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This is a lab report for experiment number 2
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Lab report done by: Aaida Hossain (2132500) Partner: Mia Alyssa Nardelli Presented to: Silja Makinen 202-NYA-05, Lab Section 14 Date of the Experiment: 27-10- Date of the Lab Report Submission: 03-11-
Procedure: Part A
A. H 2 C 2 O 4 ( aq )+^2 NaOH ( aq ) →^ Na 2 C 2 O 4 ( aq )+^2 H 2 O ( l ) B. NaOH ( aq )+^ HCl ( aq ) →^ NaCl ( aq )+ H 2 O ( l ) Table 1 : Part A : Standardization of NaOH Mass of oxalic acid dihydrate, g 1. Number of moles of oxalic acid dihydrate, mol 0. Volume of the volumetric flask, mL 50. Concentration of oxalic acid dihydrate solution, (^) mol ∙ L −^1 0. Table 2 : Part A : Determination of the unknown concentration of NaOH Titration 1 Titration 2 Titration 3 Final burette reading, mL 0.10 11.10 21. Initial burette reading, mL 11.10 21.60 32. Volume of NaOH delivered, mL 11.00 10.50 10. Volume of oxalic acid used, mL 10.00 10.00 10. Moles of oxalic acid used, mol 0.004 0.004 0. Moles of NaOH, mol 0.008 0.008 0. Concentration of NaOH, (^) mol ∙ L −^1 0.73^ 0.76^ 0. Average concentration of NaOH, (^) mol ∙ L −^1 0. Table 3: Part B: Determination of the unknown concentration of an HCl Solution Titration 1 Titration 2 Titration 3 Final burette reading, mL 10.90 20.90 31. Initial burette reading, mL 20.90 31.00 41. Volume of NaOH delivered, mL 10.00 10.10 10. Moles of NaOH, mol 0.0075 0.0076 0. Moles of acid used, mol 0.0075 0.0076 0. Volume of acid consumed, mL 10.00 10.00 10. Concentration of the acid, (^) mol ∙ L −^1 0.75^ 0.76^ 0. Average concentration of the acid, (^) mol ∙ L −^1 0. Sample calculations: Number of moles: mass molar mass = moles Number of moles of oxalic acid dihydrate =¿
¿ ¿ Moles = Molarity ∗ Volume Moles of oxalic acid used =0. mol L ∗0.4 L =0.004 mol
Stoichiometry Moles of NaOH = 0.004 mol H 2 C 2 O 4 ( aq ) 1 mol H 2 C 2 O 4 ( aq ) ∗ x moles NaOH (^) ( aq ) 2 mol NaOH ( aq ) =0.008 mol Concentration: Concentration = moles volume Concentration of oxalic acid dihydrate solution = 0.02 mol 0.05 L =0.40 mol ∙ L − 1 Volume of the NaOH delivered: Final reading − Initial reading = Volume of NaOH dilivered Volume of NaOH delivered =11.10 mL −0.10 mL =11.00 mL Average concentration ( C (^) ¿¿ 1 + C 2 + C 3 ) 3 = Cavg ¿ Average concentration of NaOH =
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