9. A sodium thiosulfate solution is standardized using pure copper as the primary standard. A sample of copper weighing 0.2624 g is dissolved in acid, excess Kl is added, and the liberated iodine was titrated with 42.18 mL of sodium thiosulfate solution. Calculate the molarity of sodium thiosulfate solution. 2Cu2* + 41 → 2Cul + 12 I2 + 252032 - 21+ S4062

9. A sodium thiosulfate solution is standardized using pure copper as the primary standard. A sample of copper weighing 0.2624 g is dissolved in acid, excess Kl is added, and the liberated iodine was titrated with 42.18 mL of sodium thiosulfate solution. Calculate the molarity of sodium thiosulfate solution. 2Cu2* + 41 → 2Cul + 12 I2 + 252032 - 21+ S4062

Chemistry & Chemical Reactivity

10th Edition

ISBN:9781337399074

Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Publisher:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel

Chapter4: Stoichiometry: Quantitative Information About Chemical Reactions

Section4.9: Spectrophotometry

Problem 2.1ACP: Excess KI is added to a 100.0-mL sample of a soft drink that had been contaminated with bleach,...

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3. A 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the end point. Report the purity of the Na2CO3 sample.
You collected the following data from a titration experiment using a 0.129M standardized NaOH solution to titrate a 26.55 mL solution with an unknown Molarity concentration (M) of sulfuric acid (H2SO4). Initial Burette Reading (mL) Final Burette Reading (mL) Vol Delivered (mL) Trial 1 0.44 19.69 ?? Trial 2 0.18 17.2 Trial 3 0.50 19.94 For just Trial 1, determine the amount of NaOH delivered for the titration with appropriate significant digits. Do not include units.
The solubility of borax, which is made up of sodium tetraborate (Na2B4O5(OH)4 8H2O), was analyzed. The dissolution of borax is: Na2B4O5(OH)4 • 8H2O(s) ⇌ 2 Na+(aq) + B4O5(OH)42–(aq) + 8 H2O(l) A 50 mL saturated solution was prepared. After filtration of solution, 5 mL aliquot was transferred to a flask and titrated using 0.432 M HCl. The endpoint was found to be 4.73 mL of the titrant. Tetraborate anion (B4O5(OH)42-) is a weak base which reacts with HCl like the following reaction: B4O5(OH)42–(aq) + 2 H+(aq) + 3 H2O(l) ⇌ 4 H3BO3(aq) What is Ksp expression for the dissolution? What is the tetraborate ions concentration in the filtrate? What is the molar solubility and Ksp of borax if the titration was done at room temperature (298 K)?
The tartaric acid, H2C4H4O6 (150.087 g/mol ) in 100.0 mL wine sample was determined using sodium hydroxide standard solution. In the experiment, 35.21 mL of the base was initially standardized against 3.001 grams primary standard of KHP (204.22 g/mol). What is the standard concentration of the prepared NaOH solution?
An impure sample of calcium carbonate with a mass of 7.95 g was reacted with 50.00 cm3 of 1.00 mol dm hydrochloric acid (an excess). The resulling solution was transferred to a volumetric flask and titrated with 11.10cm3 of 0.300 mol dm-3 sodium hydroxide solution. Determine the percentage purity by mass of the calcium carbonate sample.CaCO3 + HCl -> CaCl2 + H2O + CO2 HCl + NaOH -> NaCl +H2O a. Determine how many moles of hydrochloric acid were used.b. Determine how many moles of excess HCI was titratedc. Determine how much in moles calcium carbonate present in the sample.d. Calculate the mass of calcium carbonate presente. Determine the percentwge calcium carbonate is in the sample.
Aspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..
Use the following atomic masses (in g/mol): Mg = 24.31; H = 1; S = 32.06; O = 16; Na = 23; Cl = 35. 45; Ca = 40.08; C = 12.01; N = 14.01; Mg = 24.31 5.A sample of Chlorpheniramine Maleate (99.82%) weighing 0.502g was assayed by non-aqueous titrimetry and was found to be equivalent to 22.2mL of perchloric acid. Calculate for the normality of perchloric acid. Each mL of 0.1N perchloric acid is equivalent to 19.54mg of Chlorpheniramine Maleate.
A sodium thiosulfate solution is standardized using pure copper as the primary standard. A sample of copper weighing 0.2624 g is dissolved in acid, excess KI is added, and the liberated iodine was titrated with 42.18 mL of sodium thiosulfate solution. Calculate the molarity of sodium thiosulfate solution. 2Cu2+ + 4I- → 2CuI + I2 I2 + 2S2O32- → 2I- + S4O62-
3I- + OCl- + 2H+ ----> I3- + Cl- + H2: A 25.00 ml sample of liquid bleach was diluted to 1000 ml in a graduated flask. A 25 ml portion of the diluted sample was pipetted into an Erlenmeyer flask and treated with excess KI to oxidize OCl- to Cl- and I3- was produced at the end of the reaction. The released I3 was titrated with 0.09892 M Na2S2O3 and 8.96 ml was spent to reach the turning point besides the starch indicator. What is the percent by weight/volume of NaOCl in the bleach sample? (NaOCl:74.44 g/mol)
The thiourea in a 1.455 g sample of organic material was extracted into a dilute sulfuric acid solution and titrated with 37.31 mL of 0.009372 M Hg2+ via reaction: 4(NH2)2CS + Hg2+ →[(NH2)2CS]4 Hg2+ Is this an example of total analysis technique or concentration technique? Explain. Calculate the percent (NH2)2CS ( 76.12 g/mol) in the sample. What is classification of the analysis based on the amount of sample and amount of analytes present? Explain.
Gravimetric analysis of Fe3O4 (MW = 232 g/mole) may be undertaken with the following reactions: Fe3O4 → Fe2O3 → Fe (OH)3. Weight of sample containing 8.00% Fe3O4 that must be taken to obtain a precipitate of Fe(OH)3 (MW = 107 g/mole) that weighs 150 mg is . a. 0.108 g b. 0.325 g c. 1.355 g d. 4.065 g Amount of Fe2O3 (MW = 160 g/mole) from which 150 mg of Fe(OH)3 (MW = 107 g/mole) may be obtained is . a. 0.112 g b. 0.224 g c. 0.448 g d. none of the other choices
A 11.00 gram sample of Vodka was analysed for its alcohol content. The sample was placed in a300 mL volumetric flask and made to the mark with distilled water. A 20.0 mL aliquot of the dilutedVodka was taken from the volumetric flask and placed in a conical flask containing some distilled water from a previous washing.The 20.0 mL aliquots were titrated against a 0.1302 M Cr2O72-solution and an average titre of13.25 mL was obtained.The reaction between the alcohol and the dichromate solution is given below2Cr2O72–(aq) + 3CH3CH2OH(aq) + 16H+(aq)  4Cr3+(aq) + 3CH3COOH(aq) + 11H2O(l)a) Calculate the amount in mol of Cr2O72– in the average titre. b) Calculate the percentage, by mass, of ethanol in the vodka.