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Lab Report
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Lab Report Assignment (Lab Report Sample)


Report on laboratory assesment of reactions


Acid-Base Practical
1] 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.
A balanced chemical reaction between potassium hydrogen phthalate and sodium hydroxide is depicted as follows:
KHC8H4O4 + NaOH NaKC8H4O4 + H2O
Mole ratio of potassium hydrogen phthalate: sodium hydroxide = 1:1
Moles of potassium hydrogen phthalate that reacted with NaOH = mass of KHC8H4O4/ Mr
= 0.8592 g/204.2 g mol-1
= 4.2 × 10^-3 mol
Using 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 mol
Volume of sodium hydroxide used to standardize KHC8H4O4 is 32.70 mL
Concentration 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.
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 = 1:1
Moles of NaOH = Molarity × volume
= 0.13 molL-1 × 0.01365 L
= 1.77 × 10^-3 mol
Moles 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?
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 practical
4] 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 practical
5] What is the structure of the complex anion of potassium tris(ethanedioato)aluminate produced in the practical?
(Shriver & Atkins 1991)
(2 marks)
6] Giving reasons for your answers, would you expect to find analogous complexes if aluminium was replaced by:
1 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)
2 a transition metal?
An analogous complex can be formed if a transition metal is used because most transitio...
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