Chapter 5, Problem 5F.6BE

(i)

Interpretation Introduction

Interpretation: The mass of ${\text{KNO}}_{\text{3}}$ to be added to a $0.110\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $1.00$  has to be calculated.

Concept introduction: The ionic strength is introduced in the Debye-Huckel limiting law which relates the activity coefficient as a function of the ionic strength of the solution.  It is dependent on the concentration of all the ions present in the solution.  It is a dimensionless quantity.

Answer to Problem 5F.6BE

The mass of ${\text{KNO}}_{\text{3}}$ to be added to a $0.110\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $1.0$ is $\underset{\_}{44.94g}$.

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)

Where

• $z_i$ is the charge present on the ion.
• $b_i$ is the molality of the ion in the solution.

The molality of ${\text{KNO}}_3\left(\text{aq}\right)$ is $\text{0}\text{.110}\text{mol}{\text{kg}}^{\text{-1}}$.

The reaction for the dissociation of ${\text{KNO}}_3\left(\text{aq}\right)$ is shown as,

  ${\text{KNO}}_3\to {\text{K}}^{+}+{\text{NO}}_3^{-}$

Hence,

The molality of cation, ${\text{K}}^{\text{+}}$ $\left(b_{{\text{K}}^{\text{+}}}\right)$ is $\text{0}\text{.110}\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of anion, ${\text{NO}}_3^{-}$ $\left(b_{{\text{NO}}_3^{-}}\right)$ is $\text{0}\text{.110}\text{mol}{\text{kg}}^{\text{-1}}$.

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

The charge present on anion, ${\text{NO}}_3^{-}$ $\left(z_{{\text{NO}}_3^{-}}\right)$ is $-1$.

Let the molality of ${\text{KNO}}_{\text{3}}$ to be added be b.

The reaction for the dissociation of ${\text{KNO}}_3\left(\text{aq}\right)$ is shown as,

  ${\text{KNO}}_3\to {\text{K}}^{+}+{\text{NO}}_3^{-}$

Hence,

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

The molality of anion, ${\text{NO}}_3^{-}$ $\left(b_{{\text{NO}}_3^{-}}\right)$ is $\text{b}\text{mol}{\text{kg}}^{\text{-1}}$.

The charge present on cation, ${\text{K}}^{\text{+}}$ $\left(z_{{\text{K}}^{\text{+}}}\right)$ is $+2$.

The charge present on anion, ${\text{NO}}_3^{-}$ $\left(z_{{\text{NO}}_3^{-}}\right)$ is $-1$.

The ionic strength of the solution $\left(I\right)$ has to be $1.0$.

Substitute the values of $b_{{\text{K}}^{\text{+}}}$, $b_{{\text{NO}}_3^{-}}$, $z_{{\text{K}}^{\text{+}}}$, in equation (1)

$\begin{array}{c}{I}_{\text{NaCl}}=\frac{1}{2}\left[\left(b_{{\text{K}}^{\text{+}}}\times z_{{\text{K}}^{\text{+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)+\left(b_{{\text{K}}^{\text{+}}}\times z_{{\text{K}}^{\text{+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)\right]\\ 1.0=\frac{1}{2}\left[\begin{array}{l}\left(0.110\text{mol}{\text{kg}}^{\text{-1}}\times {\left(1\right)}^2\right)+\left(0.110\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)+\\ \left(\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {(1)}^2\right)+\left(\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)\end{array}\right]\\ 2.0=\left[0.220+2\text{b}\right]\\ \text{b}=0.89\text{mol}{\text{kg}}^{\text{-1}}\end{array}$

Thus, the molality of ${\text{KNO}}_{\text{3}}$ is $0.89\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of a solution is given by the formula,

    $\text{Molality}=\frac{\text{Moles}\text{of}{\text{KNO}}_{\text{3}}}{\text{Mass}\text{of}\text{solvent}}$ (2)

The mass of the solvent is $\text{500}\text{g}$.

The conversion of $\text{g}$ to $\text{kg}$ is done as,

    $\text{1000}\text{g}=\text{1}\text{kg}$

Therefore the conversion of $\text{500}\text{g}$ to $\text{kg}$ is done as,

    $\text{500}\text{g=}\text{0}\text{.500}\text{kg}$

Substitute the molality of ${\text{KNO}}_{\text{3}}$ and mass of the solvent in equation (2).

    $\begin{array}{c}\text{0}\text{.89}\text{mol}{\text{kg}}^{-1}\text{=}\frac{\text{Moles}\text{of}{\text{KNO}}_{\text{3}}}{\text{0}\text{.5}\text{kg}}\\ \text{Moles}\text{of}{\text{KNO}}_{\text{3}}=0.445\text{mol}\end{array}$

The number of moles of ${\text{KNO}}_{\text{3}}$ is given by the formula,

     $\text{Moles}\text{of}{\text{KNO}}_{\text{3}}=\frac{\text{Mass}\text{of}{\text{KNO}}_{\text{3}}}{\text{Molar}\text{mass}\text{of}{\text{KNO}}_{\text{3}}}$                                                   (3)

The molar mass of ${\text{KNO}}_{\text{3}}$ is $\text{101}\text{g}{\text{mol}}^{\text{-1}}$.

Substitute the number of moles of ${\text{KNO}}_{\text{3}}$ and molar mass of ${\text{KNO}}_{\text{3}}$ in equation (3).

    $\begin{array}{c}\text{0}\text{.445}\text{mol}=\frac{\text{Mass}\text{of}{\text{KNO}}_{\text{3}}}{\text{101}\text{g}\text{mol}}\\ \text{Mass}\text{of}{\text{KNO}}_{\text{3}}=\underset{\_}{44.94g}\end{array}$

Thus the mass of ${\text{KNO}}_{\text{3}}$ to be added to a $0.110\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ is $\underset{\_}{44.94g}$.

(ii)

Interpretation Introduction

Interpretation: The mass of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ to be added to a $0.110\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $1.0$ has to be calculated..

Concept introduction: The ionic strength is introduced in the Debye-Huckel limiting law which relates the activity coefficient as a function of the ionic strength of the solution.  It is dependent on the concentration of all the ions present in the solution.  It is a dimensionless quantity.

Answer to Problem 5F.6BE

The mass of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ to be added to a $0.110\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $1.0$ is $\underset{\_}{38.67g}$.

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)

Where

• $z_i$ is the charge present on the ion.
• $b_i$ is the molality of the ion in the solution.

The molality of ${\text{KNO}}_3\left(\text{aq}\right)$ is $\text{0}\text{.110}\text{mol}{\text{kg}}^{\text{-1}}$.

The reaction for the dissociation of ${\text{KNO}}_3\left(\text{aq}\right)$ is shown as,

  ${\text{KNO}}_3\to {\text{K}}^{+}+{\text{NO}}_3^{-}$

Hence,

The molality of cation, ${\text{K}}^{\text{+}}$ $\left(b_{{\text{K}}^{\text{+}}}\right)$ is $\text{0}\text{.110}\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of anion, ${\text{NO}}_3^{-}$ $\left(b_{{\text{NO}}_3^{-}}\right)$ is $\text{0}\text{.110}\text{mol}{\text{kg}}^{\text{-1}}$.

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

The charge present on anion, ${\text{NO}}_3^{-}$ $\left(z_{{\text{NO}}_3^{-}}\right)$ is $-1$.

Let the molality of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ to be added be b.

The reaction for the dissociation of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is shown as,

  $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}\to {\text{Ba}}^{2+}+2{\text{NO}}_3^{-}$

Hence,

The molality of cation, ${\text{Ba}}^{\text{2+}}$ $\left(b_{{\text{Ba}}^{\text{2+}}}\right)$ is $\text{b}\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of anion, ${\text{NO}}_3^{-}$ $\left(b_{{\text{NO}}_3^{-}}\right)$ is $\text{2}\times \text{b}\text{mol}{\text{kg}}^{\text{-1}}$.

The charge present on cation, ${\text{Ba}}^{\text{2+}}$ $\left(z_{{\text{Ba}}^{\text{2+}}}\right)$ is $+2$.

The charge present on anion, ${\text{NO}}_3^{-}$ $\left(z_{{\text{NO}}_3^{-}}\right)$ is $-1$.

The ionic strength of the solution $\left(I\right)$ has to be $0.250$.

Substitute the values of $b_{{\text{K}}^{\text{+}}}$, $b_{{\text{NO}}_3^{-}}$, $z_{{\text{K}}^{\text{+}}}$, $z_{{\text{NO}}_3^{-}}$, $b_{{\text{Ba}}^{\text{2+}}}$, $b_{{\text{NO}}_3^{-}}$, $z_{{\text{NO}}_3^{-}}$ and $z_{{\text{Ba}}^{\text{2+}}}$, $z_{{\text{Cl}}^{-}}$ in equation (1)

    $\begin{array}{c}{I}_{\text{NaCl}}=\frac{1}{2}\left[\left(b_{{\text{K}}^{\text{+}}}\times z_{{\text{K}}^{\text{+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)+\left(b_{{\text{Ba}}^{\text{2+}}}\times z_{{\text{Ba}}^{\text{2+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)\right]\\ 1=\frac{1}{2}\left[\begin{array}{l}\left(0.110\text{mol}{\text{kg}}^{\text{-1}}\times {\left(1\right)}^2\right)+\left(0.110\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)+\\ \left(\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {(2)}^2\right)+\left(\text{2}\times \text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)\end{array}\right]\\ 2=\left[\text{0}\text{.220}+\text{6b}\right]\\ \text{b}=0.2967\text{mol}{\text{kg}}^{\text{-1}}\end{array}$

Thus, the molality of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $0.2967\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of a solution is given by the formula,

    $\text{Molality}=\frac{\text{Moles}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{Mass}\text{of}\text{solvent}}$                                                                  (2)

The mass of the solvent is $\text{500}\text{g}$.

The conversion of $\text{g}$ to $\text{kg}$ is done as,

    $\text{1000}\text{g}=\text{1}\text{kg}$

Therefore the conversion of $\text{500}\text{g}$ to $\text{kg}$ is done as,

    $\text{500}\text{g=}\text{0}\text{.500}\text{kg}$

Substitute the molality of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and mass of the solvent in equation (2).

    $\begin{array}{c}\text{0}\text{.2967}\text{mol}{\text{kg}}^{-1}\text{=}\frac{\text{Moles}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{0}\text{.5}\text{kg}}\\ \text{Moles}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=0.148\text{mol}\end{array}$

The number of moles of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is given by the formula,

     $\text{Moles}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\frac{\text{Mass}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{Molar}\text{mass}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$                                     (3)

The molar mass of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ is $261.33\text{g}{\text{mol}}^{\text{-1}}$.

Substitute the number of moles of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ and molar mass of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ in equation (3).

    $\begin{array}{c}\text{0}\text{.148}\text{mol}=\frac{\text{Mass}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{261.33\text{g}\text{mol}}\\ \text{Mass}\text{of}\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\underset{\_}{38.67g}\end{array}$

Thus the mass of $\text{Ba}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}$ to be added to a $0.110\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $1.0$ is $\underset{\_}{38.67g}$.