CHEM:ATOM FOC 2E CL (TEXT)
CHEM:ATOM FOC 2E CL (TEXT)
2nd Edition
ISBN: 9780393284218
Author: Stacey Lowery Bretz, Natalie Foster, Thomas R. Gilbert, Rein V. Kirss
Publisher: WW Norton & Co
Question
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Chapter 23, Problem 23.90QA
Interpretation Introduction

To find:

The concentration ratio of  Pb+2 (aq)[Pb+2-EDTA]

Expert Solution & Answer
Check Mark

Answer to Problem 23.90QA

Solution:

The concentration ratio of free Pb2+(aq) in the blood to the much toxic PbEDTA-2 complex is  2.0 ×10-11.

Explanation of Solution

1) Concept:

The complexation reaction of free lead in the blood with the EDTA solution to form PbEDTA-2 complex is

Pb+2 aq+EDTA-4 aq  PbEDTA-2 (aq)

The formation constant of the PbEDTA-2 complex is written as

Kf= [PbEDTA-2]Pb+2[EDTA-4]

Since we need to find the ratio [Pb+2][PbEDTA-2], we can calculate it from the above relation as we have been provided with the value of formation constant of PbEDTA-2 and concentration of EDTA, i.e., [EDTA-4].

2) Formula:

Kf= [PbEDTA-2]Pb+2[EDTA-4]

3) Given:

i) Formation constant Kf=2.0 ×1018 M-1

ii) Concentration of EDTA solution EDTA-4=2.5 ×10-8 M

4) Calculations:

We know that formation constant of PbEDTA-2 complex is

Kf= [PbEDTA-2]Pb+2[EDTA-4]

[PbEDTA-2][Pb+2]= Kf × [EDTA-4]

Thus, the concentration ratio of free lead in the blood to the much less toxic PbEDTA-2 complex is

Pb+2PbEDTA-2= 1Kf × EDTA-4=12.0 ×1018 M-1 × 2.5 ×10-8 M

 Pb+2PbEDTA-2=15.0 ×1010  =2×10-11

Therefore, the concentration ratio is 2×10-11.

Conclusion:

The formation constant for PbEDTA-2 complex ion is quite large, so that a small concentration of EDTA is enough to form the complex, which lowers the toxicity due to free lead ions in the body.

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Chapter 23 Solutions

CHEM:ATOM FOC 2E CL (TEXT)

Ch. 23 - Prob. 23.11VPCh. 23 - Prob. 23.12VPCh. 23 - Prob. 23.13QACh. 23 - Prob. 23.14QACh. 23 - Prob. 23.15QACh. 23 - Prob. 23.16QACh. 23 - Prob. 23.17QACh. 23 - Prob. 23.18QACh. 23 - Prob. 23.19QACh. 23 - Prob. 23.20QACh. 23 - Prob. 23.21QACh. 23 - Prob. 23.22QACh. 23 - Prob. 23.23QACh. 23 - Prob. 23.24QACh. 23 - Prob. 23.25QACh. 23 - Prob. 23.26QACh. 23 - Prob. 23.27QACh. 23 - Prob. 23.28QACh. 23 - Prob. 23.29QACh. 23 - Prob. 23.30QACh. 23 - Prob. 23.31QACh. 23 - Prob. 23.32QACh. 23 - Prob. 23.33QACh. 23 - Prob. 23.34QACh. 23 - Prob. 23.35QACh. 23 - Prob. 23.36QACh. 23 - Prob. 23.37QACh. 23 - Prob. 23.38QACh. 23 - Prob. 23.39QACh. 23 - Prob. 23.40QACh. 23 - Prob. 23.41QACh. 23 - Prob. 23.42QACh. 23 - Prob. 23.43QACh. 23 - Prob. 23.44QACh. 23 - Prob. 23.45QACh. 23 - Prob. 23.46QACh. 23 - Prob. 23.47QACh. 23 - Prob. 23.48QACh. 23 - Prob. 23.49QACh. 23 - Prob. 23.50QACh. 23 - Prob. 23.51QACh. 23 - Prob. 23.52QACh. 23 - Prob. 23.53QACh. 23 - Prob. 23.54QACh. 23 - Prob. 23.55QACh. 23 - Prob. 23.56QACh. 23 - Prob. 23.57QACh. 23 - Prob. 23.58QACh. 23 - Prob. 23.59QACh. 23 - Prob. 23.60QACh. 23 - Prob. 23.61QACh. 23 - Prob. 23.62QACh. 23 - Prob. 23.63QACh. 23 - Prob. 23.64QACh. 23 - Prob. 23.65QACh. 23 - Prob. 23.66QACh. 23 - Prob. 23.67QACh. 23 - Prob. 23.68QACh. 23 - Prob. 23.69QACh. 23 - Prob. 23.70QACh. 23 - Prob. 23.71QACh. 23 - Prob. 23.72QACh. 23 - Prob. 23.73QACh. 23 - Prob. 23.74QACh. 23 - Prob. 23.75QACh. 23 - Prob. 23.76QACh. 23 - Prob. 23.77QACh. 23 - Prob. 23.78QACh. 23 - Prob. 23.79QACh. 23 - Prob. 23.80QACh. 23 - Prob. 23.81QACh. 23 - Prob. 23.82QACh. 23 - Prob. 23.83QACh. 23 - Prob. 23.84QACh. 23 - Prob. 23.85QACh. 23 - Prob. 23.86QACh. 23 - Prob. 23.87QACh. 23 - Prob. 23.88QACh. 23 - Prob. 23.89QACh. 23 - Prob. 23.90QACh. 23 - Prob. 23.91QACh. 23 - Prob. 23.92QACh. 23 - Prob. 23.93QACh. 23 - Prob. 23.94QACh. 23 - Prob. 23.95QACh. 23 - Prob. 23.96QA
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