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- The accompanying data (1.00-cm cells) were obtained for the spectrophotometric titration 10.00 mL of Pd(II) with 2.44 10-4 M Nitroso R(O. W Rollins and M. M. Oldham, Anal. chem .,1971, 43, 262, DOI: 10.1021/ac60297a026). Calculate the concentration of the Pd(II) solution, given that the ligand-to-cation ratio in the colored product is 2:1Sketch a photometric titration curve for the titration of Sn2+ with MnO4 . What color radiation should be used for this titration? Explain.The concentration of ammonia in a cleaning product was determined by back titration.Firstly, 10.00 cm3 of the cleaning product was pipetted into a large conical flask,containing 250.00cm3 of 0.50 mol/l HCl to give Solution A.Following a period of reaction and shaking, 50.00cm3 of Solution A was removed anddiluted to 250 cm3 with water in a volumetric flask to give Solution B.20 cm3 samples of Solution B were titrated against 0.05 mol/l Na2CO3 solution, givingan average titre of 12.45 cm3. i) Write equations for the reactions that have taken place.ii) Determine the concentration of NH3 in the original cleaning product in mol/l,g/l, ppm, and % w/v.
- A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA. Calculate mmol Ni in the 50.00 mL aliquot. Calculate mmol Br- in the 50.00 mL aliquot. Calculate the percentage of NaBr (102.894) in the 1.000 g sample.A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA The other 50.00 mL remaining solution was also analyzed for its Br- content by potentiometry using a metallic electrode of the second kind. Ecell recorded in running the solution using the potentiometric set-up was Ecell = 0.0286 V. (E0Ag/AgBr = 0.095 V) g) Compute for Eind. h) Compute pBr in the 50.00 mL aliquot. i) Compute for % NaBr ( in the potentiometric technique). j) Calculate the error between the obtained % NaBr fr the EDTA titration technique and the % NaBr from the potentiometric technique.The amount of iron in a meteorite was determined by a redox titration using KMnO4 as the titrant. A 0.4185 -g sample was dissolved in acid and the liberated Fe3+ quantitatively reduced to Fe2+, using a reductor column. Titrating with 0.0044 M KMnO4 requires 28.76 mL to reach the end point. Determine the %w/w Fe2O3 in the sample of meteorite. (Fe = 55.845 amu, Fe2O3 = 159.69 g/mol
- A 0.3045 g of CaCO3 primary standard was dissolved using concentrated HCl, evaporated to incipient dryness and quantitatively transferred to a 250 mL volumetric flask, and diluted to mark. A 10 mL of aliquot was then transferred to an Erlenmeyer flask, together with 5mL buffer and 5 drops of EBT indicator, and was used to standardize the EDTA titrant. The solution turned blue after the addition of 24.10 mL of EDTA. A blank correction was performed and was determined to be 0.4 mL. What is the concentration of EDTA obtained (MW CaCO3 = 100.0869 g/mol)?A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br-Nickel ion required 11.70 mL of 0.002146 M EDTAWhat is the percentage of NaBr (102.894) in the 1.000 g sample?A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br-50.00 mL remaining solution was analyzed for its Br- content by potentiometry using a metallic electrode of the second kind. a) Write the cell notation of the potentiometric set-up with SCE as the reference electrode. b) Write the Nernst equation that describes the indicator electrode set-up. Ecell recorded in running the solution using the potentiometric set-up was Ecell = 0.0286 V. (E0Ag/AgBr = 0.095 V) c) Compute for Eind. d) Compute pBr in the 50.00 mL aliquot. e) Compute for % NaBr ( in the potentiometric technique).
- A 0.3045 g of CaCO3 primary standard was dissolved using concentrated HCl, evaporated to incipient dryness, and quantitatively transferred to a 250 mL volumetric flask and diluted to mark. A 10 mL of aliquot was then transferred to an Erlenmeyer flask, together with 5mL buffer and 5 drops of EBT indicator, and was used to standardize the EDTA titrant. The solution turned blue after the addition of 24.10 mL of EDTA. A blank correction was made and was determined to be 0.4 mL. [Analysis] A 30 mL unknown water sample was treated with 37.6 mL of EDTA. A blank correction was made and was determined to be 0.4 mL. What is the concentration of Ca2+ (40.0780 g/mol) in ppm?A second calibration standard solution of an iron(III) salicylate complex was prepared in two steps. First, 10.0 mL of a 0.100 M stock solution was added to 90.0 mL of solvent to make 100.0 mL of the first calibration standard solution and, secondly, 80 mL of that first calibration solution plus 20.0 mL of solvent were mixed to make the second calibration standard. What is the concentration of the second calibration standard solution?The bismuth (AW 208.98) in 0.6805 g of an alloy was precipitated as BiOCl (FW 260.43) and separated from the solution by filtration. The washed precipitate was dissolved in nitric acid and treated with 15.02 mL of 0.1498 M AgNO3, causing the precipitation of AgCl. The excess AgNO3 required 14.04 mL of 0.1008 M KSCN for titration. Calculate %Bi in the sample.The bismuth (AW 208.98) in 0.6805 g of an alloy was precipitated as BiOCl (FW 260.43) and separated from the solution by filtration. The washed precipitate was dissolved in nitric acid and treated with 15.02 mL of 0.1498 M AgNO3, causing the precipitation of AgCl. The excess AgNO3 required 14.04 mL of 0.1008 M KSCN for titration. Calculate %Bi in the sample.
