Solutions of cobalt (II) ions have a magenta color with a molar absorptivity constant (e) value at 511 nm of 5.87 M-1cm-1. A 0.500 mL sample is transferred from a 50.0 mL stock solution of Co(NO3)2 and diluted with water to a final volume of 7.50 mL. The dilution had an absorbance value at 511 nm in a standard c

Solutions of cobalt (II) ions have a magenta color with a molar absorptivity constant (e) value at 511 nm of 5.87 M-1cm-1. A 0.500 mL sample is transferred from a 50.0 mL stock solution of Co(NO3)2 and diluted with water to a final volume of 7.50 mL. The dilution had an absorbance value at 511 nm in a standard c

Principles of Instrumental Analysis

7th Edition

ISBN:9781305577213

Author:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Publisher:Douglas A. Skoog, F. James Holler, Stanley R. Crouch

Chapter13: An Introduction To Ultraviolet-visible Molecular Absorption Spectrometry

Section: Chapter Questions

Problem 13.11QAP: The equilibrium constant for the conjugate acid-base pair HIn+H2OH3O++In is 8.00 10-5. From the...

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Mixing the chelating reagent B with Ni(II) forms the highly colored NiB22+, whose solutions obey Beer’s lawat 395 nm overa wide range. Provided the analytical concentration of the chelating reagent exceeds that of Ni(II) by a factor of 5 (or more), the cation exists, within the limits of observation, entirely in the form of the complex. Use the accompanying data to evaluate the formation constant Kf for the process Ni2++2BNiB22+
The equilibrium constant for the conjugate acid-base pair HIn+H2OH3O++In is 8.00 10-5. From the additional information in the following table, (a) calculate the absorbance at 430 nmand 600 nm for the following indicator concentrations: 3.00 10-4M,2.00 10-4M, 1.00 10-4M, 0.500 10-4 M, and 0.250 10-4M. (b) plot absorbance as a function of indicator concentration.
A foot powder sample containing Zn was dissolved on 50.00 mL water and was titrated to the end point color with 22.57 mL of 0.01639 M EDTA at pH=4. (α= 3.61x109, KKzny2 = 3.2 x 1016). The pZn in the sample is:
0 The acid-base indicator HIn undergoes the following reaction in dilute aqueous solution:HIn color1 m H1 1 In2color2The following absorbance data were obtained for a 5.00 3 1024M solution of HIn in 0.1 M NaOH and0.1 M HCl. Measurements were made at wavelengths of 485 nm and 625 nm with 1.00-cm cells.0.1 M NaOH A485 5 0.075 A625 5 0.9040.1 M HCl A485 5 0.487 A625 5 0.181In the NaOH solution, essentially all of the indicator is present as In2; in the acidic solution, it is essentiallyall in the form of HIn.(a) Calculate molar absorptivities for In2 and HIn at 485 and 625 nm.(b) Calculate the acid dissociation constant for the indicator if a pH 5.00 buffer containing a small amountof 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…
A 370.00 mL solution of 0.00190 M AB5 is added to a 200.00 mL solution of 0.00165 M CD2. What is pQsp for AD5?
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 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?
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.
Calculate the hardness in terms of parts per million calcium carbonate (100.0869 g/mol) in a 50.00 mL aliquot of milk sample. The sample was treated with a small amount of ammonia buffer to bring the pH to 10 followed by the addition of 10 drops EBT indicator. The milk sample required 10.00 mL of 8.000 x 10-3 M EDTA for titration. 160 ppm 4 80 320
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.