Principles of Instrumental Analysis, 6th Edition
6th Edition
ISBN: 9788131525579
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cenage Learning
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Question
Chapter 14, Problem 14.8QAP
Interpretation Introduction
(a)
Interpretation:
The molar concentration of nickel and cobalt in a solution should be determined.
Concept introduction:
The Beer-Lambert Law
A − absorbance
⁊ - molar absorptivity
l − length of the solution light passes through (cm)
c − concentration of solution (mol/L)
Interpretation Introduction
(b)
Interpretation:
The molar concentration of nickel and cobalt in a solution should be determined.
Concept introduction:
The Beer-Lambert Law
A − absorbance
⁊ - molar absorptivity
l − length of the solution light passes through (cm)
c − concentration of solution (mol/L)
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AG/F-2° V
3. Before proceeding with this problem you may want to glance at p. 466 of your textbook
where various oxo-phosphorus derivatives and their oxidation states are summarized.
Shown below are Latimer diagrams for phosphorus at pH values at 0 and 14:
-0.93
+0.38
-0.50
-0.51 -0.06
H3PO4 →H4P206 →H3PO3 →→H3PO₂ → P → PH3
Acidic solution
Basic solution
-0.28
-0.50
3--1.12
-1.57
-2.05 -0.89
PO HPO H₂PO₂ →P → PH3
-1.73
a) Under acidic conditions, H3PO4 can be reduced into H3PO3 directly (-0.28V), or via the
formation and reduction of H4P206 (-0.93/+0.38V). Calculate the values of AG's for both
processes; comment.
(3 points)
0.5
PH
P
0.0
-0.5
-1.0-
-1.5-
-2.0
H.PO,
-2.3+
-3 -2
-1
1
2
3
2
H,PO,
b) Frost diagram for phosphorus under acidic
conditions is shown. Identify possible
disproportionation and comproportionation processes;
write out chemical equations describing them. (2 points)
H,PO
4
S
Oxidation stale, N
4. For the following complexes, draw the structures and give a d-electron count of the
metal:
a) Tris(acetylacetonato)iron(III)
b) Hexabromoplatinate(2-)
c) Potassium diamminetetrabromocobaltate(III)
(6 points)
2. Calculate the overall formation constant for [Fe(CN)6]³, given that the overall formation
constant for [Fe(CN)6] 4 is ~1032, and that:
Fe3+ (aq) + e
= Fe²+ (aq)
E° = +0.77 V
[Fe(CN)6]³ (aq) + e¯ = [Fe(CN)6] (aq) E° = +0.36 V
(4 points)
Chapter 14 Solutions
Principles of Instrumental Analysis, 6th Edition
Ch. 14 - Prob. 14.1QAPCh. 14 - A 0.4740-g pesticide sample was decomposed by wet...Ch. 14 - Sketch a photometric titration curve for the...Ch. 14 - Prob. 14.4QAPCh. 14 - Prob. 14.5QAPCh. 14 - The accompanying data (1.00-cm cells) were...Ch. 14 - A 3.03-g petroleum specimen was decomposed by wet...Ch. 14 - Prob. 14.8QAPCh. 14 - Prob. 14.9QAPCh. 14 - The acid-base indicator HIn undergoes the...
Ch. 14 - Prob. 14.11QAPCh. 14 - Prob. 14.12QAPCh. 14 - Copper(II) forms a 1:1 complex with the organic...Ch. 14 - Aluminum forms a 1:1 complex with...Ch. 14 - Prob. 14.15QAPCh. 14 - Prob. 14.16QAPCh. 14 - Prob. 14.17QAPCh. 14 - Prob. 14.18QAPCh. 14 - Prob. 14.19QAPCh. 14 - Given the Information that...Ch. 14 - Prob. 14.21QAPCh. 14 - Mixing the chelating reagent B with Ni(II) forms...Ch. 14 - Prob. 14.23QAP
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