Chapter 15, Problem 15.24P

(a)

Interpretation Introduction

Interpretation:

Expression for ion selective electrode for ${\text{La}}^{3+}$ has to be written.

Concept Introduction:

Liquid- ion-selective electrode is composed of the hydrophobic ion-selective membrane that is essentially an organic polymer. This thin membrane is filled with a viscous ion exchanger solution. The purpose of this solution is to selectively bond the analyte ion. Selectivity of the bound analyte can be mathematically measured with the use of selectivity coefficient.

Potential difference for ion selective electrode is evaluated by expression as follows:

  $E=\frac{0.05916}{n}\left[\log \left({\left[{\text{C}}^{+}\right]}_{\text{outer 2}}\right)-\log \left({\left[{\text{C}}^{+}\right]}_{\text{outer 1}}\right)\right]$

Here,

${\left[{\text{C}}^{+}\right]}_{\text{outer 2}}$ denotes concentration of analyte solution 2.

${\left[{\text{C}}^{+}\right]}_{\text{inner}}$ denotes concentration of analyte solution 1.

$n$ denotes charge on analyte.

$E$ denotes potential difference.

Explanation of Solution

Expression to compute potential of ion selective electrode is given as follows:

  $E=\text{constant}+\text{β}\frac{\left(0.05916\right)}{n}\log \left[{\text{X}}^{n+}\right]$        (1)

Substitute ${\text{La}}^{3+}$ for ${\text{X}}^{n+}$ and 3 for $n$ in equation (1).

  $E=\text{constant}+\text{β}\frac{\left(0.05916\right)}{3}\log \left[{\text{La}}^{3+}\right]$        (2)

(b)

Interpretation Introduction

Interpretation:

Change in voltage in millivolts when electrode is transferred from $1.00\times {10}^{-4}\text{ M}$ solution to $1.00\times {10}^{-3}\text{ M}$ solution has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression to compute potential difference for ion selective electrode is evaluated by expression as follows:

  $E=\frac{0.05916}{n}\left[\log \left({\left[{\text{C}}^{+}\right]}_{\text{outer 2}}\right)-\log \left({\left[{\text{C}}^{+}\right]}_{\text{outer 1}}\right)\right]$        (3)

Substitute 3 for $n$, $1.00\times {10}^{-3}\text{ M}$ for ${\left[{\text{C}}^{+}\right]}_{\text{outer 2}}$ and $1.00\times {10}^{-4}\text{ M}$ for ${\left[{\text{C}}^{+}\right]}_{\text{outer 1}}$ in equation (3).

  $\begin{array}{c}E=\frac{0.05916}{3}\left[\log \left(1.00\times {10}^{-3}\right)-\log \left(1.00\times {10}^{-4}\right)\right]\\ =\frac{0.05916}{3}\left(1\right)\\ =0.01972\text{ V}\end{array}$

Conversion factor to convert $\text{V}$ to $\text{mV}$ is as follows:

  $1\text{ mV}={\text{10}}^{-3}\text{ V}$

Hence potential of $0.01972\text{ V}$ is converted $\text{mV}$ as follows:

  $\begin{array}{c}\text{Voltage}=\left(0.01972\text{ V}\right)\left(\frac{1\text{ mV}}{{10}^{-3}\text{ V}}\right)\\ =19.7\text{ mV}\end{array}$

Hence voltage change is $19.7\text{ mV}$.

(c)

Interpretation Introduction

Interpretation:

Change in voltage in millivolts when electrode is transferred from $2.36\times {10}^{-4}\text{ M}$ solution to $4.44\times {10}^{-3}\text{ M}$ solution has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Expression to compute potential difference for ion selective electrode is evaluated by expression as follows:

  $E=\frac{0.05916}{n}\left[\log \left({\left[{\text{C}}^{+}\right]}_{\text{outer 2}}\right)-\log \left({\left[{\text{C}}^{+}\right]}_{\text{outer 1}}\right)\right]$        (3)

Substitute 3 for $n$, $4.44\times {10}^{-3}\text{ M}$ for ${\left[{\text{C}}^{+}\right]}_{\text{outer 2}}$ and $2.36\times {10}^{-4}\text{ M}$ for ${\left[{\text{C}}^{+}\right]}_{\text{outer 1}}$ in equation (3).

  $\begin{array}{c}E=\frac{0.05916}{3}\left[\log \left(4.44\times {10}^{-3}\right)-\log \left(2.36\times {10}^{-4}\right)\right]\\ =0.0251\text{ V}\end{array}$

Conversion factor to convert $\text{V}$ to $\text{mV}$ is as follows:

  $1\text{ mV}={\text{10}}^{-3}\text{ V}$

Hence potential of $0.0251\text{ V}$ is converted $\text{mV}$ as follows:

  $\begin{array}{c}\text{Voltage}=\left(0.0251\text{ V}\right)\left(\frac{1\text{ mV}}{{10}^{-3}\text{ V}}\right)\\ =25.1\text{ mV}\end{array}$

Hence voltage when ${\text{Fe}}^{3+}$ is added is $25.1\text{ mV}$.

(d)

Interpretation Introduction

Interpretation:

Voltage when ${\text{Fe}}^{3+}$ is added has to be calculated.

Concept Introduction:

Refer to part (a).

Explanation of Solution

Substitute $1.00\times {10}^{-4}\text{ M}$ for $\left[{\text{La}}^{3+}\right]$ , 1 for $\text{β}$ and $0.1\text{ V}$ for $E$ in equation (1).

  $\begin{array}{c}0.1=\text{constant}+\text{β}\frac{\left(0.05916\right)}{3}\log \left(1\times {10}^{-4}\right)\\ \text{constant}=0.17888\end{array}$

Expression to compute response of ion selective electrode due to ${\text{Fe}}^{3+}$ interference is given as follows:

  $E=\text{constant}+\text{β}\frac{\left(0.05916\right)}{3}\log \left(\left[{\text{La}}^{3+}\right]+k_{{\text{La}}^{3+},{\text{ Fe}}^{3+}}\left[{\text{Fe}}^{3+}\right]\right)$        (4)

Substitute0.17888 for constant, $1.00\times {10}^{-4}\text{ M}$ for $\left[{\text{La}}^{3+}\right]$, $\text{0}\text{.010 M}$ for $\left[{\text{Fe}}^{3+}\right]$,1/1200 for $k_{{\text{La}}^{3+},{\text{ Fe}}^{3+}}$ and 1 for $\text{β}$ in equation (4).

  $\begin{array}{c}E=0.17888+\frac{0.05916}{3}\log \left(1.00\times {10}^{-4}+\left(\frac{1}{1200}\right)\left(0.01\text{ M}\right)\right)\\ =+0.10068\text{ V}\\ \approx 0.1007\text{ V}\end{array}$                                

Hence potential of $0.1007\text{ V}$ is converted $\text{mV}$ as follows:

  $\begin{array}{c}\text{Voltage}=\left(0.1007\text{ V}\right)\left(\frac{1\text{ mV}}{{10}^{-3}\text{ V}}\right)\\ =100.7\text{ mV}\end{array}$

Hence voltage when ${\text{Fe}}^{3+}$ is added is $100.7\text{ mV}$.