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The acid-base indicator Hln undergoes the following reaction in dilute

Principles of Instrumental Analysis | 6th Edition | ISBN: 9780495012016 | Authors: Douglas A. Skoog F. James Holler Stanley R. Crouch ISBN: 9780495012016 317

Solution for problem 14-10 Chapter 14

Principles of Instrumental Analysis | 6th Edition

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Principles of Instrumental Analysis | 6th Edition | ISBN: 9780495012016 | Authors: Douglas A. Skoog F. James Holler Stanley R. Crouch

Principles of Instrumental Analysis | 6th Edition

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Problem 14-10

The acid-base indicator Hln undergoes the following reaction in dilute aqueoussolution:The following absorbance data were ohtained for a 5.00 X IW' M solution ofHIn in 0.1 M NaOH and 0.1 M HCI. Measurements were made at wavelengths of485 nm and 625 nm with 1.00cm cells.0.1 MNaOH0.1 MHCIAm = 0.075A.S) = 0.487A62; = 0.904A62j = 0.181[n the NaOH solution, essentially all of the indicator is present as In ; in theacidic solution, it is essentiallv all in the form of HIn.(a) Calculate molar absorpti~ities for In - and HIn at 485 and 625 nm.(h) Calculate the acid dissociation constant for the indicator if a pH 5.00 buffercontaining a small amount of the indicator exhibits an absorbance of 0.567 at.+85nm 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 thatexhihits an absorbance of 0.492 at 485 nm and 0.2'+5 at 635 nm (l.OO-cmcells)'!(d) A 25.00-mL aliquot of a solution of purified weak organic acid HX requiredexactly 2.+.20mL of a standard solution of a strong hase to reach a phenolphthaleinend point. When exactly 12.10 mL of the hase was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the indicatorunder consideration, the absorbance was found to be 0.333 at 485 nmand 0.655 at 625 nm (I.OO-cmcells). Calculate the pH of the solution and K,for the weak acid.(e) What would be the absorhance o[ a solution at 485 and 625 nm (1.50-cmcells) that was 2.00 x 10-' M in the indicator and was buffered to a pH of6.000"

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Elements and the structure of the atom  Atom o The smallest most fundamental unit of matter  Element o A substance composed of one type of atom  Compound o Made up of 2 or more elements Nucleus  Tight group of protons and neutrons in the center of atom o Electrons are outside of nucleus Atomic mass ...

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Chapter 14, Problem 14-10 is Solved
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Textbook: Principles of Instrumental Analysis
Edition: 6
Author: Douglas A. Skoog F. James Holler Stanley R. Crouch
ISBN: 9780495012016

This full solution covers the following key subjects: . This expansive textbook survival guide covers 34 chapters, and 619 solutions. This textbook survival guide was created for the textbook: Principles of Instrumental Analysis , edition: 6. The full step-by-step solution to problem: 14-10 from chapter: 14 was answered by , our top Chemistry solution expert on 03/02/18, 06:21PM. The answer to “The acid-base indicator Hln undergoes the following reaction in dilute aqueoussolution:The following absorbance data were ohtained for a 5.00 X IW' M solution ofHIn in 0.1 M NaOH and 0.1 M HCI. Measurements were made at wavelengths of485 nm and 625 nm with 1.00cm cells.0.1 MNaOH0.1 MHCIAm = 0.075A.S) = 0.487A62; = 0.904A62j = 0.181[n the NaOH solution, essentially all of the indicator is present as In ; in theacidic solution, it is essentiallv all in the form of HIn.(a) Calculate molar absorpti~ities for In - and HIn at 485 and 625 nm.(h) Calculate the acid dissociation constant for the indicator if a pH 5.00 buffercontaining a small amount of the indicator exhibits an absorbance of 0.567 at.+85nm 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 thatexhihits an absorbance of 0.492 at 485 nm and 0.2'+5 at 635 nm (l.OO-cmcells)'!(d) A 25.00-mL aliquot of a solution of purified weak organic acid HX requiredexactly 2.+.20mL of a standard solution of a strong hase to reach a phenolphthaleinend point. When exactly 12.10 mL of the hase was added to a second 25.00-mL aliquot of the acid, which contained a small amount of the indicatorunder consideration, the absorbance was found to be 0.333 at 485 nmand 0.655 at 625 nm (I.OO-cmcells). Calculate the pH of the solution and K,for the weak acid.(e) What would be the absorhance o[ a solution at 485 and 625 nm (1.50-cmcells) that was 2.00 x 10-' M in the indicator and was buffered to a pH of6.000"” is broken down into a number of easy to follow steps, and 262 words. Principles of Instrumental Analysis was written by and is associated to the ISBN: 9780495012016. Since the solution to 14-10 from 14 chapter was answered, more than 455 students have viewed the full step-by-step answer.

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