Chapter 16, Problem 16B.8P

Interpretation Introduction

Interpretation:

The diffusion coefficients and the effective hydrodynamic radii of alkali metal cation in water have to be calculated.

The numbers of water molecule that are dragged along by the cations have to be calculated.

Concept introduction:

Conductivity is a property of a material that depends upon the charge carriers in a sample.  Molar conductivity is the conductivity per unit molar concentration.  According to Kohlrausch’s law, the molar conducivities of the dilute solutions containing strong electrolytes depend upon the square root of the concentration.  The conductivity of a solution when the concentration tends to zero is known as the limiting molar conductivity.  Limiting molar conductivity is the sum of contributions from individual ions.

Answer to Problem 16B.8P

The diffusion coefficients of alkali metal cations ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$ are $\underset{\_}{1.026\times 10^{-5}c{m}^2{s}^{-1}}$, $\underset{\_}{1.33\times 10^{-5}c{m}^2{s}^{-1}}$, $\underset{\_}{1.95\times 10^{-5}c{m}^2{s}^{-1}}$, and $\underset{\_}{2.03\times 10^{-5}c{m}^2{s}^{-1}}$ respectively.

The effective hydrodynamic radii of alkali metal cations ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$ are $\underset{\_}{238pm}$, $\underset{\_}{184pm}$, $\underset{\_}{125pm}$ and $\underset{\_}{120pm}$.

The number of water molecule attached to ${\text{Li}}^{+}$ cation is $\underset{\_}{4}$.

The number of water molecule attached to ${\text{Na}}^{+}$ cation is $\underset{\_}{2}$.

Explanation of Solution

The diffusion coefficient is calculated by the formula shown below

    $D=\frac{uRT}{zF}$        (1)

Where,

• $u$ is the mobility.
• $R$ is the gas constant $\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)$.
• $T$ is the temperature.
• $z$ is the charge on cations and anions.
• $F$ is the faraday constant $\left(9648\text{5}\text{C}{\text{mol}}^{-1}\right)$.

The conversion of temperature from $\text{°C}$ to kelvin is shown below.

    $\begin{array}{c}T\left(\text{K}\right)=T\left(\text{°C}\right)+273.15\\ =25+273.15\\ =298.1\text{5}\text{K}\end{array}$

For all the alkali metal cations, the value of $z$ is one.

Substitute the value of $R$, $T$, $z$, $F$ in equation (1).

    $\begin{array}{c}D=\frac{uRT}{zF}\\ =\frac{u\left(8.314\text{J}{\text{K}}^{-1}{\text{mol}}^{-1}\right)298.1\text{5}\text{K}}{1\times 9648\text{5}\text{C}{\text{mol}}^{-1}}\\ =\frac{u\times 2478.82}{96485}\frac{\text{J}}{\text{C}}\left(1\text{J}=1\text{C}\text{V}\right)\\ =u\times 2.56\times {10}^{-2}\text{V}\end{array}$        (2)

The alkali metals are ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$.

For ${\text{Li}}^{+}$, the value of $u$ is $4.01\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (2).

    $\begin{array}{c}D=u\times 2.56\times {10}^{-2}\text{V}\\ =4.01\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}\times 2.56\times {10}^{-2}\text{V}\\ =\underset{\_}{1.026\times 10^{-5}c{m}^2{s}^{-1}}\end{array}$

Therefore, the diffusion coefficients of alkali metal cation ${\text{Li}}^{+}$ is $\underset{\_}{1.026\times 10^{-5}c{m}^2{s}^{-1}}$.

For ${\text{Na}}^{+}$, the value of $u$ is $5.19\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (2).

    $\begin{array}{c}D=u\times 2.56\times {10}^{-2}\text{V}\\ =5.19\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}\times 2.56\times {10}^{-2}\text{V}\\ =\underset{\_}{1.33\times 10^{-5}c{m}^2{s}^{-1}}\end{array}$

Therefore, the diffusion coefficients of alkali metal cation ${\text{Na}}^{+}$ is $\underset{\_}{1.33\times 10^{-5}c{m}^2{s}^{-1}}$.

For ${\text{K}}^{+}$, the value of $u$ is $7.62\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (2).

    $\begin{array}{c}D=u\times 2.56\times {10}^{-2}\text{V}\\ =7.62\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}\times 2.56\times {10}^{-2}\text{V}\\ =\underset{\_}{1.95\times 10^{-5}c{m}^2{s}^{-1}}\end{array}$

Therefore, the diffusion coefficients of alkali metal cation ${\text{K}}^{+}$ is $\underset{\_}{1.95\times 10^{-5}c{m}^2{s}^{-1}}$.

For ${\text{Rb}}^{+}$, the value of $u$ is $7.92\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (2).

    $\begin{array}{c}D=u\times 2.56\times {10}^{-2}\text{V}\\ =7.92\times {10}^{-4}{\text{cm}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}\times 2.56\times {10}^{-2}\text{V}\\ =\underset{\_}{2.03\times 10^{-5}c{m}^2{s}^{-1}}\end{array}$

Therefore, the diffusion coefficients of alkali metal cation ${\text{Rb}}^{+}$ is $\underset{\_}{2.03\times 10^{-5}c{m}^2{s}^{-1}}$.

The effective hydrodynamic radii in water is calculated by the formula shown below.

    $a=\frac{ze}{6\pi \eta u}$        (3)

Where,

• $z$ is the charge on cation and anion.
• $e$ is the cahrge on electron $\left(1.602\times {10}^{-19}\text{C}\right)$.
• $\eta$ is the coefficient of viscosity of water $\left(0.891\times {10}^{-3}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\right)$
• $u$ is the mobility of alkali metal ion.

For all the alkali metal cations, the value of $z$ is one.

Substitute the value of $z$, $e$, and $\eta$ in equation (3).

    $\begin{array}{c}a=\frac{ze}{6\pi \eta u}\\ =\frac{1\times 1.602\times {10}^{-19}\text{C}}{6\times 3.14\times \left(0.891\times {10}^{-3}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}\right)\times u}\left(\text{1}\text{J}=\text{1}\text{C}\text{V}\right)\\ =\frac{160.2\times {10}^{-18}}{u\times 16.79}\frac{\text{J}}{\text{V}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}\left(1\text{J}=1\text{kg}{\text{m}}^2{\text{s}}^{-2}\right)\\ =9.54\times {10}^{-18}\frac{\text{kg}{\text{m}}^2{\text{s}}^{-2}}{\text{V}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}\end{array}$

Solve the above equation.

    $\begin{array}{c}a=\frac{9.54\times {10}^{-18}}{u}\frac{\text{kg}{\text{m}}^2{\text{s}}^{-2}}{\text{V}\text{kg}{\text{m}}^{-1}{\text{s}}^{-1}}\\ =\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{u}\end{array}$        (4)

The alkali metals are ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$.

For ${\text{Li}}^{+}$, the value of $u$ is $4.01\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (4).

    $\begin{array}{c}a=\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{u}\\ =\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{4.01\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}}\\ =2.38\times {10}^{-10}\text{m}\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ =\underset{\_}{238pm}\end{array}$

Therefore, the effective hydrodynamic radii of alkali metal cation ${\text{Li}}^{+}$ is $\underset{\_}{238pm}$.

For ${\text{Na}}^{+}$, the value of $u$ is $5.19\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (4).

    $\begin{array}{c}a=\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{u}\\ =\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{5.19\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}}\\ =1.84\times {10}^{-10}\text{m}\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ =\underset{\_}{184pm}\end{array}$

Therefore, the effective hydrodynamic radii of alkali metal cation ${\text{Na}}^{+}$ is $\underset{\_}{184pm}$.

For ${\text{K}}^{+}$, the value of $u$ is $7.62\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (4).

    $\begin{array}{c}a=\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{u}\\ =\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{7.62\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}}\\ =1.25\times {10}^{-10}\text{m}\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ =\underset{\_}{125pm}\end{array}$

Therefore, the effective hydrodynamic radii of alkali metal cation ${\text{K}}^{+}$ is $\underset{\_}{125pm}$.

For ${\text{Rb}}^{+}$, the value of $u$ is $7.92\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}$.

Substitute the value of $u$ in equation (4).

    $\begin{array}{c}a=\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{u}\\ =\frac{9.54\times {10}^{-18}{\text{m}}^3{\text{s}}^{-1}{\text{V}}^{-1}}{7.92\times {10}^{-8}{\text{m}}^{\text{2}}{\text{s}}^{-1}{\text{V}}^{-1}}\\ =1.20\times {10}^{-10}\text{m}\left(\frac{{10}^{12}\text{pm}}{1\text{m}}\right)\\ =\underset{\_}{120pm}\end{array}$

Therefore, the effective hydrodynamic radii of alkali metal cation ${\text{Rb}}^{+}$ is $\underset{\_}{120pm}$.

The ionic radii of alkali metal cations ${\text{Li}}^{+}$, ${\text{Na}}^{+}$, ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$ are given as shown below.

    $\begin{array}{l}{\text{Li}}^{+}{\text{Na}}^{+}{\text{K}}^{+}{\text{Rb}}^{+}\\ \text{Ionic}\text{radii}\left(\raisebox{1ex}{$r_{+}$}\!\left/ \!\raisebox{-1ex}{$\text{pm}$}\right.\right)59102138149\end{array}$

The hydrodynamic radii of ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$ are smaller than ionic radii of ${\text{K}}^{+}$ and ${\text{Rb}}^{+}$. Therefore, they have effective hydrodynamic radii. The effective hydrodynamic volume and ionic volumes of ${\text{Li}}^{+}$ and ${\text{Na}}^{+}$ are calculted by the formula shown below.

  $\Delta V=\frac{4}{3}\pi \left(a^3-r_{+}^3\right)$        (5)

For ${\text{Li}}^{+}$, the value value of $a$ and $r_{+}$ is $238\text{pm}$ and $59\text{pm}$ respectively.

Substitute the $a$ and $r_{+}$ in equation (5).

    $\begin{array}{c}\Delta V=\frac{4}{3}\pi \left(a^3-r_{+}^3\right)\\ =\frac{4}{3}\times 3.14\times \left({238}^3-{59}^3\right){\text{pm}}^3\\ =4.18\times 13275893\times {10}^{-36}{\text{m}}^3\\ =5.6\times {10}^{-29}{\text{m}}^3\end{array}$

The volume occupied by the single water molecule is $1.4\times {10}^{-29}{\text{m}}^3$.  Therefore, the number of water molecule attached to ${\text{Li}}^{+}$ cation is calculated as shown below.

    $\begin{array}{c}\text{Number}\text{of}\text{water}\text{molecule}\text{attached}\text{to}{\text{Li}}^{+}\text{cation}=\frac{5.6\times {10}^{-29}{\text{m}}^3}{1.4\times {10}^{-29}{\text{m}}^3}\\ =\underset{\_}{4}\end{array}$

Therefore, the number of water molecule attached to ${\text{Li}}^{+}$ cation is $\underset{\_}{4}$.

For ${\text{Na}}^{+}$, the value value of $a$ and $r_{+}$ is $184\text{pm}$ and $102\text{pm}$ respectively.

Substitute the $a$ and $r_{+}$ in equation (5).

    $\begin{array}{c}\Delta V=\frac{4}{3}\pi \left(a^3-r_{+}^3\right)\\ =\frac{4}{3}\times 3.14\times \left({184}^3-{102}^3\right){\text{pm}}^3\\ =4.18\times 5168296\times {10}^{-36}{\text{m}}^3\\ =2.16\times {10}^{-29}{\text{m}}^3\end{array}$

The volume occupied by the single water molecule is $1.4\times {10}^{-29}{\text{m}}^3$.  Therefore, the number of water molecule attached to ${\text{Na}}^{+}$ cation is calculated as shown below.

    $\begin{array}{c}\text{Number}\text{of}\text{water}\text{molecule}\text{attached}\text{to}{\text{Na}}^{+}\text{cation}=\frac{2.16\times {10}^{-29}{\text{m}}^3}{1.4\times {10}^{-29}{\text{m}}^3}\\ =1.54\\ \approx \underset{\_}{2}\end{array}$

Therefore, the number of water molecule attached to ${\text{Na}}^{+}$ cation is $\underset{\_}{2}$.