1(a)(b)Quinine in a 1.553-g antimalarial tablet was dissolved in sufficient 0.10 M HCI togive 250.0 mL. A 10.00 mL aliquot was then diluted to 50.0 mL with the acid. Thefluorescence intensity for the diluted sample at 347.5 nm provided a reading of 196on an arbitrary scale. A standard 100 ppm quinine solution registered 125 whenmeasured under conditions identical to those for the diluted sample. Calculate themass of quinine in milligrams in the tablet.Quinine in a 3.925-g antimalarial tablet was dissolved in sufficient 0.10 M HCI togive 1.000 L. Dilution of a 20.00 mL aliquot to 100.00 mL yielded a solution thatgave a reading of 415 at 347.5 nm. A second 20.00 mL aliquot was mixed with 10.0mL of 50-ppm quinine solution before dilution to 100 mL. The fluorescence intensityof this solution was 503. Calculate the percentage of quinine in the tablet.

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Answered: 1(a) (b) Quinine in a 1.553-g…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter28: Atomic Spectroscopy

Section: Chapter Questions

Problem 28.15QAP

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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?
5. (1) (ii) Ibuprofen (206.3 g/mol) absorbs at 255 nm and shows a molar absorptivity of 3010 L/mol.cm. A tablet of ibuprofen, 0.6654 g, was ground and 0.2203 g powder was dissolved in 20% methanol in a 100 mL volumetric flask (Solution A). Exactly 5.0 mL of Solution A was diluted in a 50.0 mL flask and the resulting solution showed an absorbance of 0.1228 AU in a 1.00 cm cell. (i) What is the concentration (mol/L) of ibuprofen in the 100 mL stock solution? (ii) Report the amount of ibuprofen in mg/Tablet. Ibuprofen H3C | HCCH; 1 H3C CH3 -CH-COOH
Quinine in a 1.664 g antimalarial tablet was dissolved in sufficient 0.10 M HCl to give 500 mL of solution. A 15.00 mL aliquot was then diluted to 100.0 mL with the acid. The fluorescense intensity for the diluted sample at 347.5 nm provided a reading of 288 on an arbitrary scale. A standard 100 ppm quinine solution registered 180 when measured under conditions identical to those for the diluted sample. Calculate the mass in milligrams of quinine in the tablet.
Blue Blue dye stock solution 0.293 M Absorbance at 630 nm 0.00265 Calibration curve y = 0.0833x A solution is prepared by diluting 2.81 mL of the blue dye stock solution to 25.00 mL. The measure absorbance for the prepared solution is listed in the data table. (a) What is the theoretical molar concentration? [Blue]theoretical | × 10 |M (b) What is the experimental molar concentration? [Bluelexperimental x 10 M (c) What is the percent error? Percent error (blue) =
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Suppose that you are trying to measure the concentration of SO2 in air. To do that, you followed the para-rosaniline method. The SO2 was collected in a 10-ml solution of HgCl4 2- then diluted to 25 ml after the addition of the appropriate reagents. The absorbance owas measured at 650 nm in a 2.0-cm cell, yielding a value of 0.535. A standard sample was prepared by substituting a 1.00 mL sample of a standard solution containing the equivalent of 15.00 ppm SO2 for the airsample. The absorbance of the standard is found out to be 0.191. Report the concentration of SO2 in the air in parts per million. The density of air is 1.19 g/L
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A student prepared four standard solutions of (NH4)2Fe(SO4)2 and measured their absorbance at 505 nm. She created a calibration plot with the best-fit line of y = 9372 x + 0.0539. She measured the absorbance of her diluted unknown sample solution as (4.69x10^-1) nm. If she prepared her diluted sample solution by diluting 10.00 mL of the original solution to a total volume of 50.00 mL with deionized water, what is the concentration of Fe2+ in her original solution? Answer in units of M. Note: Your answer is assumed to be reduced to the highest power possible.
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 cuvette (1.00 cm light path) of 0.382. Calculate the mass of Co(NO3)2 contained in the original 50.0 mL stock solution.

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