A 25.00 mL sample containing an unknown amount of Al³+; + and Pb2+required 15.54 mL of 0.05734 M EDTA to reach the end point. A 50.00 mL sample of the unknown was then treated with F¯ to mask the Al³+. To the 50.00 mL sample, 25.00 mL of 0.05734 M EDTA was added. The excess EDTA was then titrated with 0.02256 M Mn² +. A total of 25.7 mL was required to reach the methylthymol blue end point. Determine pAl³+ and pPb2+ in the unknown sample. PA1³ +: pPb²+ =

A 25.00 mL sample containing an unknown amount of Al³+; + and Pb2+required 15.54 mL of 0.05734 M EDTA to reach the end point. A 50.00 mL sample of the unknown was then treated with F¯ to mask the Al³+. To the 50.00 mL sample, 25.00 mL of 0.05734 M EDTA was added. The excess EDTA was then titrated with 0.02256 M Mn² +. A total of 25.7 mL was required to reach the methylthymol blue end point. Determine pAl³+ and pPb2+ in the unknown sample. PA1³ +: pPb²+ =

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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Macmillan Learning
A 25.00 mL sample containing an unknown amount of Al³+and Pb² +required 15.54 mL of 0.05734 M EDTA to reach the end
point. A 50.00 mL sample of the unknown was then treated with F to mask the Al³+. To the 50.00 mL sample, 25.00 mL of
0.05734 M EDTA was added. The excess EDTA was then titrated with 0.02256 M Mn² +. A total of 25.7 mL was required to
reach the methylthymol blue end point. Determine pAl³+ and pPb²+ in the unknown sample.
PA1³ +
=
pPb²+ =

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