Whether I − or Br − has smaller Δ H hydration is to be determined. Concept introduction: The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa. The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by Δ H hydration . It is always negative. Δ H hydration is directly related to the charge density on an ion. Higher the charge density, more will be Δ H hydration and vice-versa.

Question

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Answer 1

Question

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

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Answer 2

Question

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

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Answer 3

Question

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

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Answer 4

Question

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

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Answer 5

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 6

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 7

Chapter 13, Problem 13.35P

(a)

Interpretation Introduction

Interpretation:

Whether $I−$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 8

(a)

Expert Solution

Answer to Problem 13.35P

$I−$ will have a smaller $ΔHhydration$ .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Both bromine and iodine are present in the same group of the periodic table. But iodine lies below bromine so it will have a large size and volume than that of bromine. So $I−$ will have a lower ratio of charge to volume than that of $Br−$ and therefore it has smaller $ΔHhydration$ .

Answer 13

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 14

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 15

(b)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 16

(b)

Expert Solution

Answer to Problem 13.35P

$Ca2+$ will have a smaller $ΔHhydration$ .

Answer 17

(b)

Expert Solution

Answer 18

(b)

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The charge on $Ca2+$ is less than that of $Sc3+$ . So $Ca2+$ will have a lower ratio of charge to volume than that of $Sc3+$ and therefore it has smaller $ΔHhydration$ .

Answer 21

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 22

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 23

(c)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Br−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 24

(c)

Expert Solution

Answer to Problem 13.35P

$Br−$ will have a smaller $ΔHhydration$ .

Answer 25

(c)

Expert Solution

Answer 26

(c)

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

The anions are larger than cations so the ionic volume of $Br−$ is more than that of $K+$ and therefore $Br−$ will have a lower ratio of charge to volume than that of $K+$ and therefore it has smaller $ΔHhydration$ .

Answer 29

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 30

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 31

(d)

Interpretation Introduction

Interpretation:

Whether $Cl−$ or $S2−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 32

(d)

Expert Solution

Answer to Problem 13.35P

$Cl−$ will have a smaller $ΔHhydration$ .

Answer 33

(d)

Expert Solution

Answer 34

(d)

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

$Cl−$ has a smaller ionic charge as compared to $S2−$ . So $Cl−$ will have a lower ratio of charge to volume than that of $S2−$ and therefore it has smaller $ΔHhydration$ .

Answer 37

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 38

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 39

(e)

Interpretation Introduction

Interpretation:

Whether $Sc3+$ or $Al3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 40

(e)

Expert Solution

Answer to Problem 13.35P

$Sc3+$ will have a smaller $ΔHhydration$ .

Answer 41

(e)

Expert Solution

Answer 42

(e)

Expert Solution

Answer 43

Expert Solution

Answer 44

Explanation of Solution

The size of scandium is larger than that of aluminium so $Sc3+$ will have the larger volume as compared to that of $Al3+$ . So $Sc3+$ will have a lower ratio of charge to volume than that of $Al3+$ and therefore it has smaller $ΔHhydration$ .

Answer 45

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 46

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 47

(f)

Interpretation Introduction

Interpretation:

Whether $ClO4−$ or $SO42−$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 48

(f)

Expert Solution

Answer to Problem 13.35P

$ClO4−$ will have a smaller $ΔHhydration$ .

Answer 49

(f)

Expert Solution

Answer 50

(f)

Expert Solution

Answer 51

Expert Solution

Answer 52

Explanation of Solution

The charge on $ClO4−$ is less than that of $SO42−$ . So $ClO4−$ will have a lower ratio of charge to volume than that of $SO42−$ and therefore it has smaller $ΔHhydration$ .

Answer 53

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 54

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 55

(g)

Interpretation Introduction

Interpretation:

Whether $Fe2+$ or $Fe3+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 56

(g)

Expert Solution

Answer to Problem 13.35P

$Fe2+$ will have a smaller $ΔHhydration$ .

Answer 57

(g)

Expert Solution

Answer 58

(g)

Expert Solution

Answer 59

Expert Solution

Answer 60

Explanation of Solution

The ionic charge of $Fe2+$ is less than that of $Fe3+$ . So the ratio of ionic charge to the volume of $Fe2+$ will be lower than that of $Fe3+$ and therefore it has smaller $ΔHhydration$ .

Answer 61

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 62

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .

Conclusion

The species with a lower ratio of ionic charge to volume has smaller $ΔHhydration$ .

Answer 63

(h)

Interpretation Introduction

Interpretation:

Whether $K+$ or $Ca2+$ has smaller $ΔHhydration$ is to be determined.

Concept introduction:

The charge density is defined as the ratio of ionic charge and volume. It is directly proportional to the ionic charge and inversely proportional to the ionic volume. Volume is directly related to the ionic size. Smaller ion will have the charge spread over a small space so charge density will be more and vice-versa.

The enthalpy change of hydration is the enthalpy change when one mole of the ionic species is dissolved in water to give a solution of infinite dilution. It is represented by $ΔHhydration$ . It is always negative.

$ΔHhydration$ is directly related to the charge density on an ion. Higher the charge density, more will be $ΔHhydration$ and vice-versa.

Answer 64

(h)

Expert Solution

Answer to Problem 13.35P

$K+$ will have a smaller $ΔHhydration$ .

Answer 65

(h)

Expert Solution

Answer 66

(h)

Expert Solution

Answer 67

Expert Solution

Answer 68

Explanation of Solution

The ionic charge of $K+$ is smaller than that of $Ca2+$ . So $K+$ will have a lower ratio of charge to volume than that of $Ca2+$ and therefore it has smaller $ΔHhydration$ .