Chapter 6, Problem 6.4IA

Interpretation Introduction

Interpretation:

The value of $\text{p}{K}_{\text{w}}$ at the given temperatures has to be calculated. The standard enthalpy and entropy of autoprotolysis of water at $\text{25}\text{°C}$ has to be determined.

Concept introduction:

At a given temperature, the product of the concentration of ${\text{H}}^{\text{+}}$ and ${\text{OH}}^{-}$ is termed as the ionic product of the water at that temperature.  The negative logarithm of ionic product gives $\text{p}{K}_{\text{w}}$.  The standard cell potential of a couple is the potential of the couple with respect to the potential of the standard hydrogen electrode.  It is the potential of the cell under standard conditions of temperature, pressure and concentrations.

Answer to Problem 6.4IA

The value of $\text{p}{K}_{\text{w}}$ at the given temperatures is shown in the table below.

$\theta /\text{°C}$	$\text{p}{K}_{\text{w}}$
$\text{20}\text{°C}$	$14.22$
$\text{25}\text{°C}$	$14$
$\text{30}\text{°C}$	$11.82$

The value of standard entropy is $\underset{\_}{9.6485J{K}^{-1}mo{l}^{-1}}$.  The value of the standard enthalpy is $\underset{\_}{-98625.5Jmo{l}^{-1}}$.

Explanation of Solution

The cell is given below.

$\text{Pt}|{\text{H}}_{\text{2}}\left(\text{g,}{p}^{\Theta }\right)|\text{NaOH}\left(\text{aq},0.0100\text{mol}{\text{kg}}^{-1}\right),\text{NaCl}\left(\text{aq},0.01125\text{mol}{\text{kg}}^{-1}\right)|\text{AgCl}\left(\text{s}\right)|\text{Ag}\left(\text{s}\right)$

The equilibrium constant for the hydrolysis of water $K_{\text{w}}$ is expressed by the relation below.

    $K_{\text{w}}=\frac{\left[{\text{H}}_{\text{2}}\text{O}\right]}{\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]}$        (1)

Where,

• $\left[{\text{H}}_{\text{2}}\text{O}\right]$ is the concentration of water.
• $\left[{\text{H}}^{+}\right]$ is the concentration of ${\text{H}}^{+}$.
• $\left[{\text{OH}}^{-}\right]$ is the concentration of ${\text{OH}}^{-}$.

The value of $\text{pOH}$ is calculated by the formula given below.

    $\text{pOH}=-\text{log}{\text{OH}}^{-}$        (2)

The concentration of ${\text{OH}}^{-}$, $\left[{\text{OH}}^{-}\right]$ is $0.0100\text{mol}{\text{kg}}^{-1}$.

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in equation (2).

    $\begin{array}{c}\text{pOH}=-\text{log}\left(0.0100\right)\\ =2\end{array}$

The relation between $\text{pOH}$ and $\text{pH}$ is shown below.

    $\text{pOH}+\text{pH}=\text{14}$        (3)

Substitute the value of $\text{pOH}$ in equation (3).

    $\begin{array}{c}\text{2}+\text{pH}=\text{14}\\ \text{pH}=14-2\\ =12\end{array}$        (4)

The Nernst equation is shown below

    $E=E^{\text{ο}}-\frac{RT}{F}\ln \left[{\text{H}}^{+}\right]$

Where,

• $E_{\text{cell}}$ is the potential difference of the cell.
• $E_{\text{cell}}^{\Theta }$ is the standard cell potential.
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.
• $T$ is the temperature.
• $F$ is Faraday’s constant $\left(\text{96485}\text{C}{\text{mol}}^{-1}\right)$.

Let the value of $\left[{\text{H}}^{+}\right]$ at $\text{20}\text{°C}$ be $x$.

Let the value of $\left[{\text{H}}^{+}\right]$ at $\text{25}\text{°C}$ be $y$.

Let the value of $E_{\text{cell}}$ at $\text{20}\text{°C}$ be $E_1$.

Let the value of $E_{\text{cell}}$ at $\text{25}\text{°C}$ be $E_2$.

Let the temperature $\text{20}\text{°C}$ be $T_1$.

Let the temperature $\text{25}\text{°C}$ be $T_2$.

Therefore,

  $E_1=E^{\text{ο}}-\frac{R{T}_1}{F}\ln x$        (5)

And,

    $E_2=E^{\text{ο}}-\frac{R{T}_2}{F}\ln y$        (6)

Subtract equation (6) and (5).

$\begin{array}{l}{T}_1\ln x-T_2\ln y=\frac{\left(E_2-E_1\right)F}{R}\\ T_1\log x-T_2\log y=\frac{\left(E_2-E_1\right)F}{2.303R}\end{array}$        (7)

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of $\text{20}\text{°C}$ to Kelvin is done as,

    $\begin{array}{c}\text{20}\text{°C}=20+2\text{73}\text{.15 K}\\ =293.15\text{ K}\end{array}$

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of $\text{25}\text{°C}$ to Kelvin is done as,

    $\begin{array}{c}\text{25}\text{°C}=25+2\text{73}\text{.15 K}\\ =298.15\text{ K}\end{array}$

The value of $E_{\text{1}}$ is $\text{1}\text{.04774}\text{V}$.

The value of $E_{\text{2}}$ is $\text{1}\text{.04864}\text{V}$.

The value of $\text{pH}$ is calculated by the formula given below.

    $\text{pH}=-\text{log}{\text{H}}^{+}$

Therefore,

    $\begin{array}{c}{\text{pH}}_1=-\text{log}x\\ \text{log}x=-{\text{pH}}_1\end{array}$

And,

    $\begin{array}{c}{\text{pH}}_2=-\text{log}y\\ \text{log}y=-{\text{pH}}_2\end{array}$

Substitute the corresponding values in equation (7).

    $\begin{array}{c}-\left(\text{293}\text{.15}\text{K}\times {\text{pH}}_{\text{1}}\right)+\left(\text{298}\text{.15}\text{K}\times {\text{pH}}_{\text{2}}\right)=\frac{\left(\text{1}\text{.04864}\text{V}-\text{1}\text{.04774}\text{V}\right)\times \text{96485}{\text{mol}}^{-1}}{2.303\times \text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{-\text{1}}}\\ -\left(\text{293}\text{.15}\text{K}\times {\text{pH}}_{\text{1}}\right)+\left(\text{298}\text{.15}\text{K}\times {\text{pH}}_{\text{2}}\right)=\frac{0.0009\text{V}\times \text{96485}\text{C}{\text{mol}}^{-1}}{2.303\times \text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{-\text{1}}}\times \frac{\text{1}\text{J}}{\text{1}\text{CV}}\\ -293.15{\text{pH}}_{\text{1}}+\text{298}\text{.15}\text{K}{\text{pH}}_{\text{2}}=4.53\\ {\text{pH}}_{\text{1}}=\frac{\text{298}\text{.15}\text{K}{\text{pH}}_{\text{2}}-4.53}{293.15}\end{array}$

Substitute the value of $\text{pH}$ from equation (4) in the above equation.

    $\begin{array}{c}{\text{pH}}_{\text{1}}=\frac{\text{298}\text{.15}\times 12-4.53}{293.15}\\ =12.19\end{array}$        (8)

The value of $\text{p}{K}_{\text{w}}$ is calculated by the formula given below.

    $\text{p}{K}_{\text{w}}=\text{pOH}+\text{pH}$        (9)

Substitute the value of $\text{pH}$ from equation (8) and $\text{pOH}$ in equation (9).

    $\begin{array}{c}\text{p}{K}_{\text{w}}=\text{2}+\text{12}\text{.19}\\ =\underset{\_}{14.19}\end{array}$

Therefore, the value of $\text{p}{K}_{\text{w}}$ at $\text{20}\text{°C}$ is $\underset{\_}{14.19}$.

Similarly, the value of $\text{p}{K}_{\text{w}}$ at various temperatures is calculated in the table below.

$\theta /\text{°C}$	$\text{p}{K}_{\text{w}}$
$\text{20}\text{°C}$	$14.19$
$\text{25}\text{°C}$	$14$
$\text{30}\text{°C}$	$11.82$

The standard cell potential $\left(E_{\text{cell}}^{\Theta }\right)$ is given by the Nernst equation as shown below.

    $E_{\text{cell}}^{\Theta }=E_{\text{cell}}+\frac{2.303RT}{F}\log Q$        (10)

Where,

• $E$ is the potential difference of the cell.
• $R$ is the gas constant $\left(\text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)$.
• $T$ is the temperature.
• $F$ is Faraday’s constant $\left(\text{96485}\text{C}{\text{mol}}^{-1}\right)$.
• $Q$ is the reaction quotient.

For the given reaction, the reaction quotient, $Q$ can be calculated by the formula shown below.

    $Q=\frac{\text{Concentration}\text{of}\text{NaCl}}{\text{Concentration}\text{of}\text{NaOH}}$        (11)

The concentration of $\text{NaOH}$ is $0.0100\text{mol}{\text{kg}}^{-1}$.

The concentration of $\text{NaCl}$ is $0.01125\text{mol}{\text{kg}}^{-1}$.

Substitute the concentrations of $\text{NaOH}$ and $\text{NaCl}$ in equation (11).

    $\begin{array}{c}Q=\frac{0.01125\text{mol}{\text{kg}}^{-1}}{0.0100\text{mol}{\text{kg}}^{-1}}\\ =1.125\end{array}$

The value of temperature is $25\text{°C}$.

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73 K}$

Therefore, the conversion of $\text{25}\text{°C}$ to Kelvin is done as,

    $\begin{array}{c}\text{25}\text{°C}=25+2\text{73}\text{.15 K}\\ =298.15\text{ K}\end{array}$

The value of $E_{\text{cell}}$ at $\text{25}\text{°C}$ is $\text{1}\text{.04864}\text{V}$.

Substitute the value of $E_{\text{cell}}$, $Q$, $T,R$ and $F$ in equation (11).

    $\begin{array}{c}{E}_{\text{cell}}^{\Theta }=\text{1}\text{.04864}\text{V}+\frac{2.303\times \text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{-\text{1}}\times 298.15\text{ K}}{\text{96485}\text{C}{\text{mol}}^{-1}}\log 1.125\\ =\text{1}\text{.04864}\text{V}+0.003026\text{J}{\text{C}}^{-1}\times \frac{1\text{VC}}{\text{1}\text{J}}\\ =\text{1}\text{.04864}\text{V}+0.003026\text{V}\\ =1.052\text{V}\end{array}$

The value of temperature is $20\text{°C}$.

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73}\text{.15 K}$

Therefore, the conversion of $\text{25}\text{°C}$ to Kelvin is done as,

    $\begin{array}{c}\text{20}\text{°C}=20+2\text{73}\text{.15 K}\\ =293.15\text{ K}\end{array}$

The value of $E_{\text{cell}}$ at $20\text{°C}$ is $\text{1}\text{.04774}\text{V}$.

Substitute the value of $E_{\text{cell}}$, $Q$, $T,R$ and $F$ in equation (11).

    $\begin{array}{c}{E}_{\text{cell}}^{\Theta }=\text{1}\text{.04774}\text{V}+\frac{2.303\times \text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{-\text{1}}\times 293.15\text{ K}}{\text{96485}\text{C}{\text{mol}}^{-1}}\log 1.125\\ =\text{1}\text{.04774}\text{V}+0.002975\text{J}{\text{C}}^{-1}\times \frac{1\text{VC}}{\text{1}\text{J}}\\ =\text{1}\text{.04774}\text{V}+0.002975\text{V}\\ =1.051\text{V}\end{array}$

The value of temperature is $30\text{°C}$.

The conversion of Celsius to Kelvin is done as,

    ${\text{0 }}^{\text{ο}}\text{C}=2\text{73}\text{.15 K}$

Therefore, the conversion of $30\text{°C}$ to Kelvin is done as,

    $\begin{array}{c}\text{30}\text{°C}=30+2\text{73}\text{.15 K}\\ =303.15\text{ K}\end{array}$

The value of $E_{\text{cell}}$ at $30\text{°C}$ is $\text{1}\text{.04942}\text{V}$.

Substitute the value of $E_{\text{cell}}$, $Q$, $T,R$ and $F$ in equation (11).

    $\begin{array}{c}{E}_{\text{cell}}^{\Theta }=\text{1}\text{.04942}\text{V}+\frac{2.303\times \text{8}\text{.314}\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{-\text{1}}\times 303.15\text{ K}}{\text{96485}\text{C}{\text{mol}}^{-1}}\log 1.125\\ =\text{1}\text{.04942}\text{V}+0.003077\text{J}{\text{C}}^{-1}\times \frac{1\text{VC}}{\text{1}\text{J}}\\ =\text{1}\text{.04942}\text{V}+0.003077\text{V}\\ =1.053\text{V}\end{array}$

The $\Delta E_{\text{cell}}^{\Theta }$ value is calculated below.

    $\begin{array}{c}\Delta E_{\text{cell}}^{\Theta }=\text{1}\text{.052}\text{V}-\text{1}\text{.051}\text{V}\\ =\text{0}\text{.001}\text{V}\end{array}$

The difference in temperature is calculated below.

    $\begin{array}{c}\Delta T=303.1\text{K}-293.15\text{K}\\ \text{=10}\text{K}\end{array}$

The value of $\left(\frac{\partial E_{\text{cell}}^{\Theta }}{\partial T}\right)$ is calculated below.

    $\begin{array}{c}\left(\frac{\partial E_{\text{cell}}^{\Theta }}{\partial T}\right)=\frac{\text{0}\text{.001}\text{V}}{10\text{K}}\\ =1.0\times {10}^{-4}\text{V}{\text{K}}^{-1}\end{array}$

The value of entropy ($\Delta S^{\Theta }$) is calculated by the formula given below.

    $\Delta S^{\Theta }=vF\left(\frac{\partial E_{\text{cell}}^{\Theta }}{\partial T}\right)$        (12)

Where,

• $\nu$ is the number of electrons.

The number of electrons in the above cell is $1$.

Substitute the corresponding values in equation (12).

     $\begin{array}{c}\Delta S^{\Theta }=1\times 96485\text{C}{\text{mol}}^{-1}\times 1.0\times {10}^{-4}\text{V}{\text{K}}^{-1}\\ =9.6485\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}\\ =9.6485\text{C}\text{V}{\text{K}}^{-1}{\text{mol}}^{-1}\times \frac{1\text{J}}{\text{1}\text{C}\text{V}}\\ =\underset{\_}{9.6485J{K}^{-1}mo{l}^{-1}}\end{array}$

Therefore, the value of standard entropy is $\underset{\_}{9.6485J{K}^{-1}mo{l}^{-1}}$.

The standard Gibbs energy $\left(\Delta G^{\Theta }\right)$ is given by the expression,

    $\Delta G^{\Theta }=-\nu F{E}_{\text{cell}}^{\Theta }$        (13)

The number of electrons in the given cell is $1$.

The value of $E_{\text{cell}}^{\Theta }$ is $1.052\text{V}$.

Substitute the value of $\nu$, $E_{\text{cell}}^{\Theta }$, and $F$ in equation (13).

    $\begin{array}{c}\Delta G^{\Theta }=-\left(1\times 96485\text{C}{\text{mol}}^{-1}\times 1.052\text{V}\right)\\ =-\left(101502.2\text{V}\text{C}{\text{mol}}^{-1}\right)\\ =-\left(101502.2\text{V}\text{C}{\text{mol}}^{-1}\times \frac{1\text{J}}{1\text{V}\text{C}}\right)\\ =-101502.2\text{J}{\text{mol}}^{-1}\end{array}$

The standard enthalpy ($\Delta H^{\Theta }$) is calculated by the formula given below.

    $\Delta H^{\Theta }=\Delta G^{\Theta }+T\Delta S^{\Theta }$        (14)

The value of $\Delta S^{\Theta }$ is $9.6485\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}$

The value of $\Delta G^{\Theta }$ is $-101502.2\text{J}{\text{mol}}^{-1}$.

The value of temperature ($T$) is $298.15\text{K}$.

Substitute the value of $\Delta S^{\Theta }$, $\Delta G^{\Theta }$, and $T$ in equation (14).

    $\begin{array}{c}\Delta H^{\Theta }=-101502.2\text{J}{\text{mol}}^{-1}+\left(298.15\text{K}\times 9.6485\text{J}{\text{K}}^{\text{-1}}{\text{mol}}^{\text{-1}}\right)\\ =-101502.2\text{J}{\text{mol}}^{-1}+2876.700\text{J}{\text{mol}}^{-1}\\ =\underset{\_}{-98625.5Jmo{l}^{-1}}\end{array}$

Therefore, the value of the standard enthalpy is $\underset{\_}{-98625.5Jmo{l}^{-1}}$.