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A solution prepared by mixing 50.2 mL of 0.280 M AgNO3 and 50.2 mL of 0.280...

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Chemistry

A solution prepared by mixing 50.2 mL of 0.280 M AgNO3 and 50.2mL of 0.280 M TlNO3 was titrated with 0.560 M NaBr in a cellcontaining a silver indicator electrode and a reference electrodeof constant potential 0.175 V. The reference electrode is attachedto the positive terminal of the potentiometer, and the silverelectrode is attached to the negative terminal. The solubilityconstant of TlBr is Ksp = 3.6 × 10–6 and the solubility constant ofAgBr is Ksp = 5.0 × 10–13..

What is the cell voltage when the following volumes of 0.560 MNaBr have been added?

(b) 1.0 mL (c) 16.1 mL (d) 24.1 mL (e) 25.0 mL (f) 25.4 mL (g)41.2 mL (h) 50.2 mL (i) 57.2 mL

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Answer & Explanation Solved by verified expert
3.9 Ratings (459 Votes)

b)added Br- concentration=0.560M*1ml/101.4ml=5.52*10-3M

[Ag+]=0.28M*50.2ml/101.4ml=0.139M

[Ti+]=0.28M*50.2ml/101.4ml=0.139M

AgBr ksp=[Ag+][Br-]

TiBr ksp=[Ti+][Br-]

necessary minimum br- concentration=3.6-6/0.139=2.59*10-5

therefor [Ag+] in solution=5*10-13/2.59*10-5=1.93*10-8

E=E0+(RT/NF)ln[Ag+]

E=E0+0.0059log[Ag+]

=0.175+0.0059 log1.93*10-8=0.129v

c)added Br- concentration=0.560M*16.1ml/116.5ml=77.4*10-3M

[Ag+]=0.28M*50.2ml/116.5ml=0.121M

[Ti+]=0.28M*50.2ml/116.5ml=0.121M

AgBr ksp=[Ag+][Br-]

TiBr ksp=[Ti+][Br-]

necessary minimum br- concentration=3.6*10-6/0.121=2.975*10-5

therefor [Ag+] in solution=5*10-13/2.975*10-5=1.68*10-8

E=E0+(RT/NF)ln[Ag+]

E=E0+0.0059log[Ag+]

=0.175+0.0059 log1.68*10-8=0.129v

d)added Br- concentration=0.560M*24.1ml/124.5ml=0.108M

[Ag+]=0.28M*50.2ml/124.5ml=0.113M

[Ti+]=0.28M*50.2ml/124.5ml=0.113M

AgBr ksp=[Ag+][Br-]

TiBr ksp=[Ti+][Br-]

necessary minimum br- concentration=3.6*10-6/0.113=3.18*10-5

therefor [Ag+] in solution=5*10-13/3.18*10-5=1.57*10-8

E=E0+(RT/NF)ln[Ag+]

E=E0+0.0059log[Ag+]

=0.175+0.0059 log1.57*10-8=0.129v
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