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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 2:1Given the Information that Fe3++Y4FeYKi=1.01025Cu2++Y4CuY2Ki=6.31018 and the further information that, among the several reactants and products, only CuY2- absorbs radiationat 750 nm, describe how Cu(II) could be used as an indicator for the photometric titration of Fe(III) with H2Y2-. Reaction:+ Fe3++ H2Y2- FeY- + 2H+.At 580 nm, which is the wavelength of its maximum absorption, the complex Fe(SCN)2+ has a molar absorptivity of 7.00 I03L cm-1 mol-1.Calculate (a) the absorbance of a 4.47 10-5 M solution of the complex at 580 nm in a 1.00-cm cell. (b) the absorbance of a solution in a 2.50-cm cell in which the concentration of the complex is one half that in(a). (c) the percent transmittance of the solutions described in (a) and (b). (d) the absorbance of a solution that has half the transmittance of that described in (a).
- When I was a boy, I watched Uncle Wilbur measure the iron content of runoff from his banana ranch. He acidified a 25.0-mL sample with HNO3 and treated it with excess KSCN to form a red complex. (KSCN itself is colorless.) He then diluted the solution to 100.0 mL and put it in a variablepathlength cell. For comparison, he treated a 10.0-mL reference sample of 6.80 3 1024 M Fe31 with HNO3 and KSCN and diluted it to 50.0 mL. The reference was placed in a cell with a 1.00-cm pathlength. Runoff had the same absorbance as the reference when the pathlength of the runoff cell was 2.48 cm. What was the concentration of iron in Uncle Wilbur’s runoff ?Please help me fill the remaining spaces. The spaces marked with NA don't need to be calculate. Name Molecular Weight Amounts used mmol Equivalents M.P or BP (C) Density (g/mL) Indene 116.16 .5mL MP=-2 BP=181-182 .996 Borane-tetrahydrofuran Complex 85.94 2.8mL 1 MP= -108 BP=66 NA 30% Hydrogen Peroxide 34.01 1mL MP= -33 BP=108 NA 3 M NaOH(aq) 40 .7mL MP=318 BP=1390 NA Diethyl Ether 74.12 NA NA NA BP=75-78 .902 Hexanes 86.17 NA NA NA BP=65.5-68.3 .659 Potassium Permanganate 158.03 NA NA NA BP=100 1.0 1-Indanol (Product) 134.13 NA MP=50-54 NA 2-Indanol (Product) Same as Above Same as Above Same as Above NA MP=68-71 NA5. Calculate the Ionic strength of solution prepared by mixing of 0.2 M NaNO3and 0.02M of K2Cr2O7:a) s = [Ksp / 27]1/5 b) s = Ksp1//5c) s = Ksp ½ d) s = [Ksp/108]1/5e) s = 108 Ksp
- EDTA abstracts bismuth(III) from its thiourea complex:Bi1tu2631 1 H2Y22 S BiY2 1 6tu 1 2H1where tu is the thiourea molecule (NH2)2CS. Predict the shape of a photometric titration curve based on thisprocess, given that the Bi(III) or thiourea complex is the only species in the system that absorbs light at 465nm, the wavelength selected for the analysis.The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution: HIncolor1H++Incolor2 The following absorbance data were obtained for a 5.00 I0-4 M solution of HIn in 0.1 M NaOH and 0.1 M HC1. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells. 0.1 M NaOH A485 = 0.075 A625 = 0.904 0.1 M HC1 A485 = 0.487 A625 = 0.181 In the NaOH solution, essentially all of the indicator is present as In-; in the acidic solution, it is essentially all in the form of HIn. (a) Calculate molar absorptivities for In- and HIn at 485 and 625 nm. (b) Calculate the acid dissociation constant for the indicator ¡fa pH 5.00 buffer containing a small amount of the indicator exhibits an absorbance of 0.567 at 485 nm and 0.395 at 625 nm (1.00-cm cells). (c) What is the pH of a solution containing a small amount of the indicator that exhibits an absorbance of0.492 at 485 nm and 0.245 at 635 nm (1.00-cm cells)? (d) A 25.00-mL aliquot of a solution of purified weak organic acid HX required exactly 24.20 mL of a standard solution of a strong base to reach a phenolphthalein end point. When exactly 12.10 mL of the base was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the Indicator under consideration, the absorbance was found to be 0.333 at 485 nm and 0.655 at 625 nm (1.00-cmcells). Calculate the pH of the solution and Ka for the weak acid. (e) What would be the absorbance of a solution at 485 and 625 nm (1.50-cm cells) that was 2.00 10-4 M in the indicator and was buffered to a pH of 6.000?Zinc(II) and the ligand L form a 1:1 complex that absorbs strongly at 600 nm. As long as the molar concentration of L exceeds that of zinc(II) by a factor of 5, the absorbance depends only on the cation concentration. Neither zinc(II) nor L absorbs at 600 nm. A solution that is 1.59 10-4 M in zinc(II) and 1.00 10-3 M in L has an absorbance of 0.352 in a 1.00-cm cell at 600 nm. Calculate (a) the percent transmittance of this solution. (b) the percent transmittance of this solution in a 2.50-cm cell. (c) the molar absorptivity of the complex.