Lab Report Assignment (Lab Report Sample)

PJ1100 LAB RE-ASSESSMENT
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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.
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.
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?
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 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?
Solution:
(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...