Question

A solution prepared by mixing 50.2 mL of 0.280 M AgNO3 and 50.2 mL of 0.280 M TlNO3 was titrated with 0.560 M NaBr in a cell containing a silver indicator electrode and a reference electrode of constant potential 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 TlBr is Ksp = 3.6 × 10–6 and the solubility constant of AgBr is Ksp = 5.0 × 10–13..

What is the cell voltage when the following volumes of 0.560 M NaBr 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

Please show step by step

Answer 1

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