A 25.00 mL sample containing an unknown amount of A13+ and Pb2+ required 20.51 mL of 0.04633 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 0.04633 M EDTA was added. The excess EDTA was then titrated with 0.02353 M Mn2+. A total of 14.2 mL was required to reach the methylthymol blue end point. Determine pAl3+ and pPb2+ in the unknown sample. PA13+ %3D pPb2+ %3D

A 25.00 mL sample containing an unknown amount of A13+ and Pb2+ required 20.51 mL of 0.04633 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 0.04633 M EDTA was added. The excess EDTA was then titrated with 0.02353 M Mn2+. A total of 14.2 mL was required to reach the methylthymol blue end point. Determine pAl3+ and pPb2+ in the unknown sample. PA13+ %3D pPb2+ %3D

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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Question

**Titration Analysis of Al³⁺ and Pb²⁺ Ions**

A 25.00 mL sample containing an unknown amount of Al³⁺ and Pb²⁺ ions required 20.51 mL of 0.04633 M EDTA to reach the end point. A separate 50.00 mL sample of the unknown was treated with F⁻ to mask the Al³⁺ ions. To this sample, 25.00 mL of 0.04633 M EDTA was added. The excess EDTA was then titrated with 0.02353 M Mn²⁺. A total of 14.2 mL was required to reach the methylthymol blue end point.

**Objective:** Determine the concentrations (pAl³⁺ and pPb²⁺) in the unknown sample.

**Calculations:**

**pAl³⁺ =** [Enter value here]

**pPb²⁺ =** [Enter value here]

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