Chapter 5, Problem 5F.8AE

Interpretation Introduction

Interpretation: On the basis of the given activity coefficients of $\text{HBr}$ in three dilute solutions at $25°\text{C}$, the value of $B$ in the Davies equation has to be estimated.

Concept introduction: Davies equation is an extended form of the Debye-Huckel limiting law which is used when ionic strength of the solution is too high for the Debye-Huckel limiting law to be applicable.

Answer to Problem 5F.8AE

On the basis of the given activity coefficients of $\text{HBr}$ in three dilute solutions at $25°\text{C}$, the value of $B$ in the Davies equation is calculated as $\underset{\_}{2.0097}$.

Explanation of Solution

The ionic strength $\left(I\right)$ of a solution is given by the equation,

    $I=\frac{1}{2}{\displaystyle \sum _i{z}_i{}^2\left(b_i/b^{°}\right)}$                                                                                        (1)

The mean activity coefficient of $\text{5}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$ $\text{HBr}$ is $0.930$.

The molality of $\text{HBr}$ is $\text{5}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$.

The conversion of $\text{mmol}{\text{kg}}^{-1}$ to $\text{mol}{\text{kg}}^{-1}$ is done as,

    $\text{1}\text{mmol}{\text{kg}}^{-1}{\text{=10}}^{\text{-3}}\text{ mol}{\text{kg}}^{-1}$

Therefore, the conversion of $\text{5}\text{.0}\text{mmol}{\text{kg}}^{-1}$ to $\text{mol}{\text{kg}}^{-1}$ is done as,

  $5.0\text{mmol}{\text{kg}}^{-1}=\text{5}{\text{.0×10}}^{\text{-3}}\text{ mol}{\text{kg}}^{-1}$

The dissociation of $\text{HBr}$ is represented by the reaction.

    $\text{HBr}\to {\text{H}}^{\text{+}}{\text{+Br}}^{\text{-}}$

Hence,

The molality of cation, ${\text{H}}^{\text{+}}$ $\left(b_{+}\right)$ is $\text{5}{\text{.0×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$.

The molality of anion, ${\text{Br}}^{-}$ $\left(b_{-}\right)$ is $\text{5}{\text{.0×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$.

The charge present on cation, ${\text{H}}^{\text{+}}$ $\left(z_{+}\right)$ is $+1$.

The charge present on anion, ${\text{Br}}^{-}$ $\left(z_{-}\right)$ is $-1$.

Substitute the value of $b_{+}$, $b_{-}$, $z_{+}$ and $z_{-}$ in equation (1)

    $\begin{array}{c}{I}_{\text{HBr}}=\frac{1}{2}\left[\left(b_{+}\times z_{+}^2\right)+\left(b_{-}\times z_{-}^2\right)\right]\\ =\frac{1}{2}\left[\left(\text{5}{\text{.0×10}}^{\text{-3}}\times {\left(1\right)}^2\right)+\left(\text{5}{\text{.0×10}}^{\text{-3}}\times {\left(-1\right)}^2\right)\right]\\ =\text{5}{\text{.0×10}}^{\text{-3}}\end{array}$

The Davies equation is given as,

    $\log {\gamma }_{\pm }=-\frac{A\left|z_{+}{z}_{-}\right|I^{1/2}}{1+B{I}^{1/2}}+CI$                                                                            (2)

Where

• ${\gamma }_{\pm }$ is the activity coefficient.
• $A$ is $0.509$ for an aqueous solution at $25°\text{C}$.
• $I$ is the dimensionless ionic strength.
• $z_{+}$ is the charge on the positive cation.
• $z_{-}$ is the charge on the negative cation.
• $B$ is a dimensionless constant.
• $C$ is a dimensionless constant.

The mean activity coefficient of $\text{5}{\text{.0×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$ $\text{HBr}$ $\left({\gamma }_{\pm }\right)$ is $0.930$.

The ionic strength of $\text{5}{\text{.0×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$ $\text{HBr}$$\left(I\right)$ is $\text{5}{\text{.0×10}}^{\text{-3}}$.

The charge present on cation, ${\text{H}}^{\text{+}}$ $\left(z_{+}\right)$ is $+1$.

The charge present on anion, ${\text{Br}}^{-}$ $\left(z_{-}\right)$ is $-1$.

The value of $C$ for $\text{HBr}$ is $0$.

Substitute the value of $I_{\text{HBr}}$, ${\gamma }_{\pm }$, $A$, $z_{+}$, $z_{-}$ and $C$ in equation (2).

    $\begin{array}{c}\log \left(0.930\right)=-\frac{0.509\times \left|1\times 1\right|\times {\left(5\times {10}^{-3}\right)}^{1/2}}{1+B{\left(5\times {10}^{-3}\right)}^{1/2}}+0\times {\left(5\times {10}^{-3}\right)}^{1/2}\\ -0.0315=-\frac{0.509\times 0.0707}{1+B{\left(5\times {10}^{-3}\right)}^{1/2}}+0\\ -0.0315=\frac{0.0359}{1+0.0707B}\\ B=2.0055\end{array}$

The mean activity coefficient of $\text{10}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$ $\text{HBr}$ is $0.907$.

The molality of $\text{HBr}$ is $\text{10}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$.

The conversion of $\text{mmol}{\text{kg}}^{-1}$ to $\text{mol}{\text{kg}}^{-1}$ is done as,

    $\text{1}\text{mmol}{\text{kg}}^{-1}{\text{=10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$

Therefore, the conversion of $\text{10}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$ to $\text{mol}{\text{kg}}^{-1}$ is done as,

  $\begin{array}{c}\text{10}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}\text{=10}\text{.0}\times {\text{10}}^{\text{-3}}\text{ mol}{\text{kg}}^{-1}\\ ={10}^{-4}\text{mol}{\text{kg}}^{-1}\end{array}$

The dissociation of $\text{HBr}$ is represented by the reaction.

    $\text{HBr}\to {\text{H}}^{\text{+}}{\text{+Br}}^{\text{-}}$

Hence,

The molality of cation, ${\text{H}}^{\text{+}}$ $\left(b_{+}\right)$ is ${10}^{-4}\text{mol}{\text{kg}}^{-1}$.

The molality of anion, ${\text{Br}}^{-}$ $\left(b_{-}\right)$ is ${10}^{-4}\text{mol}{\text{kg}}^{-1}$.

The charge present on cation, ${\text{H}}^{\text{+}}$ $\left(z_{+}\right)$ is $+1$.

The charge present on anion, ${\text{Br}}^{-}$ $\left(z_{-}\right)$ is $-1$.

Substitute the value of $b_{+}$, $b_{-}$, $z_{+}$ and $z_{-}$ in equation (1)

    $\begin{array}{c}{I}_{\text{HBr}}=\frac{1}{2}\left[\left(b_{+}\times z_{+}^2\right)+\left(b_{-}\times z_{-}^2\right)\right]\\ =\frac{1}{2}\left[\left({\text{10}}^{\text{-4}}\times {\left(1\right)}^2\right)+\left({\text{10}}^{\text{-4}}\times {\left(-1\right)}^2\right)\right]\\ ={\text{10}}^{\text{-4}}\end{array}$

The Davies equation is given as,

    $\log {\gamma }_{\pm }=-\frac{A\left|z_{+}{z}_{-}\right|I^{1/2}}{1+B{I}^{1/2}}+CI$                                                                            (2)

Where

• ${\gamma }_{\pm }$ is the activity coefficient.
• $A$ is $0.509$ for an aqueous solution at $25°\text{C}$.
• $I$ is the dimensionless ionic strength.
• $z_{+}$ is the charge on the positive cation.
• $z_{-}$ is the charge on the negative cation.
• $B$ is a dimensionless constant.
• $C$ is a dimensionless constant.

The mean activity coefficient of ${10}^{-4}\text{mol}{\text{kg}}^{-1}$ $\text{HBr}$ $\left({\gamma }_{\pm }\right)$ is $0.907$.

The ionic strength of ${10}^{-4}\text{mol}{\text{kg}}^{-1}$ $\text{HBr}$$\left(I\right)$ is ${\text{10}}^{\text{-4}}$.

The charge present on cation, ${\text{H}}^{\text{+}}$ $\left(z_{+}\right)$ is $+1$.

The charge present on anion, ${\text{Br}}^{-}$ $\left(z_{-}\right)$ is $-1$.

The value of $C$ for $\text{HBr}$ is $0$.

Substitute the value of $I_{\text{HBr}}$, ${\gamma }_{\pm }$, $A$, $z_{+}$, $z_{-}$ and $C$ in equation (2).

    $\begin{array}{c}\log \left(0.907\right)=-\frac{0.509\times \left|1\times 1\right|\times {\left({10}^{-4}\right)}^{1/2}}{1+B{\left({10}^{-4}\right)}^{1/2}}+0\times {\left({10}^{-4}\right)}^{1/2}\\ -0.042=-\frac{0.509\times {10}^{-2}}{1+{\left({10}^{-4}\right)}^{1/2}B}+0\\ -0.042=-\frac{0.00509}{1+\left({10}^{-2}\right)B}+0\\ B=2.0047\end{array}$

The mean activity coefficient of $\text{20}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$ $\text{HBr}$ is 0.879.

The molality of $\text{HBr}$ is $\text{20}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$.

The conversion of $\text{mmol}{\text{kg}}^{-1}$ to $\text{mol}{\text{kg}}^{-1}$ is done as,

    $\text{1}\text{mmol}{\text{kg}}^{-1}{\text{=10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$

Therefore, the conversion of $\text{20}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}$ to $\text{mol}{\text{kg}}^{-1}$ is done as,

  $\text{20}\text{.0}\text{mmol}{\text{kg}}^{\text{-1}}{\text{=20×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$

The dissociation of $\text{HBr}$ is represented by the reaction.

    $\text{HBr}\to {\text{H}}^{\text{+}}{\text{+Br}}^{\text{-}}$

Hence,

The molality of cation, ${\text{H}}^{\text{+}}$ $\left(b_{+}\right)$ is ${\text{20×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$.

The molality of anion, ${\text{Br}}^{-}$ $\left(b_{-}\right)$ is ${\text{20×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$.

The charge present on cation, ${\text{H}}^{\text{+}}$ $\left(z_{+}\right)$ is $+1$.

The charge present on anion, ${\text{Br}}^{-}$ $\left(z_{-}\right)$ is $-1$.

Substitute the value of $b_{+}$, $b_{-}$, $z_{+}$ and $z_{-}$ in equation (1)

    $\begin{array}{c}{I}_{\text{HBr}}=\frac{1}{2}\left[\left(b_{+}\times z_{+}^2\right)+\left(b_{-}\times z_{-}^2\right)\right]\\ =\frac{1}{2}\left[\left(\text{20}{\text{.0×10}}^{\text{-3}}\times {\left(1\right)}^2\right)+\left(\text{20}{\text{.0×10}}^{\text{-3}}\times {\left(-1\right)}^2\right)\right]\\ =20.{\text{0×10}}^{\text{-3}}\end{array}$

The Davies equation is given as,

    $\log {\gamma }_{\pm }=-\frac{A\left|z_{+}{z}_{-}\right|I^{1/2}}{1+B{I}^{1/2}}+CI$                                                                            (2)

Where

• ${\gamma }_{\pm }$ is the activity coefficient.
• $A$ is $0.509$ for an aqueous solution at $25°\text{C}$.
• $I$ is the dimensionless ionic strength.
• $z_{+}$ is the charge on the positive cation.
• $z_{-}$ is the charge on the negative cation.
• $B$ is a dimensionless constant.
• $C$ is a dimensionless constant.

The mean activity coefficient of ${\text{20×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$ $\text{HBr}$ $\left({\gamma }_{\pm }\right)$ is $0.879$.

The ionic strength of ${\text{20×10}}^{\text{-3}}\text{mol}{\text{kg}}^{-1}$ $\text{HBr}$$\left(I\right)$ is ${\text{20×10}}^{\text{-3}}$.

The charge present on cation, ${\text{H}}^{\text{+}}$ $\left(z_{+}\right)$ is $+1$.

The charge present on anion, ${\text{Br}}^{-}$ $\left(z_{-}\right)$ is $-1$.

The value of $C$ for $\text{HBr}$ is $0$.

Substitute the value of $I_{\text{HBr}}$, ${\gamma }_{\pm }$, $A$, $z_{+}$, $z_{-}$ and $C$ in equation (2).

    $\begin{array}{c}\log \left(0.879\right)=-\frac{0.509\times \left|1\times 1\right|\times {\left(20\times {10}^{-3}\right)}^{1/2}}{1+B{\left(20\times {10}^{-3}\right)}^{1/2}}+0\times {\left(20\times {10}^{-3}\right)}^{1/2}\\ -0.0560=-\frac{0.509\times 0.1414}{1+{\left(20\times {10}^{-3}\right)}^{1/2}B}+0\\ -0.0560=-\frac{0.0719}{1+0.1414B}\\ B=2.0189\end{array}$

The mean value of $B$ is the average of above values.

  $\begin{array}{c}B=\frac{2.0055+2.0047+2.0189}{3}\\ =\underset{\_}{2.0097}\end{array}$

Thus, on the basis of the given activity coefficients of $\text{HBr}$ in three dilute solutions at $25°\text{C}$, the value of $B$ in the Davies equation is calculated as $\underset{\_}{2.0097}$.