PHYSICAL CHEMISTRY-WEBASSIGN
PHYSICAL CHEMISTRY-WEBASSIGN
11th Edition
ISBN: 9780357087411
Author: ATKINS
Publisher: CENGAGE L
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

PHYSICAL CHEMISTRY-WEBASSIGN

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