Chapter 5, Problem 5F.6AE

(i)

Interpretation Introduction

Interpretation: The mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ added to a $0.150\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ 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 depends upon the concentration of all the ions present in the solution.  It is a dimensionless quantity.

Answer to Problem 5F.6AE

The mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ added to a $0.150\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ is $\underset{\_}{2.75g}$.

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 $\text{0}\text{.150}\text{mol}{\text{kg}}^{\text{-1}}$.

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

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

The molality of cation, ${\text{K}}^{\text{+}}$ $\left(b_{{\text{K}}^{\text{+}}}\right)$ is $\text{0}\text{.150}\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of anion, ${\text{NO}}_3^{-}$ $\left(b_{{\text{NO}}_3^{-}}\right)$ is $\text{0}\text{.150}\text{mol}{\text{kg}}^{\text{-1}}$.

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 $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ to be added be b.

The reaction for the dissociation of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)$ is shown as,

  $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\to {\text{Ca}}^{2+}+2{\text{NO}}_{\text{3}}^{-}$

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

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

The charge present on cation, ${\text{Ca}}^{\text{2+}}$ $\left(z_{{\text{Ca}}^{\text{2+}}}\right)$ 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 value of $b_{{\text{K}}^{\text{+}}}$, $b_{{\text{NO}}_3^{-}}$, $z_{{\text{K}}^{\text{+}}}$, $b_{{\text{Ca}}^{\text{2+}}}$ and $z_{{\text{Ca}}^{\text{2+}}}$ in equation (1)

    $\begin{array}{c}{I}_{\text{NaCl}}=\frac{1}{2}\left[\left(b_{{\text{K}}^{\text{+}}}\times z_{{\text{K}}^{\text{+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)+\left(b_{{\text{Ca}}^{\text{2+}}}\times z_{{\text{Ca}}^{\text{2+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)\right]\\ 0.250=\frac{1}{2}\left[\begin{array}{l}\left(0.150\text{mol}{\text{kg}}^{\text{-1}}\times {\left(1\right)}^2\right)+\left(0.150\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)+\\ \left(\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {(2)}^2\right)+\left(2\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)\end{array}\right]\\ 0.500=\left[0.3+6\text{b}\right]\\ \text{b}=0.033\text{mol}{\text{kg}}^{\text{-1}}\end{array}$

Thus, the molality of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)$ is $0.033\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of a solution is given by the formula,

    $\text{Molality}=\frac{\text{Moles}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{Mass}\text{of}\text{solvent}}$                                                                  (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 $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)$ and mass of the solvent in equation (2).

    $\begin{array}{c}\text{0}\text{.033}\text{mol}{\text{kg}}^{-1}\text{=}\frac{\text{Moles}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{0}\text{.5}\text{kg}}\\ \text{Moles}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=0.0167\text{mol}\end{array}$

The number of moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2\left(\text{aq}\right)$ is given by the formula,

     $\text{Moles}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2=\frac{\text{Mass}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{Molar}\text{mass}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}$                                     (3)

The molar mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ is $\text{164}\text{.78}\text{g}{\text{mol}}^{\text{-1}}$.

Substitute the number of moles of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ and molar mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ in equation (3).

    $\begin{array}{c}\text{0}\text{.0167}\text{mol}=\frac{\text{Mass}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}}{\text{164}\text{.78}\text{g}\text{mol}}\\ \text{Mass}\text{of}\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_{\text{2}}=\underset{\_}{2.75g}\end{array}$

Thus the mass of $\text{Ca}{\left({\text{NO}}_{\text{3}}\right)}_2$ to be added to a $0.150\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ is $\underset{\_}{2.75g}$.

(ii)

Interpretation Introduction

Interpretation: The mass of $\text{NaCl}$ to be added to a $0.150\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ 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.6AE

The mass of $\text{NaCl}$ to be added to a $0.150\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ is $\underset{\_}{2.92g}$.

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 $\text{0}\text{.150}\text{mol}{\text{kg}}^{\text{-1}}$.

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 $\text{0}\text{.150}\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of anion, ${\text{NO}}_3^{-}$ $\left(b_{{\text{NO}}_3^{-}}\right)$ is $\text{0}\text{.150}\text{mol}{\text{kg}}^{\text{-1}}$.

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 $\text{NaCl}$ to be added be b.

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

  $\text{NaCl}\to {\text{Na}}^{2+}+{\text{Cl}}^{-}$

Hence,

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

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

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

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

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

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

    $\begin{array}{c}{I}_{\text{NaCl}}=\frac{1}{2}\left[\left(b_{{\text{K}}^{\text{+}}}\times z_{{\text{K}}^{\text{+}}}^2\right)+\left(b_{{\text{NO}}_{\text{3}}^{\text{-}}}\times z_{{\text{NO}}_{\text{3}}^{\text{-}}}^2\right)+\left(b_{{\text{Na}}^{\text{+}}}\times z_{{\text{Na}}^{\text{+}}}^2\right)+\left(b_{{\text{Cl}}^{\text{-}}}\times z_{{\text{Cl}}^{\text{-}}}^2\right)\right]\\ 0.250=\frac{1}{2}\left[\begin{array}{l}\left(0.150\text{mol}{\text{kg}}^{\text{-1}}\times {\left(1\right)}^2\right)+\left(0.150\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)+\\ \left(\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {(1)}^2\right)+\left(\text{b}\text{mol}{\text{kg}}^{\text{-1}}\times {\left(-1\right)}^2\right)\end{array}\right]\\ 0.500=\left[\text{0}\text{.3}+\text{2b}\right]\\ \text{b}=0.1\text{mol}{\text{kg}}^{\text{-1}}\end{array}$

Thus, the molality of $\text{NaCl}$ is $0.1\text{mol}{\text{kg}}^{\text{-1}}$.

The molality of a solution is given by the formula,

    $\text{Molality}=\frac{\text{Moles}\text{of}\text{NaCl}}{\text{Mass}\text{of}\text{solvent}}$                                                                          (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 $\text{NaCl}$ and mass of the solvent in equation (2).

    $\begin{array}{c}\text{0}\text{.1}\text{mol}{\text{kg}}^{-1}\text{=}\frac{\text{Moles}\text{of}\text{NaCl}}{\text{0}\text{.5}\text{kg}}\\ \text{Moles}\text{of}\text{NaCl}=0.05\text{mol}\end{array}$

The number of moles of $\text{NaCl}$ is given by the formula,

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

The molar mass of $\text{NaCl}$ is $58.5\text{g}{\text{mol}}^{\text{-1}}$.

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

    $\begin{array}{c}\text{0}\text{.05}\text{mol}=\frac{\text{Mass}\text{of}\text{NaCl}}{58.5\text{g}\text{mol}}\\ \text{Mass}\text{of}\text{NaCl}=\underset{\_}{2.92g}\end{array}$

Thus the mass of $\text{NaCl}$ to be added to a $0.150\text{mol}{\text{kg}}^{\text{-1}}$ solution of ${\text{KNO}}_{\text{3}}\text{(aq)}$ to increase its ionic strength to $0.250$ is $\underset{\_}{2.92g}$.