Quantitative Chemical Analysis
Quantitative Chemical Analysis
9th Edition
ISBN: 9781464135385
Author: Daniel C. Harris
Publisher: W. H. Freeman
Question
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Chapter 15, Problem 15.AE
Interpretation Introduction

Interpretation:

The cell voltage at each volume of NaBr has to be calculated and plotted.

Concept Introduction:-

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

Ecell=E+-E-

The unreacted concentration in the solution is determined by

[X]=(fractionof[X]remaining)(initialconcentrationofX)(dilutionfactor)

Ion-selective electrode:

An ion-selective electrode (ISE) best known as a specific ion electrode (SIE), is a sensor (or transducer) that transforms the activity of a specific ion dissolved in a solution into an electrical potential.

Ion-selective electrode obeys following equation.

E=constant-0.05916log[CN-]

Expert Solution & Answer
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Explanation of Solution

Given:

The titration reaction is a reduction reaction.

Ag++e-Ag(s)

The cell voltage of the reaction is written as

Cell Voltage:

Cell voltage is the difference between cathode potential and anode potential.

Ecell=E+-E- =E+-E(S.C.E)=E+0.241

Where,

E+=(0.799-0.05916log1[Ag+])

E+=(0.799-0.05916log1[Ag+])0.241

E =0.558+0.05916log[Ag+]

Another titration reaction is

Br-+Ag+AgBr(s)

The solubility constant for AgBr(s) is 5.0×10-13 .

The given equivalence point is Ve=25.0 which is between 0to25mL

The unreacted Ag+ in the solution.

[X]=(fractionof[X]remaining)(initialconcentrationofX)(dilutionfactor)

At 1.0mL: [Ag+]=(24.0mL25.0mL)(0.100M)(50.0mL51.0mL)=0.0941M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[0.0941]=0.497V

At 12.5mL: [Ag+]=(12.5mL25.0mL)(0.100M)(50.0mL62.5mL)=0.0400M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[0.0400]=0.475V

At 24.0mL: [Ag+]=(1.0mL25.0mL)(0.100M)(50.0mL74.0mL)=0.00270M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[0.002700]=0.406V

At 24.9mL: [Ag+]=(0.10mL25.0mL)(0.100M)(50.0mL74.9mL)=2.670×10-4M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[2.6×10-4M]=0.347V

After 25mL , all AgBr has precipitated and there is excess Br- in solution.

At 25.1mL: [Br-]=(0.10mL75.1mL)(0.200M)=2.670×10-4M

[Ag+]=KSP[Br-]=(5.0×10-13)(2.67×10-4)=1.88×10-9M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[1.88×10-9M]=+0.0042V

At 26.0mL: [Br-]=(0.10mL76.0mL)(0.200M)=2.66×10-4M

[Ag+]=KSP[Br-]=(5.0×10-13)(2.66×10-4)=1.90×10-9M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[1.90×10-9M]=-0.0017V

At 35.0mL: [Br-]=(0.10mL85.0mL)(0.200M)=0.0235M

[Ag+]=KSP[Br-]=(5.0×10-13)(0.0235M)=1.90×10-9M

E=0.558+0.05916log[Ag+]

E=0.558+0.05916log[0.0235]=-0.073V

All the obtained potential are plotted as,

Quantitative Chemical Analysis, Chapter 15, Problem 15.AE

Figure 1

Conclusion

The cell voltage at each volume of NaBr has to be calculated and plotted.

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