Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
Question
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Chapter 5, Problem 5F.6AE

(i)

Interpretation Introduction

Interpretation: The mass of Ca(NO3)2 added to a 0.150molkg-1 solution of KNO3(aq) to increase its ionic strength to 0.250 has to be calculated.

Concept introduction: The ionic strength is introduced in the Debye-Huckel limiting law which relates the activity coefficient as a function of the ionic strength of the solution.  It depends upon the concentration of all the ions present in the solution.  It is a dimensionless quantity.

(i)

Expert Solution
Check Mark

Answer to Problem 5F.6AE

The mass of Ca(NO3)2 added to a 0.150molkg-1 solution of KNO3(aq) to increase its ionic strength to 0.250 is 2.75g_.

Explanation of Solution

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

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

Where,

  • zi is the charge present on the ion.
  • bi is the molality of the ion in the solution.

The molality of KNO3(aq) is 0.150molkg-1.

The reaction for the dissociation of KNO3(aq) is shown as,

  KNO3K++NO3

The molality of cation, K+ (bK+) is 0.150molkg-1.

The molality of anion, NO3 (bNO3) is 0.150molkg-1.

The charge present on cation, K+ (zK+) is +1.

The charge present on anion, NO3 (zNO3) is 1.

Let the molality of Ca(NO3)2 to be added be b.

The reaction for the dissociation of Ca(NO3)2(aq) is shown as,

  Ca(NO3)2Ca2++2NO3

The molality of cation, Ca2+ (bCa2+) is bmolkg-1.

The molality of anion, NO3 (bNO3) is 2×bmolkg-1=2bmolkg-1.

The charge present on cation, Ca2+ (zCa2+) is +2.

The charge present on anion, NO3 (zNO3) is 1.

The ionic strength of the solution (I) has to be 0.250.

Substitute the value of bK+, bNO3, zK+, bCa2+ and zCa2+ in equation (1)

    INaCl=12[(bK+×zK+2)+(bNO3-×zNO3-2)+(bCa2+×zCa2+2)+(bNO3-×zNO3-2)]0.250=12[(0.150molkg-1×(1)2)+(0.150molkg-1×(1)2)+(bmolkg-1×(2)2)+(2bmolkg-1×(1)2)]0.500=[0.3+6b]b=0.033molkg-1

Thus, the molality of Ca(NO3)2(aq) is 0.033molkg-1.

The molality of a solution is given by the formula,

    Molality=MolesofCa(NO3)2Massofsolvent                                                                  (2)

The mass of the solvent is 500g.

The conversion of g to kg is done as,

    1000g=1kg

Therefore the conversion of 500g to kg is done as,

    500g=0.500kg

Substitute the molality of Ca(NO3)2(aq) and mass of the solvent in equation (2).

    0.033molkg1=MolesofCa(NO3)20.5kgMolesofCa(NO3)2=0.0167mol

The number of moles of Ca(NO3)2(aq) is given by the formula,

     MolesofCa(NO3)2=MassofCa(NO3)2MolarmassofCa(NO3)2                                     (3)

The molar mass of Ca(NO3)2 is 164.78gmol-1.

Substitute the number of moles of Ca(NO3)2 and molar mass of Ca(NO3)2 in equation (3).

    0.0167mol=MassofCa(NO3)2164.78gmolMassofCa(NO3)2=2.75g_

Thus the mass of Ca(NO3)2 to be added to a 0.150molkg-1 solution of KNO3(aq) to increase its ionic strength to 0.250 is 2.75g_.

(ii)

Interpretation Introduction

Interpretation: The mass of NaCl to be added to a 0.150molkg-1 solution of KNO3(aq) to increase its ionic strength to 0.250 has to be calculated..

Concept introduction: The ionic strength is introduced in the Debye-Huckel limiting law which relates the activity coefficient as a function of the ionic strength of the solution.  It is dependent on the concentration of all the ions present in the solution.  It is a dimensionless quantity.

(ii)

Expert Solution
Check Mark

Answer to Problem 5F.6AE

The mass of NaCl to be added to a 0.150molkg-1 solution of KNO3(aq) to increase its ionic strength to 0.250 is 2.92g_.

Explanation of Solution

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

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

Where

  • zi is the charge present on the ion.
  • bi is the molality of the ion in the solution.

The molality of KNO3(aq) is 0.150molkg-1.

The reaction for the dissociation of KNO3(aq) is shown as,

  KNO3K++NO3

Hence,

The molality of cation, K+ (bK+) is 0.150molkg-1.

The molality of anion, NO3 (bNO3) is 0.150molkg-1.

The charge present on cation, K+ (zK+) is +1.

The charge present on anion, NO3 (zNO3) is 1.

Let the molality of NaCl to be added be b.

The reaction for the dissociation of NaCl(aq) is shown as,

  NaClNa2++Cl

Hence,

The molality of cation, Na+ (bNa+) is bmolkg-1.

The molality of anion, Cl (bCl) is bmolkg-1.

The charge present on cation, Na+ (zNa+) is +1.

The charge present on anion, Cl (zCl) is 1.

The ionic strength of the solution (I) has to be 0.250.

Substitute the value of bK+, bNO3, zK+, zNO3, bNa+, bCl, zNa+ and zCl in equation (1)

    INaCl=12[(bK+×zK+2)+(bNO3-×zNO3-2)+(bNa+×zNa+2)+(bCl-×zCl-2)]0.250=12[(0.150molkg-1×(1)2)+(0.150molkg-1×(1)2)+(bmolkg-1×(1)2)+(bmolkg-1×(1)2)]0.500=[0.3+2b]b=0.1molkg-1

Thus, the molality of NaCl is 0.1molkg-1.

The molality of a solution is given by the formula,

    Molality=MolesofNaClMassofsolvent                                                                          (2)

The mass of the solvent is 500g.

The conversion of g to kg is done as,

    1000g=1kg

Therefore the conversion of 500g to kg is done as,

    500g=0.500kg

Substitute the molality of NaCl and mass of the solvent in equation (2).

    0.1molkg1=MolesofNaCl0.5kgMolesofNaCl=0.05mol

The number of moles of NaCl is given by the formula,

    MolesofNaCl=MassofNaClMolarmassofNaCl                                                        (3)

The molar mass of NaCl is 58.5gmol-1.

Substitute the number of moles of NaCl and molar mass of NaCl in equation (3).

    0.05mol=MassofNaCl58.5gmolMassofNaCl=2.92g_

Thus the mass of NaCl to be added to a 0.150molkg-1 solution of KNO3(aq) to increase its ionic strength to 0.250 is 2.92g_.

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Chapter 5 Solutions

Atkins' Physical Chemistry

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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