A solution prepared by mixing 60.0 mL of 0.360 M AgNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIB is Kp = 3.6 x 10-6 and the solubility constant of AgBr is Kp = 5.0 x 10-13. Which precipitate forms first? TIBr AgBr Which of the expressions shows how the cell potential E denends on [Agt1?
A solution prepared by mixing 60.0 mL of 0.360 M AgNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIB is Kp = 3.6 x 10-6 and the solubility constant of AgBr is Kp = 5.0 x 10-13. Which precipitate forms first? TIBr AgBr Which of the expressions shows how the cell potential E denends on [Agt1?
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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Transcribed Image Text:29.0 mL
E =
V
29.9 mL
E =
V
30.3 mL
E =
V
49.0 mL
E =
V
60.0 mL
E =
V
68.0 mL
E =
V
![A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M
NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The
reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the
negative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is
= 5.0 x 10-13.
Ksp
Which precipitate forms first?
O TIBI
O AgBr
Which of the expressions shows how the cell potential, E, depends on [Ag*]?
E = 0.175 V – [0.799 V – 0.05916log ([Ag+])]
1
E = |0.799 V – 0.05916log
- 0.175 V
[Ag*)
E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 V
E = 0.175 V - |0.799 V – 0.05916log
[Ag*
* (ল)
Calculate the first and second equivalence points of the titration.
first equivalence point:
mL
second equivalence point:
mL
What is the cell potential after each of the given volumes of 0.720 M NaBr have been added?
1.0 mL
E =
V
19.0 mL
E =
V](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F610a3831-2871-40ad-92e5-7e802d678e71%2Fe4f66720-e690-487b-9856-921a3a5c4420%2Fp11pelm_processed.png&w=3840&q=75)
Transcribed Image Text:A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M
NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The
reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the
negative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is
= 5.0 x 10-13.
Ksp
Which precipitate forms first?
O TIBI
O AgBr
Which of the expressions shows how the cell potential, E, depends on [Ag*]?
E = 0.175 V – [0.799 V – 0.05916log ([Ag+])]
1
E = |0.799 V – 0.05916log
- 0.175 V
[Ag*)
E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 V
E = 0.175 V - |0.799 V – 0.05916log
[Ag*
* (ল)
Calculate the first and second equivalence points of the titration.
first equivalence point:
mL
second equivalence point:
mL
What is the cell potential after each of the given volumes of 0.720 M NaBr have been added?
1.0 mL
E =
V
19.0 mL
E =
V
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