Student Solutions Manual to accompany Atkins' Physical Chemistry 11th  edition
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
11th Edition
ISBN: 9780198807773
Author: ATKINS
Publisher: Oxford University Press
Question
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Chapter 5, Problem 5F.8AE
Interpretation Introduction

Interpretation: On the basis of the given activity coefficients of HBr in three dilute solutions at 25°C, the value of B in the Davies equation has to be estimated.

Concept introduction: Davies equation is an extended form of the Debye-Huckel limiting law which is used when ionic strength of the solution is too high for the Debye-Huckel limiting law to be applicable.

Expert Solution & Answer
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Answer to Problem 5F.8AE

On the basis of the given activity coefficients of HBr in three dilute solutions at 25°C, the value of B in the Davies equation is calculated as 2.0097_.

Explanation of Solution

The ionic strength (I) of a solution is given by the equation,

    I=12izi2(bi/b°)                                                                                        (1)

The mean activity coefficient of 5.0mmolkg-1 HBr is 0.930.

The molality of HBr is 5.0mmolkg-1.

The conversion of mmolkg1 to molkg1 is done as,

    1mmolkg1=10-3 molkg1

Therefore, the conversion of 5.0mmolkg1 to molkg1 is done as,

  5.0mmolkg1=5.0×10-3 molkg1

The dissociation of HBr is represented by the reaction.

    HBrH++Br-

Hence,

The molality of cation, H+ (b+) is 5.0×10-3molkg1.

The molality of anion, Br (b) is 5.0×10-3molkg1.

The charge present on cation, H+ (z+) is +1.

The charge present on anion, Br (z) is 1.

Substitute the value of b+, b, z+ and z in equation (1)

    IHBr=12[(b+×z+2)+(b×z2)]=12[(5.0×10-3×(1)2)+(5.0×10-3×(1)2)]=5.0×10-3

The Davies equation is given as,

    logγ±=A|z+z|I1/21+BI1/2+CI                                                                            (2)

Where

  • γ± is the activity coefficient.
  • A is 0.509 for an aqueous solution at 25°C.
  • I is the dimensionless ionic strength.
  • z+ is the charge on the positive cation.
  • z is the charge on the negative cation.
  • B is a dimensionless constant.
  • C is a dimensionless constant.

The mean activity coefficient of 5.0×10-3molkg1 HBr (γ±) is 0.930.

The ionic strength of 5.0×10-3molkg1 HBr(I) is 5.0×10-3.

The charge present on cation, H+ (z+) is +1.

The charge present on anion, Br (z) is 1.

The value of C for HBr is 0.

Substitute the value of IHBr, γ±, A, z+, z and C in equation (2).

    log(0.930)=0.509×|1×1|×(5×103)1/21+B(5×103)1/2+0×(5×103)1/20.0315=0.509×0.07071+B(5×103)1/2+00.0315=0.03591+0.0707BB=2.0055

The mean activity coefficient of 10.0mmolkg-1 HBr is 0.907.

The molality of HBr is 10.0mmolkg-1.

The conversion of mmolkg1 to molkg1 is done as,

    1mmolkg1=10-3molkg1

Therefore, the conversion of 10.0mmolkg-1 to molkg1 is done as,

  10.0mmolkg-1=10.0×10-3 molkg1=104molkg1

The dissociation of HBr is represented by the reaction.

    HBrH++Br-

Hence,

The molality of cation, H+ (b+) is 104molkg1.

The molality of anion, Br (b) is 104molkg1.

The charge present on cation, H+ (z+) is +1.

The charge present on anion, Br (z) is 1.

Substitute the value of b+, b, z+ and z in equation (1)

    IHBr=12[(b+×z+2)+(b×z2)]=12[(10-4×(1)2)+(10-4×(1)2)]=10-4

The Davies equation is given as,

    logγ±=A|z+z|I1/21+BI1/2+CI                                                                            (2)

Where

  • γ± is the activity coefficient.
  • A is 0.509 for an aqueous solution at 25°C.
  • I is the dimensionless ionic strength.
  • z+ is the charge on the positive cation.
  • z is the charge on the negative cation.
  • B is a dimensionless constant.
  • C is a dimensionless constant.

The mean activity coefficient of 104molkg1 HBr (γ±) is 0.907.

The ionic strength of 104molkg1 HBr(I) is 10-4.

The charge present on cation, H+ (z+) is +1.

The charge present on anion, Br (z) is 1.

The value of C for HBr is 0.

Substitute the value of IHBr, γ±, A, z+, z and C in equation (2).

    log(0.907)=0.509×|1×1|×(104)1/21+B(104)1/2+0×(104)1/20.042=0.509×1021+(104)1/2B+00.042=0.005091+(102)B+0B=2.0047

The mean activity coefficient of 20.0mmolkg-1 HBr is 0.879.

The molality of HBr is 20.0mmolkg-1.

The conversion of mmolkg1 to molkg1 is done as,

    1mmolkg1=10-3molkg1

Therefore, the conversion of 20.0mmolkg-1 to molkg1 is done as,

  20.0mmolkg-1=20×10-3molkg1

The dissociation of HBr is represented by the reaction.

    HBrH++Br-

Hence,

The molality of cation, H+ (b+) is 20×10-3molkg1.

The molality of anion, Br (b) is 20×10-3molkg1.

The charge present on cation, H+ (z+) is +1.

The charge present on anion, Br (z) is 1.

Substitute the value of b+, b, z+ and z in equation (1)

    IHBr=12[(b+×z+2)+(b×z2)]=12[(20.0×10-3×(1)2)+(20.0×10-3×(1)2)]=20.0×10-3

The Davies equation is given as,

    logγ±=A|z+z|I1/21+BI1/2+CI                                                                            (2)

Where

  • γ± is the activity coefficient.
  • A is 0.509 for an aqueous solution at 25°C.
  • I is the dimensionless ionic strength.
  • z+ is the charge on the positive cation.
  • z is the charge on the negative cation.
  • B is a dimensionless constant.
  • C is a dimensionless constant.

The mean activity coefficient of 20×10-3molkg1 HBr (γ±) is 0.879.

The ionic strength of 20×10-3molkg1 HBr(I) is 20×10-3.

The charge present on cation, H+ (z+) is +1.

The charge present on anion, Br (z) is 1.

The value of C for HBr is 0.

Substitute the value of IHBr, γ±, A, z+, z and C in equation (2).

    log(0.879)=0.509×|1×1|×(20×103)1/21+B(20×103)1/2+0×(20×103)1/20.0560=0.509×0.14141+(20×103)1/2B+00.0560=0.07191+0.1414BB=2.0189

The mean value of B is the average of above values.

  B=2.0055+2.0047+2.01893=2.0097_

Thus, on the basis of the given activity coefficients of HBr in three dilute solutions at 25°C, the value of B in the Davies equation is calculated as 2.0097_.

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Chapter 5 Solutions

Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition

Ch. 5 - Prob. 5A.4DQCh. 5 - Prob. 5A.5DQCh. 5 - Prob. 5A.1AECh. 5 - Prob. 5A.1BECh. 5 - Prob. 5A.2AECh. 5 - Prob. 5A.2BECh. 5 - Prob. 5A.3AECh. 5 - Prob. 5A.3BECh. 5 - Prob. 5A.4AECh. 5 - Prob. 5A.4BECh. 5 - Prob. 5A.5AECh. 5 - Prob. 5A.5BECh. 5 - Prob. 5A.6AECh. 5 - Prob. 5A.6BECh. 5 - Prob. 5A.7AECh. 5 - Prob. 5A.7BECh. 5 - Prob. 5A.8AECh. 5 - Prob. 5A.8BECh. 5 - Prob. 5A.9AECh. 5 - Prob. 5A.9BECh. 5 - Prob. 5A.10AECh. 5 - Prob. 5A.10BECh. 5 - Prob. 5A.11AECh. 5 - Prob. 5A.11BECh. 5 - Prob. 5A.1PCh. 5 - Prob. 5A.3PCh. 5 - Prob. 5A.4PCh. 5 - Prob. 5A.5PCh. 5 - Prob. 5A.6PCh. 5 - Prob. 5A.7PCh. 5 - Prob. 5B.1DQCh. 5 - Prob. 5B.2DQCh. 5 - Prob. 5B.3DQCh. 5 - Prob. 5B.4DQCh. 5 - Prob. 5B.5DQCh. 5 - Prob. 5B.6DQCh. 5 - Prob. 5B.7DQCh. 5 - Prob. 5B.1AECh. 5 - Prob. 5B.1BECh. 5 - Prob. 5B.2AECh. 5 - Prob. 5B.2BECh. 5 - Prob. 5B.3AECh. 5 - Prob. 5B.3BECh. 5 - Prob. 5B.4AECh. 5 - Prob. 5B.4BECh. 5 - Prob. 5B.5AECh. 5 - Prob. 5B.5BECh. 5 - Prob. 5B.6AECh. 5 - Prob. 5B.6BECh. 5 - Prob. 5B.7AECh. 5 - Prob. 5B.7BECh. 5 - Prob. 5B.8AECh. 5 - Prob. 5B.8BECh. 5 - Prob. 5B.9AECh. 5 - Prob. 5B.9BECh. 5 - Prob. 5B.10AECh. 5 - Prob. 5B.10BECh. 5 - Prob. 5B.11AECh. 5 - Prob. 5B.11BECh. 5 - Prob. 5B.12AECh. 5 - Prob. 5B.12BECh. 5 - Prob. 5B.1PCh. 5 - Prob. 5B.2PCh. 5 - Prob. 5B.3PCh. 5 - Prob. 5B.4PCh. 5 - Prob. 5B.5PCh. 5 - Prob. 5B.6PCh. 5 - Prob. 5B.9PCh. 5 - Prob. 5B.11PCh. 5 - Prob. 5B.13PCh. 5 - Prob. 5C.1DQCh. 5 - Prob. 5C.2DQCh. 5 - Prob. 5C.3DQCh. 5 - Prob. 5C.1AECh. 5 - Prob. 5C.1BECh. 5 - Prob. 5C.2AECh. 5 - Prob. 5C.2BECh. 5 - Prob. 5C.3AECh. 5 - Prob. 5C.3BECh. 5 - Prob. 5C.4AECh. 5 - Prob. 5C.4BECh. 5 - Prob. 5C.1PCh. 5 - Prob. 5C.2PCh. 5 - Prob. 5C.3PCh. 5 - Prob. 5C.4PCh. 5 - Prob. 5C.5PCh. 5 - Prob. 5C.6PCh. 5 - Prob. 5C.7PCh. 5 - Prob. 5C.8PCh. 5 - Prob. 5C.9PCh. 5 - Prob. 5C.10PCh. 5 - Prob. 5D.1DQCh. 5 - Prob. 5D.2DQCh. 5 - Prob. 5D.1AECh. 5 - Prob. 5D.1BECh. 5 - Prob. 5D.2AECh. 5 - Prob. 5D.2BECh. 5 - Prob. 5D.3AECh. 5 - Prob. 5D.3BECh. 5 - Prob. 5D.4AECh. 5 - Prob. 5D.4BECh. 5 - Prob. 5D.5AECh. 5 - Prob. 5D.5BECh. 5 - Prob. 5D.6AECh. 5 - Prob. 5D.1PCh. 5 - Prob. 5D.2PCh. 5 - Prob. 5D.3PCh. 5 - Prob. 5D.4PCh. 5 - Prob. 5D.5PCh. 5 - Prob. 5D.6PCh. 5 - Prob. 5D.7PCh. 5 - Prob. 5E.1DQCh. 5 - Prob. 5E.2DQCh. 5 - Prob. 5E.3DQCh. 5 - Prob. 5E.4DQCh. 5 - Prob. 5E.1AECh. 5 - Prob. 5E.1BECh. 5 - Prob. 5E.2AECh. 5 - Prob. 5E.2BECh. 5 - Prob. 5E.3AECh. 5 - Prob. 5E.3BECh. 5 - Prob. 5E.4AECh. 5 - Prob. 5E.4BECh. 5 - Prob. 5E.5AECh. 5 - Prob. 5E.5BECh. 5 - Prob. 5E.1PCh. 5 - Prob. 5E.2PCh. 5 - Prob. 5E.3PCh. 5 - Prob. 5F.1DQCh. 5 - Prob. 5F.2DQCh. 5 - Prob. 5F.3DQCh. 5 - Prob. 5F.4DQCh. 5 - Prob. 5F.5DQCh. 5 - Prob. 5F.1AECh. 5 - Prob. 5F.1BECh. 5 - Prob. 5F.2AECh. 5 - Prob. 5F.2BECh. 5 - Prob. 5F.3AECh. 5 - Prob. 5F.3BECh. 5 - Prob. 5F.4AECh. 5 - Prob. 5F.4BECh. 5 - Prob. 5F.5AECh. 5 - Prob. 5F.5BECh. 5 - Prob. 5F.6AECh. 5 - Prob. 5F.6BECh. 5 - Prob. 5F.7AECh. 5 - Prob. 5F.7BECh. 5 - Prob. 5F.8AECh. 5 - Prob. 5F.8BECh. 5 - Prob. 5F.1PCh. 5 - Prob. 5F.2PCh. 5 - Prob. 5F.3PCh. 5 - Prob. 5F.4PCh. 5 - Prob. 5.1IACh. 5 - Prob. 5.2IACh. 5 - Prob. 5.3IACh. 5 - Prob. 5.4IACh. 5 - Prob. 5.5IACh. 5 - Prob. 5.6IACh. 5 - Prob. 5.8IACh. 5 - Prob. 5.9IACh. 5 - Prob. 5.10IA
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