Volume of acid(cm3) 0.1m 21.45cm3 0.5m 78.9cm3 0.75m 100+cm3 1.0m 100+cm3 2.0m 100+cm3 We decided from the results that the best measurable concentration of alkali would be the 0.1m concentrations. Investigation:      Test 1     Test 2     Test 3     Average Manufacturers’     10.27cm3     10.3cm3     9.73cm3     10.11cm3 Shop 1     10.3cm3     9.5cm3     10.25cm3     10.11cm3 Shop 2     18.4cm3     17.9cm3     18.4cm3     18.23cm3 Shop 3     40.1cm3     38.45cm3     35.7cm3     38.08cm3 Analysis: These results show that the vinegar from Shop 2 and Shop 3 have been significantly diluted with water or some other substance. The results obtained from shop1 show that its is very similar in concentration to the manufacturers’. Two of the results show that it is less concentrated than the manufacturers’, but one shows that it is more concentrated, this is impossible so its is most likely due to human error and not taking the readings properly. Concentrations: Manufacturers’= 1 ¸10.11 = 0.0989119683 Shop1 = 1 ¸10.11 = 0.0989119683 Shop2 = 1 ¸18.23 = 0.054854635 Shop3 = 1 ¸38.08 = 0.026260504 These results show an obvious difference in the concentration with the manufacturers’ being the most concentrated and shop 3’s being the least concentrated. Evaluation: The results from this experiment can be considered reasonably accurate we took the results quite accurately.
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