The equilibrium constant for the following reaction is 1.0 x 10²3. Cr³+ (aq) + H₂EDTA² (aq) ⇒ CrEDTA¯(aq) + 2H+ (aq) 0₂C-CH₂ CH2CO2 EDTA+ 0₂C-CH₂ N-CH₂-CH₂-N M CH₂ -CO₂ Ethylenediaminetetraacetate EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt Na₂H₂ EDTA, are used to treat heavy metal poisoning. Calculate [Cr³+] at equilibrium in a solution originally 0.0050 M in Cr³+ and 0.060 M in H₂ EDTA²- and buffered at pH = 6.00. Assume that H₂EDTA² form dominates at this pH. [Cr³+] = [

The equilibrium constant for the following reaction is 1.0 x 10²3. Cr³+ (aq) + H₂EDTA² (aq) ⇒ CrEDTA¯(aq) + 2H+ (aq) 0₂C-CH₂ CH2CO2 EDTA+ 0₂C-CH₂ N-CH₂-CH₂-N M CH₂ -CO₂ Ethylenediaminetetraacetate EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt Na₂H₂ EDTA, are used to treat heavy metal poisoning. Calculate [Cr³+] at equilibrium in a solution originally 0.0050 M in Cr³+ and 0.060 M in H₂ EDTA²- and buffered at pH = 6.00. Assume that H₂EDTA² form dominates at this pH. [Cr³+] = [

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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Coordination Chemistry

It is a part of chemistry, in which the study of compounds that have central atoms surrounded by molecules or ions, known as ligand, is known as coordination chemistry.

Question

The equilibrium constant for the following reaction is 1.0 x 102³.
Cr³+ (aq) + H₂ EDTA² (aq) = CrEDTA¯(aq) + 2H+ (aq)
0₂C-CH₂-
CH2–CO2
CH2CO,
=
EDTA+
O₂C-CH₂
Ethylenediaminetetraacetate
EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt Na₂ H₂ EDTA, are used to treat
heavy metal poisoning. Calculate [Cr³+] at equilibrium in a solution originally 0.0050 M in Cr³+ and 0.060 M in H₂EDTA² and buffered at
3+
pH
6.00. Assume that H₂ EDTA² form dominates at this pH.
[Cr³+] =
N- -CH₂-CH₂-N
M

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