How much free lead (II) ion is present in a 95.4 mL of a solution that contains 136.8 microgram of Pb2* per deciliter of solution upon addition of 95.4 mL 0.0100 M EDTA? Give just the numerical portion of the answer: X 10 22 M Pb2+ Kr for [Pb(EDTA)2 is 2.0 X 1018 As a side note: For adults, blood lead concentrations > 89 ug/dL require treatment for lead poisoning even if the patient is asymptomatic.

How much free lead (II) ion is present in a 95.4 mL of a solution that contains 136.8 microgram of Pb2* per deciliter of solution upon addition of 95.4 mL 0.0100 M EDTA? Give just the numerical portion of the answer: X 10 22 M Pb2+ Kr for [Pb(EDTA)2 is 2.0 X 1018 As a side note: For adults, blood lead concentrations > 89 ug/dL require treatment for lead poisoning even if the patient is asymptomatic.

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter17: Electrochemistry And Its Applications

Section: Chapter Questions

Problem 101QRT

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