Chapter 14, Problem 14B.4P

(a)

Interpretation Introduction

Interpretation:

The favourable orientation of the ${\text{H}}_{\text{2}}\text{O}$ molecule when approaching an anion has to be stated.  The magnitude of the electric field experienced by the anion when the water dipole is $\text{1}\text{.0}\text{nm}$ from the ion has to be calculated.

Concept introduction:

Polarized bonds are a result of the electronegativity difference between bonding atoms.  The more electronegative atom acquires a partial negative charge and the less electronegative atom acquires a partial positive charge.  This results in the formation of a dipole in the bond between the two atoms.

Answer to Problem 14B.4P

The magnitude of the electric field experienced by the anion when the water dipole is $\text{1}\text{.0}\text{nm}$ from the ion is $\underset{\_}{1.11\times 10^{8}V{m}^{-1}}$.

Explanation of Solution

The positive end of the dipole will lie closer to the negative anion when the ${\text{H}}_{\text{2}}\text{O}$ molecule approaches an anion.

The expression for the electric field ($E$) generated at a distance $r$ is shown below.

    $E=\frac{\mu }{2\text{π}{\epsilon }_0{r}^3}$        (1)

Where,

• ${\epsilon }_0$ is the vacuum permittivity. ($8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}$)
• $\mu$ is the dipole moment.

The value of $\mu$ is $\text{1}\text{.85}\text{D}$.

The conversion of $\text{D}$ to $\text{C}\text{m}$ is done as shown below.

    $1\text{D}=3.34\times {10}^{-30}\text{C}\text{m}$

Therefore, the conversion of $\text{1}\text{.85}\text{D}$ to $\text{C}\text{m}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{.85}\text{D}=\text{1}\text{.85}\times 3.34\times {10}^{-30}\text{C}\text{m}\\ =6.179\times {10}^{-30}\text{C}\text{m}\end{array}$

The value of $r$ is $\text{1}\text{.0}\text{nm}$.

The conversion of $\text{nm}$ to $\text{m}$ is shown below.

    $1\text{nm}={10}^{-9}\text{m}$

Substitute the value of ${\epsilon }_0$ , $r$ and $\mu$ in equation (1).

    $\begin{array}{c}E=\frac{6.179\times {10}^{-30}\text{C}\text{m}}{2\times \text{3}\text{.14}\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times {\left({10}^{-9}\text{m}\right)}^3}\\ =\frac{6.179\times {10}^{-30}\text{C}\text{m}}{5.5603\times {10}^{-38}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^2}\\ =1.11\times {10}^8\text{J}{\text{C}}^{-1}{\text{m}}^{-1}\left(1\text{J}{\text{C}}^{-1}=1\text{V}\right)\\ =\underset{\_}{1.11\times 10^{8}V{m}^{-1}}\end{array}$

Therefore, the magnitude of the electric field experienced by the anion is $\underset{\_}{1.11\times 10^{8}V{m}^{-1}}$.

(b)

Interpretation Introduction

Interpretation:

The magnitude of the electric field experienced by the anion when the water dipole is $\text{0}\text{.3}\text{nm}$ from the ion has to be calculated.

Concept introduction:

Same as concept introduction in part (a).

Answer to Problem 14B.4P

The magnitude of the electric field experienced by the anion when the water dipole is $\text{0}\text{.3}\text{nm}$ from the ion is $\underset{\_}{4.12\times 10^{9}V{m}^{-1}}$.

Explanation of Solution

The positive end of the dipole will lie closer to the negative anion when the ${\text{H}}_{\text{2}}\text{O}$ molecule approaches an anion.

The expression for the electric field ($E$) generated at a distance $r$ is shown below.

    $E=\frac{\mu }{2\text{π}{\epsilon }_0{r}^3}$        (1)

Where,

• ${\epsilon }_0$ is the vacuum permittivity. ($8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}$)
• $\mu$ is the dipole moment.

The value of $\mu$ is $\text{1}\text{.85}\text{D}$.

The conversion of $\text{D}$ to $\text{C}\text{m}$ is done as shown below.

    $1\text{D}=3.34\times {10}^{-30}\text{C}\text{m}$

Therefore, the conversion of $\text{1}\text{.85}\text{D}$ to $\text{C}\text{m}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{.85}\text{D}=\text{1}\text{.85}\times 3.34\times {10}^{-30}\text{C}\text{m}\\ =6.179\times {10}^{-30}\text{C}\text{m}\end{array}$

The value of $r$ is $\text{0}\text{.3}\text{nm}$.

The conversion of $\text{nm}$ to $\text{m}$ is shown below.

    $1\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $\text{0}\text{.3}\text{nm}$ to $\text{m}$ is shown below.

    $\text{0}\text{.3}\text{nm}=\text{0}\text{.3}\times {10}^{-9}\text{m}$

Substitute the value of ${\epsilon }_0$ , $r$ and $\mu$ in equation (1).

    $\begin{array}{c}E=\frac{6.179\times {10}^{-30}\text{C}\text{m}}{2\times \text{3}\text{.14}\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times {\left(\text{0}\text{.3}\times {10}^{-9}\text{m}\right)}^3}\\ =\frac{6.179\times {10}^{-30}\text{C}\text{m}}{1.5013\times {10}^{-39}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^2}\\ =4.12\times {10}^9\text{J}{\text{C}}^{-1}{\text{m}}^{-1}\left(1\text{J}{\text{C}}^{-1}=1\text{V}\right)\\ =\underset{\_}{4.12\times 10^{9}V{m}^{-1}}\end{array}$

Therefore, the magnitude of the electric field experienced by the anion is $\underset{\_}{4.12\times 10^{9}V{m}^{-1}}$.

(c)

Interpretation Introduction

Interpretation:

The magnitude of the electric field experienced by the anion when the water dipole is $\text{30}\text{nm}$ from the ion has to be calculated.

Concept introduction:

Same as concept introduction in part (a).

Answer to Problem 14B.4P

The magnitude of the electric field experienced by the anion when the water dipole is $\text{30}\text{nm}$ from the ion is $\underset{\_}{4.12\times 10^{3}V{m}^{-1}}$.

Explanation of Solution

The positive end of the dipole will lie closer to the negative anion when the ${\text{H}}_{\text{2}}\text{O}$ molecule approaches an anion.

The expression for the electric field ($E$) generated at a distance $r$ is shown below.

    $E=\frac{\mu }{2\text{π}{\epsilon }_0{r}^3}$        (1)

Where,

• ${\epsilon }_0$ is the vacuum permittivity. ($8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}$)
• $\mu$ is the dipole moment.

The value of $\mu$ is $\text{1}\text{.85}\text{D}$.

The conversion of $\text{D}$ to $\text{C}\text{m}$ is done as shown below.

    $1\text{D}=3.34\times {10}^{-30}\text{C}\text{m}$

Therefore, the conversion of $\text{1}\text{.85}\text{D}$ to $\text{C}\text{m}$ is done as shown below.

    $\begin{array}{c}\text{1}\text{.85}\text{D}=\text{1}\text{.85}\times 3.34\times {10}^{-30}\text{C}\text{m}\\ =6.179\times {10}^{-30}\text{C}\text{m}\end{array}$

The value of $r$ is $\text{30}\text{nm}$.

The conversion of $\text{nm}$ to $\text{m}$ is shown below.

    $1\text{nm}={10}^{-9}\text{m}$

Therefore, the conversion of $\text{30}\text{nm}$ to $\text{m}$ is shown below.

    $\begin{array}{c}\text{30}\text{nm}=\text{30}\times {10}^{-9}\text{m}\\ =\text{3}\times {10}^{-8}\text{m}\end{array}$

Substitute the value of ${\epsilon }_0$ , $r$ and $\mu$ in equation (1).

    $\begin{array}{c}E=\frac{6.179\times {10}^{-30}\text{C}\text{m}}{2\times \text{3}\text{.14}\times 8.854\times {10}^{-12}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^{-1}\times {\left(\text{3}\times {10}^{-8}\text{m}\right)}^3}\\ =\frac{6.179\times {10}^{-30}\text{C}\text{m}}{1.5013\times {10}^{-33}{\text{J}}^{-1}{\text{C}}^2{\text{m}}^2}\\ =4.12\times {10}^3\text{J}{\text{C}}^{-1}{\text{m}}^{-1}\left(1\text{J}{\text{C}}^{-1}=1\text{V}\right)\\ =\underset{\_}{4.12\times 10^{3}V{m}^{-1}}\end{array}$

Therefore, the magnitude of the electric field experienced by the anion is $\underset{\_}{4.12\times 10^{3}V{m}^{-1}}$.