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A 44.00 mL mixture containing 0.0631 M TI* and 0.0631 M Hg3+ is titrated with 0.0877 M NaI resulting in the formation of the precipitates TII and Hg,I,. Calculate pI¯ at the given points in the titration. The K4p values for TII and Hg,L, are 5.54 x 10-8 and 2.9 × 10-29 , respectively. 46.8 mL pI¯ = 78.2 mL pI¯ = second equivalence point pl- = 103.2 mL pI¯:

A 44.00 mL mixture containing 0.0631 M TI* and 0.0631 M Hg3+ is titrated with 0.0877 M NaI resulting in the formation of the precipitates TII and Hg,I,. Calculate pI¯ at the given points in the titration. The K4p values for TII and Hg,L, are 5.54 x 10-8 and 2.9 × 10-29 , respectively. 46.8 mL pI¯ = 78.2 mL pI¯ = second equivalence point pl- = 103.2 mL pI¯:

Chemistry
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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A 44.00 mL mixture containing 0.0631 M TI* and 0.0631 M Hg3+ is titrated with 0.0877 M NaI resulting in the formation of the precipitates TII and Hg,I,. Calculate pI¯ at the given points in the titration. The K4p values for TII and Hg,I, are 5.54 x 10-8 and 2.9 × 10-29, respectively. 46.8 mL pI- = 78.2 mL pI- = second equivalence point pI¯ = 103.2 mL pI =
Transcribed Image Text:A 44.00 mL mixture containing 0.0631 M TI* and 0.0631 M Hg3+ is titrated with 0.0877 M NaI resulting in the formation of the precipitates TII and Hg,I,. Calculate pI¯ at the given points in the titration. The K4p values for TII and Hg,I, are 5.54 x 10-8 and 2.9 × 10-29, respectively. 46.8 mL pI- = 78.2 mL pI- = second equivalence point pI¯ = 103.2 mL pI =
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