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APA
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Literature & Language
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Lab Report
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English (U.S.)
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Lab Report Assignment (Lab Report Sample)

Instructions:

Report on laboratory assesment of reactions

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Content:

PJ1100 LAB RE-ASSESSMENT
Name
Institution
Course
Tutor
Date
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 Ã&mda...
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