2 pages/≈550 words
Literature & Language
Lab Report Assignment (Lab Report Sample)
Report on laboratory assesment of reactionssource..
PJ1100 LAB RE-ASSESSMENT
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 Ã— 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 Ã— 10^-3 mol
Volume of sodium hydroxide used to standardize KHC8H4O4 is 32.70 mL
Concentration of sodium hydroxide solution = moles/volume
= (4.2 Ã— 10^-3 mol Ã— 1000 mL)/ 32.70 Ml
= 0.13 M
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 Ã— volume
= 0.13 molL-1 Ã— 0.01365 L
= 1.77 Ã— 10^-3 mol
Moles of monoprotic acid, HA = 1.77 Ã&mda...
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