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Analysing aspirin tablets.

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Topic: Analysing aspirin tablets. I familiarised myself with the Material Safety Data Sheets of toxic substances. PLANNING (A) Aspirin1, or acetylsalicylic acid is a drug in the family of salicylates, often used as an analgesic (to relieve minor aches and pains), antipyretic (to reduce fever), and as an anti-inflammatory. It also has an antiplatelet ("blood-thinning") effect and is used in long-term, low doses to prevent heart attacks and cancer. figure 1 figure 2 Aspirin tablets consist mainly of 2-ethanoylhydroxybenzoic acid (acetylsalicylic acid, CH3COOC6H4COOH, figure 3). figure 3 Although the acidic conditions found in stomach do not affect aspirin, the alkaline juices in the intestines hydrolyse it to ethanoate (acetate) ions and 2-hydroxybenzoate ions. CH3COOC6H4COOH + 2OH- � CH3COO- + HOC6H4COO- + H2O Aim: To determine the percentage of CH3COOC6H4COOH in aspirin tablets. Hypothesis: When a known amount of standard sodium hydroxide solution is used in excess to hydrolyse a known mass of aspirin tablets, we may determine the percentage of acetylsalicylic acid in those tablets. Prediction: The quantity of acetylsalicylic acid will be smaller than the one given on the packaging and in the leaflet enclosed. PLANNING (B) Requirements: - 3 aspirin tablets (Acidium acetylsalicylium 500 mg) - pipette (25 cm3) ...read more.


DATA COLLECTION Mass of 3 aspirin tablets/ g 1.8 Mean mass of one tablet/ g 0.6 Titration number Rough Accurate 1 Accurate 2 Final burette reading/ cm3 1 1.4 1.3 Initial burette reading/ cm3 0 0 0 Volume of 0.10 mol dm-3 HCl added/ cm3 24.0 23.6 23.7 Mean volume of HCl added 23.8 V c n NaOH 25 cm3 = 0.025 dm3 ? ? HCl 23.8 cm3 = 0.023.8 dm3 0.10 mol dm-3 0.00238 mol DATA PROCESSING AND PRESENTATION * Standardisation of the approx. 1.0 mol dm-3 NaOH During titration the reaction below takes place: HCl(aq) + NaOH(aq) � NaCl(aq) + H20(l) That is why the solution changes its colour from pink into transparent. The ratio between HCl and NaOH is 1/1, so one mol of HCl reacts with one mol of NaOH. n = c � V NaOH VNaOH = 25 cm3 = 0.025 dm3 cNaOH = ? nNaOH = nHCl HCl VHCl = 23.8 cm3 = 0.0238 dm3 cNaOH = 0.10 mol dm-3 nNaOH = 0.10 mol dm-3 � 0.0238 dm3 = 0.00238 mol 1. From the balanced chemical equation find the mole ratio NaOH:HCl 1:1 2. Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10-3 moles at the equivalence point 3. ...read more.


burette: 0.5% 4. calibrated flask: 0.2% Total: 0.92 % CONCLUSION AND EVALUATION As we can see from the results above, the prediction made at the very beginning of this lab was correct. Neither type of acid or base nor the concentration of acid does not have influence on the enthalpy of neutralisation. Hence we may assume that the enthalpy of neutralisation is equal to the enthalpy change for H+ + OH-� H2O. The enthalpy change for this reaction, however, is -57.9 kJ mol-1. The differences between my results and the theoretical value may come from the fact that the measurements were not very accurate. The temperatures of the acids, bases and mixtures might have been influenced by cool beakers. Therefore the temperatures were a bit lower than they should have been. If the ?T was higher by 3oC, the enthalpy of neutralisation would be almost the same as in the sources. I do not know how to improve the experiment so that data gathered will be similar to theoretical values. I reckon in classroom conditions such mistake is not a serious one. SOURCES: 1. Green J, Damji S. 2001. Chemistry. Second edition. IBID Press, Victioria, Australia. 2. http://en.wikipedia.org/wiki/Standard_enthalpy_change_of_neutralisation http://en.wikipedia.org/wiki/Aspirin 1 The definition comes from http://en.wikipedia.org/wiki/Aspirin Figures 1 and 2 come from http://en.wikipedia.org/wiki/Aspirin Figure 3 comes from http://home.caregroup.org/clinical/altmed/interactions/Images/Drugs/aspirin.gif ANALYSING ASPIRIN TABLETS 1 ...read more.

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