Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
Quantitative Chemical Analysis 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card)
9th Edition
ISBN: 9781319090241
Author: Daniel C. Harris, Sapling Learning
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 9e And Sapling Advanced Single Course For Analytical Chemistry (access Card), 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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