A 45.00 mL mixture containing 0.0746 M TI and 0.0746 M Hg* is titrated with 0.0839 M Nal resulting in the formation of the precipitates TIl and Hg, L,. Calculate pl at the given points in the titration. The K, values for TIl and Hg,I, are 5.54 x 10* and 2.9 x 10-, respectively. 62.2 ml. pl= 96.2 ml. pl = second equivalence point pl = 125.0 ml pl =
A 45.00 mL mixture containing 0.0746 M TI and 0.0746 M Hg* is titrated with 0.0839 M Nal resulting in the formation of the precipitates TIl and Hg, L,. Calculate pl at the given points in the titration. The K, values for TIl and Hg,I, are 5.54 x 10* and 2.9 x 10-, respectively. 62.2 ml. pl= 96.2 ml. pl = second equivalence point pl = 125.0 ml pl =
Chemistry
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Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
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Chapter1: Chemical Foundations
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![A 45.00 mL mixture containing 0.0746 M TI and 0.0746 M Hg* is titrated with 0.0839 M Nal resulting in the formation
of the precipitates TII and Hg, L,. Calculate pl at the given points in the titration. The K values for TII and Hg, I, are
5.54 x 10- and 2.9 x 10-29, respectively.
62.2 ml. pl
96.2 ml. pl=
second equivalence point pl =
125.0 mL pl =](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2ebe679-f904-41e0-96da-a20832da902a%2F366e00b7-7ef9-4ab1-8ce7-bb7da139feb6%2Fktwywz_processed.jpeg&w=3840&q=75)
Transcribed Image Text:A 45.00 mL mixture containing 0.0746 M TI and 0.0746 M Hg* is titrated with 0.0839 M Nal resulting in the formation
of the precipitates TII and Hg, L,. Calculate pl at the given points in the titration. The K values for TII and Hg, I, are
5.54 x 10- and 2.9 x 10-29, respectively.
62.2 ml. pl
96.2 ml. pl=
second equivalence point pl =
125.0 mL pl =
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