Chapter 23, Problem 23.27QAP

Interpretation Introduction

(a)

Interpretation:

The Ca²⁺ concentration in the inner solution of electrode 2 should be determined.

Concept introduction:

Nernst equation gives the cell potential under non-standard conditions.

$\text{E}={\text{E}}^0-\frac{2.303RT}{nF}\log Q$

E − cell potential

E⁰ − standard cell potential

R − universal gas constant

T − temperature in Kelvin

n − number of electrons transferred

F − Faraday constant

Q − Reaction quotient

${\text{E}}_{cell}={\text{E}}_{cathode}^0-{\text{E}}_{anode}^0$

Answer to Problem 23.27QAP

$4.56\times {10}^{-13}\text{ M}$

Explanation of Solution

The given reactions are represented as follows:

${\text{Ca}}^{2+}{\text{ + H}}_2{\text{Y}}^{2-}\to {\text{ CaY}}^{2-}{\text{ + 2 H}}^{+}$

${\text{Ca}}^{2+}{\text{ + Y}}^{4-}\iff {\text{ CaY}}^{2-}{\text{ K}}_f=4.47\times {10}^{10}$

The equilibrium constant can be calculated as follows:

$K_f=\frac{\left[{\text{CaY}}^{2-}\right]}{\left[{\text{Ca}}^{2+}\right]\left[{\text{Y}}^{4-}\right]}$

Here,

$\begin{array}{l}\left[{\text{CaY}}^{2-}\right]=1.00\times {10}^{-3}\text{ M}\\ \left[{\text{Y}}^{4-}\right]=\left(5.0\times {10}^{-2}-1.00\times {10}^{-3}\right)=0.049\text{ M}\end{array}$

If x mol/L of Ca-EDTA complex is dissociated.

$\begin{array}{l}{\text{ Ca}}^{2+}{\text{ + Y}}^{4-}\iff {\text{ CaY}}^{2-}\\ I\text{ 0}\text{.049 M 1}\text{.00}\times {\text{10}}^{-3}\text{ M}\\ \text{C +x +x -x}\\ \text{E +x }\left(0.049+x\right)\left(\text{1}\text{.00}\times {\text{10}}^{-3}-x\right)\end{array}$

$\begin{array}{l}4.47\times {10}^{10}=\frac{1.00\times {10}^{-3}-x}{x\times \left(0.049+x\right)}\\ x<<<<0.049\\ 4.47\times {10}^{10}=\frac{1.00\times {10}^{-3}-x}{0.049x}\\ 2.19\times {10}^{9}x=1.00\times {10}^{-3}-x\\ 2.19\times {10}^{9}x=1.00\times {10}^{-3}\\ x=4.56\times {10}^{-13}\text{ M}\end{array}$

Interpretation Introduction

(b)

Interpretation:

The ionic strength of the solution in Electrode 2 should be determined.

Concept introduction:

The ionic strength can be calculated as follows:

$\mu =\frac{1}{2}\left(c_1{Z}_1^2+c_2{Z}_2^2+c_3{Z}_3^2+\mathrm{....}\right)$

Here,

µ − ionic strength

c − molar concentrations of the ions

Z − charge on the ion.

Answer to Problem 23.27QAP

$\mu =0.4535$

Explanation of Solution

On putting the respective molar concentrations of ions and charge on the ion, the ionic strength can be calculated as follows:

$\begin{array}{l}\mu =\frac{1}{2}\left(\left[{\text{Ca}}^{2+}\right]{\left(2\right)}^2+\left[{\text{Cl}}^{-}\right]{\left(1\right)}^2+\left[{\text{Y}}^{4-}\right]{\left(4\right)}^2+\left[{\text{Na}}^{+}\right]{\left(1\right)}^2+\left[{\text{OH}}^{-}\right]{\left(1\right)}^2+\left[{\text{CaY}}^{2-}\right]{\left(1\right)}^2\right)\\ \mu =\frac{1}{2}\left(4.56\times {10}^{-13}{\left(2\right)}^2+2\times 1\times {10}^{-3}{\left(1\right)}^2+0.049{\left(4\right)}^2+6.0\times {10}^{-2}{\left(1\right)}^2+6.0\times {10}^{-2}{\left(1\right)}^2+1\times {10}^{-3}{\left(1\right)}^2\right)\\ \mu =\frac{1}{2}\left(1.824\times {10}^{-12}+2\times {10}^{-3}+0.784+6.0\times {10}^{-2}+6.0\times {10}^{-2}+1\times {10}^{-3}\right)\\ \mu =\frac{1}{2}\times 0.907\\ \mu =0.4535\end{array}$

Interpretation Introduction

(c)

Interpretation:

The activity of Ca²⁺ in electrode 2 should be determined.

Concept introduction:

The activity of coefficient of species A can be calculated as follows:

$-\log {\gamma }_A=\frac{0.509Z_A^2\sqrt{\mu }}{1+3.28{\alpha }_A\sqrt{\mu }}$

Here,

$\begin{array}{l}{\gamma }_A=\text{ activity coefficient of species A}\\ {\text{Z}}_A=\text{Charge on the species A}\\ \mu \text{ = ionic strength of the solution}\\ {\alpha }_A=\text{ the effective diameter of the hydrated ion in nanometers}\end{array}$

Answer to Problem 23.27QAP

$a_{C{a}^{2+}}=1.173\times {10}^{-13}\text{ M}$

Explanation of Solution

The activity of coefficient of species A can be calculated as follows:

$\begin{array}{l}-\log {\gamma }_{C{a}^{2+}}=\frac{0.509{\left(2\right)}^2\sqrt{0.4535}}{1+3.28\times 0.6\sqrt{0.4535}}\\ -\log {\gamma }_{C{a}^{2+}}=0.5896\\ {\gamma }_{C{a}^{2+}}=0.2572\end{array}$

Also,

$\begin{array}{l}{a}_{C{a}^{2+}}={\gamma }_{C{a}^{2+}}\left[{\text{Ca}}^{2+}\right]\\ a_{C{a}^{2+}}=0.2572\times 4.56\times {10}^{-13}\text{ M}\\ a_{C{a}^{2+}}=1.173\times {10}^{-13}\text{ M}\end{array}$

Interpretation Introduction

(d)

Interpretation:

The cell potential versus the pCa should be plotted and pCa value where the plot deviated more than 5% from linearity should be determined. For the linear portion, the slope and the intercept should be determined.

Concept introduction:

For cations,

$E_{cell}=K-\frac{0.0592}{n}pX$

Here,

K − constant

n − number of moles of electrons

X − activity of the cation.

Answer to Problem 23.27QAP

At $\text{pCa = 9}$ value, the plot deviates more than 5 % from linearity.

Slope = -0.0298

Intercept =0.185733

The plot obeys the equation 23-29.

Explanation of Solution

The data given is as follows:

Activity of Ca2+, M	pCa	Cell potential, V
1.00E-03	3	9.30E-02
1.00E-04	4	7.30E-02
1.00E-05	5	3.70E-02
1.00E-06	6	2.00E-03
1.00E-07	7	-2.30E-02
1.00E-08	8	-5.10E-02
1.00E-09	9	5.50E-02

The plot is represented as follows:

At $\text{pCa = 9}$ value, the plot deviates more than 5 % from linearity.

Let’s draw the graph eliminating this value.

According to the linear regression analysis done in a spread sheet,

Slope = -0.0298

Intercept =0.185733

According to the equation 23.29,

$\begin{array}{l}{E}_{cell}=K-\frac{0.0592}{2}p\text{Ca}\\ E_{cell}=-0.0296p\text{Ca}+K\end{array}$

Therefore, the plot obeys the equation 23-29.

Interpretation Introduction

(e)

Interpretation:

The cell potential versus the pCa should be plotted for electrode 2 and the range of linearity should be determined. For the linear portion, the slope and the intercept should be determined.

Concept introduction:

For cations,

$E_{cell}=K-\frac{0.0592}{n}pX$

Here,

K − constant

n − number of moles of electrons

X − activity of the cation.

Answer to Problem 23.27QAP

Range of linearity is from pCa 3 to 6

Slope = -0.0292

Intercept = 0.3119

This electrode obeys the equation 23-29 for higher Ca²⁺ activities

Explanation of Solution

The data given is as follows:

Activity of Ca2+, M	pCa	Cell potential, V
1.00E-03	3	2.28E-01
1.00E-04	4	1.90E-01
1.00E-05	5	1.65E-01
1.00E-06	6	1.39E-01
5.60E-07	6.25181197	1.05E-01
3.20E-07	6.49485002	6.30E-02
1.80E-07	6.74472749	3.60E-02
1.00E-07	7	2.30E-02
1.00E-08	8	1.80E-02
1.00E-09	9	1.70E-02

The graph is plotted as follows:

Range of linearity is from pCa 3 to 6

Let’s draw the graph eliminating non-linear values.

According to the linear regression analysis done in a spread sheet,

Slope = -0.0292

Intercept = 0.3119

$\begin{array}{l}{E}_{cell}=K-\frac{0.0592}{2}p\text{Ca}\\ E_{cell}=-0.0296p\text{Ca}+K\end{array}$

This electrode obeys the equation 23-29 for higher Ca²⁺ activities.

Interpretation Introduction

(f)

Interpretation:

The reason for the term super-Nernstianto be used for electrode 2 should be discussed.

Concept introduction:

Electrode 2 has an inner solution with low activities of Ca²⁺.

Answer to Problem 23.27QAP

A super-Nernstian response is a result of an inward flux of ions which are to be analyzed.

Explanation of Solution

As a result of a super-Nernstian response, an inward flux of ions, Ca²⁺ in this case needed to be analyzed into the ion-selective membrane. In the diffusion layer, it decreases Ca²⁺ ions at the sample-membrane phase boundary. When there is low Ca²⁺ ions concentration, this super-Nernstian behavior usually takes place. It is usually experienced at concentrations at or below 1 × 10-6 M.

Interpretation Introduction

(g)

Interpretation:

The cell potential versus the pCa should be plotted and the range of linearity should be determined. For the linear portion, the slope and the intercept should be determined.

Concept introduction:

For cations,

$E_{cell}=K-\frac{0.0592}{n}pX$

Here,

K − constant

n − number of moles of electrons

X − activity of the cation.

Answer to Problem 23.27QAP

Range of linearity is between pCa 3 to 6

Slope = -0.0288

Intercept = 0.2636

The plot obeys the equation 23-29 at higher concentrations.

Explanation of Solution

Activity of Ca2+, M	pCa	Cell potential, V
1.00E-03	3	1.75E-01
1.00E-04	4	1.50E-01
1.00E-05	5	1.23E-01
1.00E-06	6	8.80E-02
1.00E-07	7	7.50E-02
1.00E-08	8	7.20E-02
1.00E-09	9	7.10E-02

Range of linearity is between pCa 3 to 6.

Let’s draw the graph eliminating non-linear values.

According to the linear regression analysis done in a spread sheet,

Slope = -0.0288

Intercept = 0.2636

According to the equation 23.29,

$\begin{array}{l}{E}_{cell}=K-\frac{0.0592}{2}p\text{Ca}\\ E_{cell}=-0.0296p\text{Ca}+K\end{array}$

Therefore, the plot obeys the equation 23-29 at higher concentrations.

Interpretation Introduction

(h)

Interpretation:

Electrode 3 is said to have Ca²⁺ release. This term should be explained.

Concept introduction:

Electrode 3 has an inner solution with higher activities of Ca²⁺.

Answer to Problem 23.27QAP

When the concentration of Ca²⁺ in the solution is too low than that of inner solution of the electrode, Ca²⁺ flux outward to the solution through diffusion.

Explanation of Solution

Ca²⁺ is a result of an outward flux of ions which are to be analyzed, in this case Ca²⁺ ions out of the ion-selective membrane. This results in increase of Ca²⁺ ions in the diffusion layer at the sample-membrane phase boundary. This Ca²⁺ release usually happens when there is a low concentration of Ca²⁺ ions in the solution to be analyzed.

Interpretation Introduction

(i)

Interpretation:

The alternative explanations for experimental results given in the article should be described.

Concept introduction:

An electrochemical cell, which is used to make potentiometric measurements with a membrane electrode are known as ion selective electrodes. ISE is highly selective for a specific ion. This selectivity depends on the composition of the membrane. Main component of an ISE is inner reference solution, and outer analyte. The potential is developed at the membrane either by ion exchange process or ion transport process.

Answer to Problem 23.27QAP

The reason for super-Nernstian response is that at low sample activities, the gradient required to maintain the membrane flux is no more negligible relative to the bulk. Ca²⁺ uptake is a result of the strong coextraction of Ca(NO₃)₂ into the membrane.

Explanation of Solution

The emf observed with ISE are a logarithmic function of the bulk sample activities. Deviation from this behavior can be explained either by activity change in the organic membrane surface layer or by differences between sample activities in the bulk and the membrane surface layer. Deviation are a result of ions release from or uptake to the membrane. These ion fluxes originate from concentration polarization within the membrane, asymmetrical ion exchange and coextraction reactions at both membrane sides.

The reason for super-Nernstian response is that at low sample activities, the gradient required to maintain the membrane flux is no more negligible relative to the bulk. Ca²⁺ uptake is a result of the strong coextraction of Ca(NO₃)₂ into the membrane.