Practical - & questions sheet - Lab Report Example

PJ1100 LAB RE-ASSESSMENT Acid-Base Practical A sodium hydroxide solution is standardised by titrating 0.8592 g of a primary standard potassium hydrogen phthalate (Mr = 204.2 g mol-1) to a phenolphthalein endpoint, requiring 32.70 mL. Determine the concentration of the sodium hydroxide solution.Solution: A balanced chemical reaction between potassium hydrogen phthalate and sodium hydroxide is depicted as follows:KHC8H4O4 + NaOH NaKC8H4O4 + H2OMole ratio of potassium hydrogen phthalate: sodium hydroxide = 1:1Moles of potassium hydrogen phthalate that reacted with NaOH = mass of KHC8H4O4/ Mr = 0.

8592 g/204.2 g mol-1 = 4.2 × 10^-3 molUsing number of moles of KHC8H4O4 and the balanced chemical equation, the number of moles of NaOH present at each equivalence point = number of moles of KHC8H4O4 = 4.2 × 10^-3 molVolume of sodium hydroxide used to standardize KHC8H4O4 is 32.70 mLConcentration of sodium hydroxide solution = moles/volume = (4.

2 × 10^-3 mol × 1000 mL)/ 32.70 Ml= 0.13 M(5 marks)2] The sodium hydroxide standardised above was then titrated against 25.00 mL of a solution of an unknown monoprotic acid of unknown concentration. The titration required 13.65 mL of the sodium hydroxide to reach the endpoint. Determine the concentration of the unknown acid solution.Solution:A general reaction involved in neutralization titrations NaOH and a monoprotic acid can be depicted as follows: HA + NaOH NaA + H2O Mole ratio of NaOH : HA = 1:1Moles of NaOH = Molarity × volume = 0.

13 molL-1 × 0.01365 L= 1.77 × 10^-3 molMoles of monoprotic acid, HA = 1.77 × 10^-3 mol (mole ratio of 1:1)Concentration of HA = moles/volume = 1.77 × 10^-3 mol/ 0.025L = 0.071 M(5 marks)3] Why would phenolphthalein be used as the indicator in this titration?Solution:Phenolphthalein indicator is preferred in this titration because it changes colour over pH range of 8-10 and this helps the end point to be as close as the equivalence point.(2 marks)Heterocyclic synthesis practical4] Give a concise mechanistic sequence for the formation of 2-amino-4,6-dimethylpyrimidine formed during the practical and explain the role of all other reagents used.(5 marks)Potassium tris(ethanedioato)aluminate synthesis practical5] What is the structure of the complex anion of potassium tris(ethanedioato)aluminate produced in the practical?

Solution:(Shriver & Atkins 1991)(2 marks)6] Giving reasons for your answers, would you expect to find analogous complexes if aluminium was replaced by:a) an alkali metal?An analogous complex will not be found if Aluminium was replaced with an alkali metal because aluminium is of valence three while alkali metals are of valence 1 therefore different complexes will be formed.(2 marks)b) a transition metal?An analogous complex can be formed if a transition metal is used because most transition metals have a variety of oxidation states hence different valances (Lawrance 2010).(2 marks)7] Given that it requires 21.

70 mL of 0.02 M KMnO4 solution to react with 0.15 g of the aluminium complex formed in the practical, calculate the % ethanedioate in the product.Solution: Moles of KMnO4 = 0.02M × 0.0217 L = 4.34 × 10^ -4 molMoles of potassium tris(ethanedioato)aluminate = 0.15g/(437.15gmol-1) = 3.43× 10^ -4 molMass of ethanedioate in the complex = (264 × 0.15g)/437.15= 0.09 g% ethanedioate = (0.09g × 100%)/0.15 g = 60%(5 marks)Aspirin synthesis practical8] Draw a mechanism for the synthesis of aspirin from a reaction between salicylic acid and acetyl chloride.

Solution:(5 marks)9] What information can be obtained from the melting point of a compound?Solution:Melting point of a compound can be used to determine the purity of the compound hence its identity (Hornback 2006).(2 marks)Reference listHornback, JM 2006, Organic Chemistry, 2dn edn. Thompson Learning, Inc, Belmont.Shriver, DF Atkins, PW & Langford, CH 1991, Inorganic Chemistry, Longman, London.Lawrance, GA 2010, Introduction to Coordination Chemistry, John Wiley & Sons Ltd, West Sussex.