Chapter 8, Problem 8.AE

(a)

Interpretation Introduction

Interpretation:

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{KNO}}_{\text{3}}$ has to be calculated.

Concept introduction:

Ionic strength $\left(\mu \right)$:

The ionic strength $\left(\text{μ}\right)$ of the solution is given by

$\begin{array}{l}\text{μ}\text{=}\frac{\text{1}}{\text{2}}\left({\text{c}}_{\text{1}}{\text{z}}_{\text{1}}^{\text{2}}\text{+}{\text{c}}_{\text{2}}{\text{z}}_{\text{2}}^{\text{2}}\text{+}\mathrm{.......}\right)\text{=}\frac{\text{1}}{\text{2}}{\displaystyle \sum _{\text{i}}{\text{c}}_{\text{i}}{\text{z}}_{\text{i}}{}^{\text{2}}}\\ \text{where,}\\ {\text{c}}_{\text{i}}\text{-}\text{concentration}\text{of}\text{the}{\text{i}}^{\text{th}}\text{species}\text{and}{\text{z}}_{\text{i}}\text{-}\text{is}\text{its}\text{charge}\text{.}\end{array}$

Answer to Problem 8.AE

• The ionic strength of $\text{0}\text{.2}\text{mM}{\text{KNO}}_{\text{3}}$ is $\text{0}\text{.2}\text{mM}\text{.}$

Explanation of Solution

Given information,

$\text{0}\text{.2}\text{mM}\text{solution}\text{of}{\text{KNO}}_{\text{3}}$

Calculate the ionic strength by using the ionic strength calculation formula

$\begin{array}{l}\text{μ}\text{=}\frac{\text{1}}{\text{2}}\left[\left[{\text{K}}^{\text{+}}\right]{\left(\text{+1}\right)}^{\text{2}}\text{+}\left[{\text{NO}}_{\text{3}}{}^{\text{-}}\right]{\left(\text{-1}\right)}^{\text{2}}\right]\\ \text{=}\left(\frac{\text{1}}{\text{2}}\left[\left(\text{0}\text{.2}\right){\left(\text{+1}\right)}^{\text{2}}\text{+}\left[\text{0}\text{.2}\right]{\left(\text{-1}\right)}^{\text{2}}\right]\right)\text{mM}\\ \text{=}\left(\frac{\text{1}}{\text{2}}\left[\text{0}\text{.4}\right]\right)\text{mM}\text{=}\text{0}\text{.2}\text{mM}\text{.}\end{array}$

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{KNO}}_{\text{3}}$ is $\text{0}\text{.2}\text{mM}\text{.}$

Conclusion

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{KNO}}_{\text{3}}$ was calculated by using the concentration and charge of the ions.

(b)

Interpretation Introduction

Interpretation:

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{Cs}}_{\text{2}}{\text{CrO}}_{\text{4}}$ has to be calculated.

Concept introduction:

Ionic strength $\left(\mu \right)$:

The ionic strength $\left(\text{μ}\right)$ of the solution is given by

$\begin{array}{l}\text{μ}\text{=}\frac{\text{1}}{\text{2}}\left({\text{c}}_{\text{1}}{\text{z}}_{\text{1}}^{\text{2}}\text{+}{\text{c}}_{\text{2}}{\text{z}}_{\text{2}}^{\text{2}}\text{+}\mathrm{.......}\right)\text{=}\frac{\text{1}}{\text{2}}{\displaystyle \sum _{\text{i}}{\text{c}}_{\text{i}}{\text{z}}_{\text{i}}{}^{\text{2}}}\\ \text{where,}\\ {\text{c}}_{\text{i}}\text{-}\text{concentration}\text{of}\text{the}{\text{i}}^{\text{th}}\text{species}\text{and}{\text{z}}_{\text{i}}\text{-}\text{is}\text{its}\text{charge}\text{.}\end{array}$

Answer to Problem 8.AE

• The ionic strength of $\text{0}\text{.2}\text{mM}{\text{Cs}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\text{0}\text{.6}\text{mM}\text{.}$

Explanation of Solution

Given information,

$\text{0}\text{.2}\text{mM}\text{solution}\text{of}{\text{Cs}}_{\text{2}}{\text{CrO}}_{\text{4}}$

Calculate the ionic strength by using the ionic strength calculation formula

$\begin{array}{l}\text{μ}\text{=}\frac{\text{1}}{\text{2}}\left[\text{2}\left[{\text{Cs}}^{\text{+}}\right]{\left(\text{+1}\right)}^{\text{2}}\text{+}\left[{\text{CrO}}_{\text{4}}{}^{\text{2-}}\right]{\left(\text{-2}\right)}^{\text{2}}\right]\\ \text{=}\left(\frac{\text{1}}{\text{2}}\left[\left(\text{0}\text{.4}\right){\left(\text{+1}\right)}^{\text{2}}\text{+}\left[\text{0}\text{.2}\right]{\left(\text{-2}\right)}^{\text{2}}\right]\right)\text{mM}\\ \text{=}\left(\frac{\text{1}}{\text{2}}\left[\text{1}\text{.2}\right]\right)\text{mM}\text{=}\text{0}\text{.6}\text{mM}\text{.}\end{array}$

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{Cs}}_{\text{2}}{\text{CrO}}_{\text{4}}$ is $\text{0}\text{.6}\text{mM}\text{.}$

Conclusion

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{Cs}}_{\text{2}}{\text{CrO}}_{\text{4}}$ was calculated by using the concentration and charge of the ions.

(c)

Interpretation Introduction

Interpretation:

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{MgCl}}_{\text{2}}\text{plus}\text{0}\text{.3}\text{mM}{\text{AlCl}}_{\text{3}}$ has to be calculated.

Concept introduction:

Ionic strength $\left(\mu \right)$:

The ionic strength $\left(\text{μ}\right)$ of the solution is given by

$\begin{array}{l}\text{μ}\text{=}\frac{\text{1}}{\text{2}}\left({\text{c}}_{\text{1}}{\text{z}}_{\text{1}}^{\text{2}}\text{+}{\text{c}}_{\text{2}}{\text{z}}_{\text{2}}^{\text{2}}\text{+}\mathrm{.......}\right)\text{=}\frac{\text{1}}{\text{2}}{\displaystyle \sum _{\text{i}}{\text{c}}_{\text{i}}{\text{z}}_{\text{i}}{}^{\text{2}}}\\ \text{where,}\\ {\text{c}}_{\text{i}}\text{-}\text{concentration}\text{of}\text{the}{\text{i}}^{\text{th}}\text{species}\text{and}{\text{z}}_{\text{i}}\text{-}\text{is}\text{its}\text{charge}\text{.}\end{array}$

Answer to Problem 8.AE

• The ionic strength of $\text{0}\text{.2}\text{mM}{\text{MgCl}}_{\text{2}}\text{plus}\text{0}\text{.3}\text{mM}{\text{AlCl}}_{\text{3}}$ is $2.4\text{mM}\text{.}$

Explanation of Solution

Given information,

$\text{0}\text{.2}\text{mM}{\text{MgCl}}_{\text{2}}\text{plus}\text{0}\text{.3}\text{mM}{\text{AlCl}}_{\text{3}}$

Calculate the ionic strength by using the ionic strength calculation formula

$\begin{array}{l}\text{μ}\text{=}\frac{\text{1}}{\text{2}}\left[\left[{\text{Mg}}^{\text{2+}}\right]{\left(\text{+2}\right)}^{\text{2}}\text{+}\text{2}\left[{\text{Cl}}^{\text{-}}\right]{\left(\text{-1}\right)}^{\text{2}}\text{+}\left[{\text{Al}}^{\text{3+}}\right]{\left(\text{3}\right)}^{\text{2}}\text{+}\text{3}\left[{\text{Cl}}^{\text{-}}\right]{\left(\text{-1}\right)}^{\text{2}}\right]\\ \text{=}\left(\frac{\text{1}}{\text{2}}\left[\left[\text{0}\text{.2}\right]{\left(\text{+2}\right)}^{\text{2}}\text{+}\left[\text{0}\text{.4}\right]{\left(\text{-1}\right)}^{\text{2}}\text{+}\left[\text{0}\text{.3}\right]{\left(\text{3}\right)}^{\text{2}}\text{+}\left[\text{0}\text{.9}\right]{\left(\text{-1}\right)}^{\text{2}}\right]\right)\text{mM}\\ \text{=}\left(\frac{\text{1}}{\text{2}}\left[\text{0}\text{.8}\text{+}\text{0}\text{.4}\text{+}\text{2}\text{.7}\text{+}\text{0}\text{.9}\right]\right)\text{mM}\text{=}\text{2}\text{.4}\text{mM}\text{.}\end{array}$

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{MgCl}}_{\text{2}}\text{plus}\text{0}\text{.3}\text{mM}{\text{AlCl}}_{\text{3}}$ is $2.4\text{mM}\text{.}$

Conclusion

The ionic strength of $\text{0}\text{.2}\text{mM}{\text{MgCl}}_{\text{2}}\text{plus}\text{0}\text{.3}\text{mM}{\text{AlCl}}_{\text{3}}$ was calculated by using the concentration and charge of the ions.