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\begin{document}
\title{Aquatic Chemistry Exercises 23-34}
\author{Hugo Wahlström\and Tobias Ellingsen}%
\vspace{-1em}
\date{}
\begingroup
\let\center\flushleft
\let\endcenter\endflushleft
\maketitle
\endgroup
\sloppy
\subsection*{Exercise 23}
{\label{983568}}
The Saturation Index (\(SI\)) is a ratio that shows the
relationship between the activities of a solute and the dissolution
equilibrium constant K for that solute. When the product of the
activities of the dissolved species of the solute matches that of the
equilibrium constant~\(SI=1\) there is thermodynamic
equilibrium between solute and the solvent. If the~\(SI\) is
bigger than 1, the solution is oversaturated with the dissolved mineral,
if it is smaller than 1, the solution is undersaturated.
~
In this exercise, we calculated the~\(SI\) for the
dissolution of limestone in an aquatic system. We calculated it twice,
once taking activity factors in account and once treating the solution
as ideal (\(\gamma=1\)). We then compared and discussed the
results of the two different calculations.
~
For calculations of the~\(SI\), treating the solution as
ideal, we started by creating two row-vectors (one for each location) in
Matlab with the concentrations given in the table. The concentrations
were converted into~\(\frac{mol}{L}\)~by dividing with the species
respective molar masses. We then made use of equation 8.4 found in the
compendium, putting in the measured value for~\([Ca^{2+}]\),
and~\([CO_3^{2-}]\) rewritten in terms
of~\([HCO_3^-]\)and~\(pH\) according to eq 8.3
as~such:
\begin{equation}
SI=\frac{\left[Ca^{2+}\right]\cdot K_{HCO_3}\cdot\frac{\left[HCO_3^-\right]}{10^{-pH}}}{K_{CaCO_3}}
\end{equation}
Next, for calculations including activity factors, the ionic
strength~\(I\) of the waters in the two locations was
calculated in Matlab. We used the model for calculating ionic strength
that is given in Appendix 1, inserting row-vectors containing the
concentrations of species in~\(\frac{mol}{L}\) and their respective
charge.~
~
Activity factors for~\([HCO_3^-]\)and~\([Ca^{2+}]\)~were then
calculated with the extended Deby\selectlanguage{ngerman}é-Huckel equation with~values of the
hydrated ionic radius for ~\([HCO_3^-]\) and~\([Ca^{2+}]\)
gathered from Appendix 1. The activity factors~ were then inserted into
eq. 1 ~and multiplied with the concentrations of the two species to get
the value of the~\(SI\).~
~
The results for the two calculations of SI, along with activities, and
activity factors can be seen in table {\ref{729550}}.~
% Please add the following required packages to your document preamble:
% \usepackage{multirow}\selectlanguage{english}
\begin{table}[!htbp]
\caption{{Activity, activity factors and SI values in the Pennsylvania and Florida waters}}
\label{SI}
\begin{tabular}{c|c|c|c|c|c|}
\multicolumn{2}{l|}{\textbf{}} & \selectlanguage{greek}\textbf{Activity Factor (γ)}\selectlanguage{english} & \textbf{Activity} & \selectlanguage{greek}\textbf{SI (γ=1)}\selectlanguage{english} & \textbf{SI} \\ \hline
\multirow{3}{*}{\textbf{Pennsylvania}} & Ca(2+) & 0,685 & 0,0014 & \multirow{3}{*}{1,885} & \multirow{3}{*}{1,427} \\ \cline{2-4}
& HCO3(-) & 1,105 & 0,0051 & & \\ \cline{2-4}
& CO3(2-) & N/A & 3,91*10\textasciicircum{}-6 & & \\ \hline
\multirow{3}{*}{\textbf{Florida}} & Ca(2+) & 0,7832 & 0,007 & \multirow{3}{*}{2,538} & \multirow{3}{*}{2,118} \\ \cline{2-4}
& HCO3(-) & 1,0653 & 0,0022 & & \\ \cline{2-4}
& CO3(2-) & N/A & 1.06*10\textasciicircum{}-5 & & \\ \cline{2-6}
\end{tabular}
\end{table}
~
The data suggests that both groundwaters are oversaturated with respect
to limestone. Calculations of the saturation index including activity
factors is considered to give more realistic values because of
intermolecular forces being taken into account. The calculated activity
factors suggest that the intermolecular forces are higher in the ideal
case (\(\ \gamma=1\)).
\par\null\par\null
\textbf{File: e23.m}
\begin{verbatim}
% ion concentration in mg / L
% Ca(2+) Mg(3+) Na(+) K(+) HCO3(-) SO4(2-) Cl(-) NO3(-) pH
florMg = [34 5.6 3.2 0.5 124 2.4 4.5 0.1 8.00]; % mg/L
pennMg = [83 17 8.5 6.3 279 27 17 38 7.36]; % mg/L
%% 2
z = [2 2 1 1 -1 -2 -1 -1 1];
M = [40.078 24.305 22.900 39.098 61.0173 96.0643 35.45 62.005 1]; % molar mass
florM = (florMg * 10^-3) ./ M; % mol / L
florM(9) = 10^-(8);
pennM = (pennMg * 10^-3) ./ M; % mol / L
pennM(9) = 10^-(7.36);
iFlor = 0.5 * sum(florM .* (z.^2));
iPenn = 0.5 * sum(pennM .* (z.^2));
AfFlorCa = 10^((-0.51 * 2^2 * sqrt(iFlor)) / (1 + 0.33*6*sqrt(iFlor)));
AfFlorHCO3 = 10^((-0.51 * -1^2 * sqrt(iFlor))/(1+0.33*4*sqrt(iFlor)));
AfPennCa = 10^((-0.51 * 2^2 * sqrt(iPenn))/(1+0.33*6*sqrt(iPenn)));
AfPennHCO3 = 10^((-0.51 * -1^2 * sqrt(iPenn))/(1+0.33*4*sqrt(iPenn)));
kHCO325 = 10^(-10.312);
actFlorCa = florM(1) * AfFlorCa;
actFlorCO3 = kHCO325 * ((florM(5)* AfFlorHCO3) / 10^-(florMg(9)));
actFlorHCO3 = florM(5)* AfFlorHCO3
kHCO310 = 10^(-10.471);
actPennCa = pennM(1) * AfPennCa;
actPennCO3 = kHCO310 * ((pennM(5) * AfPennHCO3) / 10^-(pennMg(9)));
actPennHCO3 = pennM(5) * AfPennHCO3
kCaco325 = 10^(-8.48);
kCaco310 = 10^(-8.41);
SIflor2 = (actFlorCa*actFlorCO3) / kCaco325;
SIpenn2 = (actPennCa*actPennCO3) / kCaco310;
%% 1
SIpenn1 = (pennM(1) * kHCO310 * (pennM(5) / 10^-pennMg(9))) / kCaco310
SIflor1 = (florM(1) * kHCO325 * (florM(5) / 10^-florMg(9))) / kCaco325
%% 3
% horizontal order mg/L, mol/L, AF, activity, SIconc, SIactivity
table = [
pennMg(1) pennM(1) AfPennCa actPennCa SIpenn1 SIpenn2;
pennMg(5) pennM(5) AfPennHCO3 actPennHCO3 0 0;
0 0 0 actPennCO3 0 0;
florMg(1) florM(1) AfFlorCa actFlorCa SIflor1 SIflor2;
florMg(5) florM(5) AfFlorHCO3 actFlorHCO3 0 0 0 0 0 actFlorCO3 0 0;]
csvwrite('informativetable.csv',table)
\end{verbatim}
\pagebreak
\subsection*{Exercise 24}
{\label{791063}}
In this exercise, the relationship between increasing partial pressure
of \(CO_2\) and concentration of dissolved~\(CaCO_3\)
in a pure water system was investigated.
Through the dissolution equilibrium equation for~\(CaCO_3\) we
can find an expression for the concentration of~\(Ca^{2+}\) as
such
\begin{equation}
\left[Ca^{2+}\right]=\frac{K_{CaCO_3}\cdot\left[H^+\right]^2}{K_{HCO_3^-}\cdot K_{H_2CO_3}\cdot K_H\cdot P_{CO_2}}
\end{equation}
As the concentration of~\([H^+]\) is dependent on the partial
pressure~\(P_{CO_2}\) as well, a value for this is required for
input in the above equation. The algebraic equation necessary is a 4th
order algebraic equation with respect to ~\([H^+]\) shown below
\begin{equation}
\label{H+calc}
0=\left[H^+\right]^3\ +\ \frac{2K_{CaCO_3}\left[H^+\right]^4}{K_{H_2CO_3}K_{HCO_3}K_HP_{CO_2}}-\left(K_W+K_{H_2CO_3}K_HP_{CO_2}\right)\left[H^+\right]-2K_{H_2CO_3}K_{HCO_3K_H}P_{CO_2}
\end{equation}
The algebraic equation for~\([H^+]\)~was solved through the
roots-function in Matlab. The value for~\([H^+]\)~was then
inserted into the dissolution equilibrium reaction and the concentration
of ~\(Ca^{2+}\) as a function of the partial pressure
of~\(CO_2\) can then be calculated. The results are plotted as
shown in fig. {\ref{729550}}.~\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/e24plot/e24plot}
\caption{{Concentration of~\(Ca^{2+}\)~as a funciton
of~\(P_{CO_2}\)at saturation.
{\label{729550}}%
}}
\end{center}
\end{figure}
Because of the higher partial pressure of~\(CO_2\) , the
concentration of free carbonic acid in the solution will increase. This
shifts the dissolution equilibrium of~\(CaCO_3\) to more
products as more carbonic acid is consumed, creating~\(HCO_3^-\)
and .\(CO_3^{2-}\)
\textbf{File: e24.m}
\begin{verbatim}
fplot(@caconc, [10^-4 10^-1])
xlabel('P_{CO_2} / atm')
ylabel('[Ca^{2+}] / mole \cdot L^{-1}')
\end{verbatim}
\textbf{File: caconc.m}
\begin{verbatim}
function ca=caconc(pco2)
if pco2 == 0
pco2 = 10^-4;
end
kh2co3 = 10^-(6.482);
kh = 10^-(1.260);
khco3 = 10^(-10.471);
kcaco3 = 10^(-8.41);
h = hconc(pco2);
ca = (kcaco3 * h^2) / (khco3*kh2co3*kh*pco2)
end
\end{verbatim}
\pagebreak
\subsection*{Exercise 25~}
{\label{711741}}
In this exercise we look at how the solubility
of~\(CaCO_3\)~changes when
either~\(NaHCO_3\)~and~\(H_2SO_4\) are added to the
solution.
Adding~\(NaHCO_3\) and~\(H_2SO_4\)~will increase the
concentration of~\(Na^+\) and~\(SO_4^{2-}\)in the
solution,~\(Na^+\) being a cation to a weak weak base
and~\(SO_4^{2-}\) being a anion to a strong acid. The ANC-value for
the solution in each case of addition of each substance and added to
eq.~{\ref{H+calc}} found in exercise 24. The
concentration of~\(Ca^{2+}\) was then calculated for each case
and then plotted as shown in fig. {\ref{536113}}.
\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/e25plot/e25plot}
\caption{{Concentration of~\(Ca^{2+}\)~as a function of~\(P_{CO_2}\)
after addition of~\(NaHCO_3\) and~\(H_2SO_4\).
{\label{536113}}%
}}
\end{center}
\end{figure}
The results show than the solubility of~\(CaCO_3\) increases
when~\(H_2SO_4\) is added and decreases when~\(NaHCO_3\)
is added. Increasing concentration of~\(SO_4^{2-}\) decreases the
buffering capacity of the solution~ which increases the availability of
free~\(H^+\) which in turn leads to an increase of solubility
of~\(CaCO_3\). Increasing concentrations
of~\(Na^+\)increases the buffering capacity of the solution
which decreases the availability of free~\(H^+\) which in
turn decreases the solubility of~\(CaCO_3\).
\par\null
\textbf{File: e25.m}
\begin{verbatim}
fplot(@caconc, [10^-4 10^-1], '--')
hold on
fplot(@caconc251, [10^-4 10^-1])
fplot(@caconc252, [10^-4 10^-1])
xlabel('P_{CO_2} / atm')
ylabel('[Ca^{2+}] / mole \cdot L^{-1}')
legend('Nothing added', 'NaHCO_3 added', 'H_{2}SO_{4} added')
\end{verbatim}
\textbf{File: caconc251.m}
\begin{verbatim}
function ca=caconc251(pco2)
if pco2 == 0
pco2 = 10^-4;
end
kh2co3 = 10^-(6.482);
kh = 10^-(1.260);
khco3 = 10^(-10.471);
kcaco3 = 10^(-8.41);
h = hconc251(pco2);
ca = (kcaco3 * h^2) / (khco3*kh2co3*kh*pco2);
end
\end{verbatim}
\textbf{File: caconc252.m}
\begin{verbatim}
function ca=caconc252(pco2)
if pco2 == 0
pco2 = 10^-4;
end
kh2co3 = 10^-(6.482);
kh = 10^-(1.260);
khco3 = 10^(-10.471);
kcaco3 = 10^(-8.41);
h = hconc252(pco2);
ca = (kcaco3 * h^2) / (khco3*kh2co3*kh*pco2);
end
\end{verbatim}
\pagebreak
\subsection*{Exercise 26~}
{\label{711741}}
In this exercise we derive a simplified equation of the dissolution
of~\(CaCO_3\) in a pure water system and compare this simplified
equation to the one used in exercise 24 and 25.~
\par\null
The simplified equation is derived assuming that the charge balance can
be reduced to
\begin{equation}
\(2\left[Ca^{2+}\right]=\left[HCO_3^-\right]\)
\end{equation}
The next step is to express the concentration of~\(HCO_3^-\)~in
different terms as such
\begin{equation}
\(2\left[Ca^{2+}\right]=\frac{K_{H_2CO_3}K_HP_{CO_2}}{\left[H^+\right]}\)
\end{equation}
Next, we can express the denominator term as~
\begin{equation}
\(\frac{1}{\left[H^+\right]}=\frac{\left[CO_3^{2-}\right]}{K_{HCO_3}\left[HCO_3^-\right]}\)
\end{equation}
And insert in eq. (5)
\begin{equation}
\(2\left[Ca^{2+}\right]=\frac{K_{H_2CO_3}K_HP_{CO_2}\left[CO_3^{2-}\right]}{K_{HCO_3}\left[HCO_3^-\right]}\)
\end{equation}
Now we utilize the assumption found in eq. (4) and insert it as such
\begin{equation}
\(2\left[Ca^{2+}\right]=\frac{K_{H_2CO_3}K_HP_{CO_2}\left[CO_3^{2-}\right]}{K_{HCO_3}\cdot2\left[Ca^{2+}\right]}\)
\end{equation}
Next, we utilize the saturation index relationship
\begin{equation}
\(\left[CO_3^{2-}\right]=\frac{K_{CaCO_3}}{\left[Ca^{2+}\right]}\)
\end{equation}
We insert it into the equation, rearrange the numbers and arrive at our
final equation
\begin{equation}
\(2\left[Ca^{2+}\right]=\frac{K_{H_2CO_3}K_HP_{CO_2}K_{CaCO_3}}{K_{HCO_3}2\left[Ca^{2+}\right]^2}\)
\end{equation}
\begin{equation}
\(\left[Ca^{2+}\right]=\left(\frac{K_{H_2CO_3}K_HP_{CO_2}K_{CaCO_3}}{4K_{HCO_3}}\right)^{\frac{1}{3}}\)
\end{equation}
Using matlab we can plot the concentration of~\(Ca^{2+}\)using
the simplified equation and the equation used in previous exercises to
see how well they compare. The plot is shown in fig.
.{\ref{567450}}\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/e26plot/e26plot}
\caption{{Concentration of~\(Ca^{2+}\)~as a function of~\(P_{CO_2}\)
{\label{567450}}%
}}
\end{center}
\end{figure}
The plot illustrates that the two equations corresponds well together,
which indicates that the simplified equation is applicable.
\textbf{File: e26.m}
\begin{verbatim}
fplot(@caconc, [10^-4 10^-1])
hold on
fplot(@caconc26, [10^-4 10^-1], '+')
xlabel('P_{CO_2} / atm')
ylabel('[Ca^{2+}] / mole \cdot L^{-1}')
legend('Original equation', 'Simplified equation')
\end{verbatim}
\textbf{File: caconc26.m}
\begin{verbatim}
function ca=caconc26(pco2)
kh2co3 = 10^-(6.482);
kh = 10^-(1.260);
khco3 = 10^(-10.471);
kcaco3 = 10^(-8.41);
ca=((pco2*kh2co3*kh*kcaco3) / (4*khco3))^(1/3)
end
\end{verbatim}
\pagebreak
\subsection*{Exercise 27}
{\label{711741}}
In this exercise we investigate how the solubility of~\(CaCO_3\)
changes with temperature. For this we utilise the simplified equation
for the dissolution of~\(CaCO_3\)~shown below
\begin{equation}
\(\left[Ca^{2+}\right]=\left(\frac{K_{H_2CO_3}K_HP_{CO_2}K_{CaCO_3}}{4K_{HCO_3}}\right)^{\frac{1}{3}}\)
\end{equation}
We assume normal atmospheric pressure. All expressions used in this
exercise are shown below
\begin{equation}
\(K_{H_2CO_3}=10^{(14.82-(3401/T)-0.0327T)}\)
\end{equation}
\begin{equation}
\(K_H=10^{\left(-12.59+(2198/T)+0.0126T)\right)}\)
\end{equation}
\begin{equation}
\(K_{HCO_3}=10^{(6.53-(2906/T)-0.0238T)}\)
\end{equation}
\begin{equation}
\(P_{CO_2}=4\cdot10^{-4}\ pa\)
\end{equation}
\begin{equation}
\(K_{CaCO_3}=K_{T0^{ }}\cdot e^{\left(\frac{\Delta H^0}{R}\cdot\left(\frac{1}{T0}-\frac{1}{T}\right)\right)}\)
\end{equation}
\begin{equation}
\(\Delta H^0=-9.61\cdot10^3\ \frac{J}{mol}\)
\end{equation}
\begin{equation}
\(R=8.3145\ \frac{J}{K\cdot mol}\)
\end{equation}
\begin{equation}
\(K0=10^{-8.38}\)
\end{equation}
\begin{equation}
\(T0=273.15\ K\)
\end{equation}
\(K0\) is the equilibrium constant for~\(CaCO_3\)
at~\(T0\).
The solubility of~\(CaCO_3\) was plotted as a function of
temperature up to~\(100\selectlanguage{ngerman}°C\), shown in fig
{\ref{683853}}.\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/e27plot/e27plot}
\caption{{Concentration of~\(Ca^{2+}\) as a function of temperature
{\label{683853}}%
}}
\end{center}
\end{figure}
We can see from the graph that the solubility decreases with
temperature. This explains the mineral deposits on coffee machines and
sauce pans as~\(CaCO_3\) will tend to crystallise at higher
temperatures.
\textbf{File: e27.m}
\begin{verbatim}
fplot(@caconc27, [273.15 373.15])
xlabel('Temperature / [K]')
ylabel('[Ca^{2+}] / mol \cdot L^{-1}')
\end{verbatim}
\textbf{File: caconc27.m}
\begin{verbatim}
function ca=caconc27(T)
kh=10^(-12.59+(2198/T)+0.0126*T);
kh2co3=10^(14.82-(3401/T)-0.0327*T);
khco3=10^(6.53-(2906/T)-0.0238*T);
pco2=4*10^-4;
T0 = 273.15; % K
dH0 = -9.61*10^3; % J/mol
R = 8.3145; % J / (K * mol)
K0 = 10^-8.38; % kcaco3 @ T0
kcaco3 = K0 * exp((dH0/R) * ((1/T0) - (1/T)));
ca=((pco2*kh2co3*kh*kcaco3) / (4*khco3))^(1/3);
end
\end{verbatim}
\pagebreak
\subsection*{Exercise 28}
{\label{983568}}
Oxidation and reduction reactions (redox) are frequently occurring in
nature. They play important roles in both biotic and abiotic processes
that cycle nutrients, toxins and other substances in natural waters.
Redox reactions often include~\(H^+\) ions and thus affect
and are affected by water pH. Redox reactions can also transform
compounds from non-toxic to toxic forms (or reverse) or from a
water-soluable to a non-water-soluable form (or the reverse).
Understanding the conditions under which redox reactions occur are
therefore crucial to several water related disciplines such as drinking
water provision and waste water treatment. This exercise aims to examine
two redox reactions to determine their spontaneity under standard
conditions; the reduction of~\(As\left(V\right)\) to
toxic~\(As(III)\) and the reduction of solid~\(Fe\left(III\right)OOH\)
to ionic~\(Fe\left(II\right)^{2+}\), both reactions with organic carbon
(\(CH_2O\)) being oxidised.
\par\null
~The spontaneity of a reaction can be characterised by examining the
change in Gibbs free energy (\(\Delta G\)). Spontaneous reactions
have a negative~\(\Delta G\).~\(\Delta G\) is related to the
standard electrode potential (\(E^{\circ}\)) by
\begin{equation}
\label{deltage}
\Delta G = -nFE^\circ
\end{equation}
Where n is the number of moles of electrons and~\(F\) is
Faraday's constant (\(9.65\cdot10^4\) C / mol). From
({\ref{deltage}}) it an be seen that all reactions with
a positive~\(E^{\circ}\)are spontaneous. Determining the
spontaneity of a redox reaction can thus be reduced to the problem of
determining the sign of \(E^{\circ}\).
\par\null
The reduction of~\(As\left(V\right)\) to toxic~\(As(III)\) with
organic carbon is characterised as:
\begin{equation}
\label{arsenic1}
2H_2AsO^-_4 + CH_2O \rightarrow 2H_2AsO_3^- + CO_2 + H_2O
\end{equation}
This reaction can be divided into two half-cell reactions
(\(E^{\circ}\)values from course workbook Table 9.2, p. 94):
\begin{equation}
\label{ashalfred}
\frac{1}{2} H_2AsO_4^- + H^+ + e^- \rightarrow \frac{1}{2} H_2AsO_3^- + \frac{1}{2} H_2O \qquad E^\circ_{red} = 0.203 \text{V}
\end{equation}
\begin{equation}
\label{ashalfox}
\frac{1}{4} CH_2O + \frac{1}{4} H_2O \rightarrow \frac{1}{4} CO_2 + H^+ + e^- \quad \qquad \qquad E^\circ_{ox}=0.0708 \text{V}
\end{equation}
The standard electrode potentials of the half cells can then be summed
to find the reaction standard electrode potential (\(E_{tot}\)):
\begin{equation}
\label{astot}
E^\circ_{tot} = E^\circ_{red} + E^\circ_{ox} = 0.2738 \text{V}
\end{equation}
It can therefore be concluded that this reaction is spontaneous under
standard conditions.
\par\null
The reduction of solid~\(Fe\left(III\right)OOH\) to
ionic~\(Fe\left(II\right)^{2+}\)with organic carbon can be characterised as:
\begin{equation}
\label{fe1}
4FeOOH + CH_2O + 8H^+ \rightarrow 4Fe^{2+} + CO_2 + 7 H_2O
\end{equation}
This reaction can be devided into two half-cell reactions
(\(E^{\circ}\)values from course workbook Table 9.2, p. 94 and
Exercise 28 description, p. 95):
\begin{equation}
\label{fehalfred}
FeOOH + 3H^+ + e^- \rightarrow Fe^{2+} + 2 H_2O \qquad E^\circ_{red} = 0.708 \text{V}
\end{equation}
\begin{equation}
\label{fehalfox}
\frac{1}{4} CH_2O + \frac{1}{4} H_2O \rightarrow \frac{1}{4}CO_2 + H^+ + e^- \qquad E^\circ_{ox} = 0.0708 \text{V}
\end{equation}
And \(E_{tot}^{\circ}\) can be calculated as:
\begin{equation}
\label{astot}
E^\circ_{tot} = E^\circ_{red} + E^\circ_{ox} = 0.7788 \text{V}
\end{equation}
Here to it can be concluded that the reaction is spontaneous under
standard conditions.
\par\null
As can be seen from the calculations above, both reactions examined are
spontaneous under standard conditions. Both of these reactions may be
expected to have an impact on the viability/treatment requirements on
using groundwater for drinking water in areas with iron or arsenic in
the bedrock.
\par\null\par\null
\pagebreak
\subsection*{Exercise 29}
{\label{525914}}
Eutrophication is a serious problem affecting many bodies of water
around the world. While fresh water systems generally phosphorous
limited, salt water systems are generally nitrogen limited. Excessive
loading of nitrogen in water systems, especially salt water systems such
as the baltic sea can therefore have severe negative environmental
consequences. The removal of nitrogen from wastewater is therefore often
an important step in wastewater treatment. This process is called
de-nitrification and is a biotic redox process that transforms nitrates
in the water to nitrogen gas that is vented to the atmosphere. The
reaction uses organic carbon as the reduction agent as well as a carbon
source for the biotic organisms. However wastewater often does not
contain sufficient organic carbon and so carbon must be added. This is
often done in the form an addition of methanol (\(CH_3OH\)) or
ethanol (\(C_2H_5OH\)). This exercise aims to determine which of
these two carbon sources is most thermodynamically advantageous, per
mole, at standard state.
\par\null
The the denitrifying half-cell reaction can be written
\begin{equation}
\label{denithalf}
\frac{1}{5} NO_3^- + \frac{6}{5} H^+ + e^- \rightarrow \frac{1}{10} N_2 + \frac{3}{5} H_2O \qquad E^\circ_{red} = 1.245 \text{V}
\end{equation}
This half-cell reaction can then be paired with either the half-cell
reaction for the reduction of methanol
({\ref{methhalf}}) or ethanol
({\ref{ethhalf}}):
\begin{equation}
\label{methhalf}
\frac{1}{6} CO_2 + e^- + H^+ \leftrightharpoons \frac{1}{6} CH_3OH + \frac{1}{6}H_2O \qquad E^\circ_{red} = 0.02\text{V}
\end{equation}
\begin{equation}
\label{ethhalf}
\frac{1}{6} CO_2 + e^- + H^+ \leftrightharpoons \frac{1}{12} C_2H_5OH + \frac{1}{4}H_2O \qquad E^\circ_{red} = 0.136\text{V}
\end{equation}
Gibbs free energy~(\(\Delta G\)) is the thermodynamic property
that describes the spontaneity of a reaction with more negative values
of~\(\Delta G\) being more thermodynamically adventurous. The
problem of determining weather ethanol or methanol is more
thermodynamically advantageous per mole as the reducing agent in the
denitrifying reaction thus reduces to determining which of the reactions
({\ref{denithalf}}) +
(-~{\ref{methhalf}}) or
({\ref{denithalf}}) +
(-~{\ref{ethhalf}}) yields the smallest
~\(\Delta G\) per mole ethanol/methanol.
\par\null
Given
\begin{equation}
\label{deltage}
\Delta G = -nFE^\circ
\end{equation}
Where n is the number of moles of electrons and~\(F\) is
Faraday's constant (\(9.65\cdot10^4\) C / mol), this can be calculated
as follows:
\par\null
\subsubsection*{Methanol}
{\label{569713}}\par\null
\begin{equation}
\label{methdeltag}
\Delta G_{meth} = -1 \cdot 9.65 \cdot 10^4 \cdot (1.245 - 0.02) = -1.225 F
\end{equation}
This however is the ~\(\Delta G\) for~\(\frac{1}{6}\)mol
methanol. The per mole ~\(\Delta G\) is thus \(-1.225F\cdot6\ =-7.35F\).
\par\null
\subsubsection*{Ethanol}
{\label{201155}}
\begin{equation}
\label{ethdeltag}
\Delta G_{eth} = -1 \cdot 9.65 \cdot 10^4 \cdot (1.245 - 0.136) = -1.109 F
\end{equation}
This is the~\(\Delta G\) for~\(\frac{1}{12}\) mol ethanol. The
per mole ~\(\Delta G\) is thus \(-1.109F\cdot12\ =-13.308F\).
\par\null
Here it can be seen that the per mole ~\(\Delta G\) for ethanol
(-13.308F) is smaller than that for methanol (-7.35F) and thus it can be
concluded that as a source of organic carbon in denitrification ethanol
is more thermodynamically advantageous.
\par\null
It can also be seen from the stoichiometry of the half-cell reactions
that less ethanol is required per electron (half that of methanol),
meaning that also quantitatively half as much ethanol is required for
the same amount of denitrification. Given both the thermodynamic and
stoichiometric considerations ethanol would appear to be a chemically
superior source of organic carbon for the denitrification process.
\par\null
\pagebreak
\subsection*{Exercise 30}
{\label{335245}}
Natural water systems are often complex, containing a multitude of
substances in different forms. What forms these substances take can
affect their water solubility or state and thus their abundance in the
water, their toxicity or other qualities that are desirable to know from
a water engineering and environmental perspective. Many substances of
interest in water chemistry are transformed from one state to another by
redox reactions, both biotically and abiotically. It is therefore of
importance to an understanding of natural water systems to be able to
model under which circumstances different redox reactions take place.
The aim of this exercise is to examine three different scenarios and to
model under which circumstances different abundances of chemical species
occur.
\par\null
The central quantity of interest when modelling systems of redox
reactions is the redox potential (\(E_h\)). The redox
potential is a quantity that describes the~\emph{system} and can be
viewed as a measure of the number of electrons available to take part in
redox reactions. Thermodynamically previous exercies have defined the
standard electrode potential,\(\)\(E^{\circ}\) as~
\begin{equation}
\label{eq:e0def}
\Delta G^\circ = -nFE^\circ
\end{equation}
Where~\(\Delta G^{\circ}\) is the standard Gibbs free
energy,~\(n\) is the number of moles of electrones
transferred during a redox reaction and~\(F\) is Faraday's
constant (\(9.65\cdot10^4\ \ \text{C}\ \cdot\text{mol}^{-1}\)). Similarly~\(E_h\) can be
defined as
\begin{equation}
\label{ehdef}
\Delta G = -nFE_h
\end{equation}
The relationship between \(\Delta G^{\circ}\)and \(\Delta G\) is
defined as
\begin{equation}
\label{eq:gibbsdef}
\Delta G = \Delta G^\circ + RT \ln Q
\end{equation}
Where~\(T\) is the tempreature,~\(R\) is the
gas constant (\(8.3145\ \text{J}\cdot\text{K}^{-1}\cdot\text{mol}\)) and~\(Q\) is the reaction
quotient which for the reaction~\(a \cdot \text{oxidized form} + m \cdot H^+ + n\cdot e^- \rightarrow b\cdot \text{[reduced form]}\) is defined as
\begin{equation}
\label{qdef}
Q = \frac{\text{[reduced form]}^b}{\text{[oxidised form]}^a [H^+]^m}
\end{equation}
Where the notation~\(a_Z^{ }\) denotes the activity of species Z.
Combining ({\ref{eq:e0def}}),
({\ref{ehdef}}) and
({\ref{eq:gibbsdef}}) and rearranging
for~\(E_h\) yields
\begin{equation}
\label{ehdeftot}
E_h = E^\circ - \frac{RT}{nF} \ln Q
\end{equation}
Substituting in values for the constants, a temperature value of 298K,
assuming \(n=1\) and re-arranging yields
\begin{equation}
\label{eq:ehdeffin}
E_h = E^\circ - 0.0257 \ln \frac{\text{[reduced form]}^b}{\text{[oxidised form]}^a} - 0.059 \cdot m \cdot \text{pH}
\end{equation}
From ({\ref{eq:ehdeffin}}) it can be seen that for a
given redox pair at given concentrations each value
of~\(E_h\) will correspond to a pH. This fact can be used to
generate graphs that characterise the conditions (pH
and~\(E_h\)) under which a redox pair will appear
under/at/over a given concentration in the oxidised/reduced form.\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/e30/e30}
\caption{{Figure showing the~\(E_h\) vs pH relationships for the redox
pairs~\textbf{(}\emph{\textbf{left}}\textbf{)}:~\(MnO_2 / Mn^{2+}\)
and~\(NO_3^- / N_2\), concentrations of~\(0.1\text{mmol}\cdot\text{L}^{-1}\), partial
pressures 0.8 atm.~
\textbf{(}\emph{\textbf{center}}\textbf{)}:~\(N_2 / NH_4^+\)~
and~\(NO_3^- /N_2\), concentrations of~\(0.1\text{mmol}\cdot\text{L}^{-1}\), partial
pressures 0.8
atm.~\textbf{(}\emph{\textbf{right}}\textbf{)}:~~\(Fe(OH)_3/Fe^{2+}\)
and~\(CO_2 / CH_4\), ion concentrations of~\(0.1\text{mmol}\cdot\text{L}^{-1}\)
and~\(P_{CO_2} = P_{CH_4}\).
{\label{289679}}%
}}
\end{center}
\end{figure}
Figure {\ref{289679}} shows three graphs of three
scenarios A - left, B - center, C - right of the redox
potential~\(E_h\), pH and concentrations for various species
of redox couples. The figure was generated using equation
({\ref{eq:ehdeffin}}) implemented in matlab using the
script~\texttt{e30.m} and the functions~\emph{A:}~\texttt{ehmn.m}
and~\texttt{ehno3.m},~\emph{B:}~\texttt{ehnh4.m}
and~\texttt{ehno3},~\emph{C:}~\texttt{ehfeoh3.m} and~\texttt{ehco2.m} -
all of which can be viewed below. Values for~\(E^{\circ}\) used for
the calculations are from the course workbook Table 9.2, p. 94.
\subsubsection*{Scenario A}
{\label{105955}}
The exercise is to identify the~\(E_h\)/pH range
where~\(\left[Mn^{2+}\right]\)will be higher than 0.1 mmol/L
if~\(\left[NO_3^-\right]\)= 0.1 mmol/L. This can be expressed as
\begin{equation}
\label{eq:conAsum}
\begin{cases}
[NO_3^-] = 0.1 \text{mmol} \cdot \text{L}^{-1} \\
[Mn^{2+}] > 0.1 \text{mmol} \cdot \text{L}^{-1}
\end{cases}
\end{equation}
It can be seen from Figure~{\ref{289679}} (left) that
this satisfied only by~\(E_h\)/pH values lying on
the~\(\left[NO_3^-\right] = 0.1 \text{mmol} \cdot \text{L}^{-1}\) line that are also below the~\(\left[Mn^{2+}\right] = 0.1 \text{mmol} \cdot \text{L}^{-1}\)
line (pH \textless{} 1.712).
\subsubsection*{Scenario B}
{\label{894260}}
The exercise is to identify the~\(E_h\)/pH range where
neither~\(\left[NO_3^-\right]\)nor~\(\left[NH_4^+\right]\) is higher
than~\(0.1 \text{mmol} \cdot \text{L}^{-1}\) given a~\(P_{N_2}=0.8\ \text{atm}\). This can be
expressed as
\par\null
\begin{equation}
\label{eq:conBsum}
\begin{cases}
[NO_3^-] < 0.1 \text{mmol} \cdot \text{L}^{-1} \\
[NH_4^+] < 0.1 \text{mmol} \cdot \text{L}^{-1}
\end{cases}
\end{equation}
It can be seen from Figure~{\ref{289679}} (center) that
this satisfied only by all~\(E_h\)/pH values lying between
the line~\([NH_4^+] = 0.1 \text{mmol} \cdot \text{L}^{-1}\) and~\([NO_3^-] = 0.1 \text{mmol} \cdot \text{L}^{-1}\).
\subsubsection*{Scenario C}
{\label{231830}}
The exercise is to identify the~\(E_h\)/pH range
where~\(Fe^{2+}\) can reduce~\(CO_2\)
to~\(CH_4\) at equal pressure between ~\(CO_2\)
and~~\(CH_4\). This can be expressed as
\par\null
\begin{equation}
\label{eq:conCsum}
\begin{cases}
P_{CH_4} = P_{CO_2} \\
Fe^{2+} > 0.1 \text{mmol} \cdot \text{L}^{-1}
\end{cases}
\end{equation}
It can be seen from Figure~{\ref{289679}} (right) that
this satisfied only by~\(E_h\)/pH values on the
line~\(P_{CH_{4\ }}=\ P_{CO_2}\)that are also below the line~\(\left[Fe^{2+}\right]=\ 0.1 \text{mmol} \cdot \text{L}^{-1}\)
(pH \textless{} 11.563).
\par\null
As conclusion it is worth re-iterating that these graphs are valid for
T=298K. The graphs are generated from well established thermodynamic
equations and known constants and it is therefore to be expected that
they are highly valid.vv
\textbf{File:e e30.m}
\begin{verbatim}
%% A
figure(1)
fplot( @ehmn, [0 10] )
hold on
fplot( @ehno3, [0 10], '--' )
grid on
xlabel('pH')
ylabel('E_h')
axis([0 10 0 1.5])
legend('[Mn^{2+}] = 0.1 mM', '[NO_3^-] = 0.1 mM')
%% B
figure(2)
fplot( @ehnh4, [0 10] )
hold on
fplot( @ehno3, [0 10], '--' )
grid on
xlabel('pH')
ylabel('E_h')
axis([0 10 0 1.5])
legend('[NH_4^+] = 0.1 mM', '[NO_3^-] = 0.1 mM')
%% C
figure(3)
fplot( @ehfeoh3, [0 14] )
hold on
fplot( @ehco2, [0 14], '--' )
grid on
xlabel('pH')
ylabel('E_h')
axis([0 14 -1 1.4])
legend('[Fe^{2+}] = 0.1 mM', 'P_{CH_4} = P_{CO_2}')
\end{verbatim}
\textbf{File: ehmn.m}
\begin{verbatim}
function ehres=ehmn(pH)
e0 = 1.23 ; % V
redconc = 0.1*10^-3 ; % M
oxconc = 1 ;
a = 1/2 ;
b = 1/2 ;
m = 2 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\textbf{File: ehno3.m}
\begin{verbatim}
function ehres=ehno3(pH)
e0 = 1.245 ; % V
redconc = 0.1 * 10^-3 ; % M
oxconc = 0.8 ; % atm
a = 1/5 ;
b = 1/10 ;
m = 6/5 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\textbf{File: ehnh4.m}
\begin{verbatim}
function ehres=ehnh4(pH)
e0 = 0.274 ; % V
redconc = 0.1 * 10^-3 ; % M
oxconc = 0.8 ; % atm
a = 1/6 ;
b = 1/3 ;
m = 4/3 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\textbf{File: ehfeoh3.m}
\begin{verbatim}
function ehres=ehfeoh3(pH)
e0 = 1.057 ; % V
redconc = 0.1 * 10^-3 ; % M
oxconc = 1 ;
a = 1 ;
b = 1 ;
m = 3 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\textbf{File: ehco2.m}
\begin{verbatim}
function ehres=ehco2(pH)
e0 = -0.0708 ; % V
redconc = 1 ; % atm
oxconc = 1 ; % atm
a = 1/4 ;
b = 1/4 ;
m = 1 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\pagebreak
\subsection*{Exercise 31}
{\label{596759}}
Lake sediments are ideal environments for observing many redox
reactions. The sediments will tend to contain a gradient of among other
things oxygen with~\(O_2\) being more abundant close the the
sediment surface. The sediment will also tend to accumulate various
compounds. This combination has the potential to contain life that use
the gradient in redox potential and abundance of various compounds to
generate energy.~ Understanding the chemical processes happening in lake
sediment can also be of interest when attempting to model the effect of
dreading, changing water pH or increase on pollutants on the aquatic
systems. For example a shifting pH may cause a redox copule to shift
from a water insoluble to a water soluble form - quickly changing the
chemical composition of the water. The exercise aims to explain the
concentration profiles of~\(NO_3^-\),~\(Mn\)
and~\(Fe\) in the pore water of a lake sediment from an
experiment from 1979.
\par\null
Lake sediment will tend to follow a pattern of decreasing redox
potential~\(E_h\), with increased sediment
depth~\(z\). As~\(E_h\) decreases different
redox couples will transition from their oxidised to reduced form.
Couples follow a characteristic sequence as can be seen from the redox
ladder, Figure {\ref{522292}}.
\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/redox-ladder/redox-ladder}
\caption{{The redox-ladder shows the sequence in which redox couples transition
from their oxidised to reduced states with decreasing redox
potential~\(E_h\). Source: Course workbook Figure 9.1, p. 98.
{\label{522292}}%
}}
\end{center}
\end{figure}\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/sediment/sediment}
\caption{{Figure showing species concentration profiles for~\(NO_3^-,\ Mn\)
and~\(Fe\) in a lake sediment. Data source: Froelich et al.,
1979, graph source: Course workbook Exercise 31, p. 102.
{\label{301281}}%
}}
\end{center}
\end{figure}
Figure~{\ref{301281}} illustrates the sequence of
transition depicted in the redox ladder
(Figure~{\ref{522292}}). As~\(z\)
increases~\(E_h\) decreases and water
soluble~\(NO_3^-\)gets reduced to~\(N_2\) gas.
By~\(z=30\)cm~\(E_h\) has decreased sufficiently
that water insoluble~\(MnO_2\) starts to be reduced to water
soluble~\(Mn^{2+}\) and starts to be detectable in the pour
water. By~\(z=45\)cm~\(E_h\) has decreased
sufficiently that water insoluble~\(Fe(OH)_3\) has started to be
reduced to water soluble \(Fe^{2+}\).
\par\null
Decreasing~\(E_h\) and the redox ladder explain the general
shape of the species concentration profiles. It can be observed that the
concentration profiles are not completely regular, eg,
the~\(NO_3^-\) profile initially increases between the first
and the second data point and again between the 7th and 8th and again
between the 14th and 15th. This is not explained by the presented model.
However it seems reasonable that these minor variations are the result
either of errors in the sampling or due to localised
disturbances/variations in sediment and that overall the explination
model presented in this exercise is sufficient.
\par\null
\newpage
\subsection*{Exercise 32}
{\label{564865}}
Natural waters are complex and include a multitude of different
compounds in different oxidation states. Compounds can take part in
multiple different redox half-cells depending on the environmental
conditions. The concentrations of different species are thus not only
dependent on the pH and redox potential~\(E_h\), of a system,
but also of the other redox reactions that the components take part in.
This exercise aims to investigate the concentrations
of~\(Fe^{2+}\) and~\(Fe^{3+}\) ions in
various~\(E_h\)/pH conditions in which both
the~\(Fe\left(OH\right)_3+H^++e^-\rightarrow Fe^{2+}+3H_2O\) and~\(Fe^{3+}\rightarrow\ Fe^{2+}\ +\ e^-\) half-cells take place.
\par\null
Similarly as in Exercise 30, the~\(E_h\) for the reaction
\(a \cdot \text{oxidized form} + m \cdot H^+ + n\cdot e^- \rightarrow b\cdot \text{[reduced form]}\) at a given pH for a redox couple at a given
concentration of the oxidised and reduced species can be calculated as
\begin{equation}
\label{eq:ehdeffin32}
E_h = E^\circ - 0.0257 \ln \frac{\text{[reduced form]}^b}{\text{[oxidised form]}^a} - 0.059 \cdot m \cdot \text{pH}
\end{equation}
assuming T=298K and where~\(E^{\circ}\) denotes the standard
electrode potential of the half-cell. From
({\ref{eq:ehdeffin}}) it can be seen that if
\(m=0\) and \(\text{[oxidised form]}\ =\text{[reduced form]}\) then \(E_h=E^{\circ}\).\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/e32/e32}
\caption{{Figure showing the~\(E_h\)/pH relationship for the redox
couples~\(Fe\left(OH\right)_3 / Fe^{2+}\),~\([Fe^{2+}] = 0.1 \text{mmol} \cdot \text{L}^{-1}\)
and~\(Fe^{3+} / Fe^{2+}\),~\([Fe^{3+}] = [Fe^{2+}] = 0.1 \text{mmol} \cdot \text{L}^{-1}\) with the four regions created
by the intersecting lines of equi-concentration labelled A,B,C and D.
{\label{353086}}%
}}
\end{center}
\end{figure}
Figure~{\ref{353086}} shows the intersecting lines of
equi-concnetration for the~\(E_h\)/pH graph for the redox
couples~\(Fe\left(OH\right)_3 / Fe^{2+}\) and~\(Fe^{3+} / Fe^{2+}\).
Figure~{\ref{353086}} was generated in matlab using the
script \texttt{e32.m} (see below) and the function~\texttt{ehfeoh3.m}
(see above under exercise 30). The line for~
\(Fe^{3+} / Fe^{2+}\),~\([Fe^{3+}] = [Fe^{2+}] = 0.1 \text{mmol} \cdot \text{L}^{-1}\) was plotted as
\(E_h=E^{\ \circ}=0.711\ \text{V}\).
\par\null
\subsubsection*{Section A}
{\label{242521}}
Section A can be described by~
\begin{equation}
\label{secA}
\begin{cases}
[Fe^2+] < 0.1 \text{mmol} \cdot \text{L}^{-1} \\
[Fe^3+] > 0.1 \text{mmol} \cdot \text{L}^{-1}
\end{cases}
\end{equation}
Below the solid line the~\(Fe\left(OH\right)_3 / Fe^{2+}\) couple will be predominantly
in the~\(Fe^{2+}\). However the ~\(Fe^{3+} / Fe^{2+}\)couple will
predominantly be in the~\(Fe^{3+}\) form and will convert the
available~\(Fe^{2+}\)to~\(Fe^3\).
\subsubsection*{Section C}
{\label{926966}}
Section C can be described by
\begin{equation}
\label{secC}
\begin{cases}
[Fe^2+] > 0.1 \text{mmol} \cdot \text{L}^{-1} \\
[Fe^3+] < 0.1 \text{mmol} \cdot \text{L}^{-1}
\end{cases}
\end{equation}
\par\null
Both the redox couples will predominantly be in
the~\(Fe^{2+}\)form.
\subsubsection*{Section B}
{\label{794163}}
Section B can be described by
\begin{equation}
\label{secB}
\begin{cases}
[Fe^2+] < 0.1 \text{mmol} \cdot \text{L}^{-1} \\
[Fe^3+] < 0.1 \text{mmol} \cdot \text{L}^{-1}
\end{cases}
\end{equation}
The ~\(Fe^{3+} / Fe^{2+}\) couple will exist predominantly in
the~\(Fe^{2+}\) form. However the~\(Fe\left(OH\right)_3 / Fe^{2+}\) couple will
exist predominantly in the~\(Fe\left(OH\right)_3\) form and will convert the
available \(Fe^{2+}\) to \(Fe\left(OH\right)_3\).
\subsubsection*{Section D}
{\label{231798}}
Section D can be described by
\begin{equation}
\label{secD}
\begin{cases}
[Fe^2+] < 0.1 \text{mmol} \cdot \text{L}^{-1} \\
[Fe^3+] = \text{unknown}
\end{cases}
\end{equation}
Available~\(Fe^{2+}\) will predominantly be converted to
either~\(Fe\left(OH\right)_3\) or~\(Fe^{3+}\) leading to
a~~\([Fe^{2+}] < 0.1 \text{mmol} \cdot \text{L}^{-1}\). However the~\(\left[Fe^{3+}\right]\) will depend on the
initial concentration of~\(Fe^{3+}\) ions as these will
predominantly remain as~\(Fe^{3+}\).
\par\null
Figure~{\ref{353086}} illustrates the complexity
present in natural water systems. Sections A, B and C are examples of
states where the ion concentration is determined by~\(E_h\)
and pH of the system and how shifts in either~\(E_h\) or pH
can cause large transitions between different forms. This can be of
environmental interest, especially if one form of a compound is
non-water soluble or toxic. For example small changes in water pH an
cause a compound found in the sediment to become water soluble with
large environmental or health consequences. Section D shows how the
ionic concentration can be dependent on the initial state of the system.
\textbf{File: e32.m}
\begin{verbatim}
fplot( @ehfeoh3, [0 8] )
hold on
plot(linspace(0, 8), linspace(0.711, 0.711), '--')
grid on
xlabel('pH')
ylabel('E_h')
legend('[Fe^{2+}] = 0.1 mM', '[Fe^{2+}] = [Fe^{3+}] = 0.1 mM')
\end{verbatim}
\pagebreak
\subsection*{Exercise 33}
{\label{299467}}\par\null
One important application of water chemistry is the provision of safe drinking water. One common drinking water source is groundwater. Groundwater, depending on the composition of the geology can have very different levels of different compounds, some of which affect the smell, taste and/or toxicity. Most countries maintain standards for what levels of various compounds are considered safe or acceptable in drinking water. In Sweden these standards are set and maintained by Livsmedelsverket. Two compounds that have set standards are $Mn^{2+}$, which can have negative health effects, and $Fe^{2+}$ which has an unpleasant taste. The levels are 0.2 mg/L for $Fe^{2+}$ and 0.05 mg/L for are $Mn^{2+}$ \footnote{\label{livs}Source: Livsmedelsverket SLVFS 2001:30}. Both $Mn^{2+}$ and $Fe^{2+}$ take part in redox couples, namely $Fe(OH)_3 / Fe^{2+}$ and $MnO_2 / Mn^{2+}$, where the oxidised form is non-water-soluble. This means that it is possible to reduce the concentrations by changing the $E_h$ or pH so as to cause percipitation. This exercise aims to investigate which $E_h$ values need to be reached in order to ensure the $[Fe^{2+}]$ and $[Mn^{2+}]$ are below the Swedish standard at a typical pH range for groundwater of 8-8.3.
As has been deduced in previous exercises the redox
potential~\(E_h\), for the reaction~\(a \cdot \text{oxidized form} + m \cdot H^+ + n\cdot e^- \rightarrow b\cdot \text{[reduced form]}\) at a
given pH for a redox couple at a given concentration of the oxidised and
reduced species can be calculated as~
\begin{equation}
\label{eq:ehdeffin33}
E_h = E^\circ - 0.0257 \ln \frac{\text{[reduced form]}^b}{\text{[oxidised form]}^a} - 0.059 \cdot m \cdot \text{pH}
\end{equation}
assuming T=298K and where~\(E^{\circ}\) denotes the standard
electrode potential of the half-cell. Assuming an ideal water system
with only the~\(Fe(OH)_{3}/Fe^{2+}\) and~\(MnO_{2}/Mn^{2+}\) redox couples
present the~\(E_h\) and pH values that lead to specific
concentrations of~\(Mn^{2+}\) and~\(Fe^{2+}\) can then be
determined by plotting.
\par\null\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/e33/e33}
\caption{{Figure showing the~\(E_h\)/pH relationship for the redox
couples~\(Fe\left(OH\right)_3 / Fe^{2+}\),~\([Fe^{2+}]=0.2\text{mg}\cdot\text{L}^{-1}\)
and~\(MnO_2 / Mn^{2+}\),~\([Mn^{2+}] = 0.05 \text{mg} \cdot \text{L}^{-1}\).
{\label{140286}}%
}}
\end{center}
\end{figure}
Figure~{\ref{140286}} was generated in matlab using the
script~\texttt{e33.m} and the equation
({\ref{eq:ehdeffin}}) as implemented in the
functions~\texttt{ehfe2.m} and~\texttt{ehmn2.m} (see below for script
and functions).~
\par\null
The redox potential required to achieve ~\(\left[Mn^{2+}\right]<0.05\ \ \text{mg}\cdot\ \text{L}^{-1}\)
and~\(\left[Fe^{2+}\right]<\ 0.2\ \text{mg}\cdot\text{L}^{-1}\) in the pH interval 8-8.3 can be read from
Figure~{\ref{140286}} as all~\(E_h\)values
above the~\(\left[Mn^{2+}\right] = 0.05 \text{mg}\cdot\ \text{L}^{-1}\) line. Specifically for pH
8,~\(E_h>0.465\) V and for pH 8.3,~\(E_h>0.429\) V.
\par\null
It should be noted that two assumptions were made in these calculations.
Firstly it was assumed that no other redox reactions
involving~\(Mn^{2+}\) or~\(Fe^{2+}\) will affect the ion
concentrations. Examination of Figure~{\ref{353086}}
(Exercise 32) indicates that the~\(Fe^{3+} / Fe^{2+}\) redox couple will
not interact as the~\(Fe^{2+}\) form will be dominant as we are
below~\(E^{\circ}=0.711\text{V}\). However it is plausible that other~ redox
couples including~\(Fe^{2+}\) or~\(Mn^{2+}\) may
interract, chaning the required~\(E_h\) value. To rule this
out all the relevant redox couples could also be plotted. Secondly it
was assumed that iron and manganese is present in sufficient quantities
in the water to start with. As this exercise explicitly is about iron
and manganese removal this is a reasonable assumption.
\par\null
In conclusion the results of this exercise indicate what value
of~\(E_h\) should be exceeded order to ensure levels of iron
and manganese that comply with the Swedish drinking water standard. The
results are valid for an idealised system, however before being applied
for drinking water treatment the interactions with
other~\(Fe^{2+}\) and~\(Mn^{2+}\) redox couples should
be modelled.
\textbf{File: e33.m}
\begin{verbatim}
fplot( @ehfe2, [8 8.3] )
hold on
fplot( @ehmn2, [8 8.3], '--' )
grid on
xlabel('pH')
ylabel('E_h')
axis([8 8.3 -0.15 0.55])
legend('[Fe^{2+}] = 0.2 mg / L', '[Mn^{2+}] = 0.05 mg / L')
\end{verbatim}
\textbf{File: ehfe2.m}
\begin{verbatim}
function ehres=ehfe2(pH)
e0 = 1.057 ; % V
redconc = 0.2e-3 / 55.58 ; % 0.2 mg / L Fe in mol / L
oxconc = 1 ;
a = 1 ;
b = 1 ;
m = 3 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\textbf{File: ehmn2.m}
\begin{verbatim}
function ehres=ehmn2(pH)
e0 = 1.23 ; % V
redconc = 0.05e-3 / 54.94 ; % 0.05 mg / L Mn in mol / L
oxconc = 1 ;
a = 1/2 ;
b = 1/2 ;
m = 2 ;
ehres=e0 - 0.0257 * log( (redconc^b) / (oxconc^a) ) - 0.059 * m * pH ;
end
\end{verbatim}
\pagebreak
\subsection*{Exercise 34}
{\label{246766}}
Previous exercises have examined the concept of acid neutralizing
capacity, ANC. The ability of a water system to buffer against additions
of acids and bases are of great importance in understanding the
environmental implications of acid rain or chemical spills. The previous
examinations of ANC did not consider the buffering properties of iron
and how they depend on the redox potential~\(E_h\), of the
system. This exercise aims to model the ANC of a water system containing
the~\(Fe\left(OH\right)_3 / Fe^{2+}\) redox couple at different redox potentials.
\par\null
The~\(Fe\left(OH\right)_3 / Fe^{2+}\) redox couple is described by (\(E^{\circ}\)
from course workbook Table 9.2, p. 94)
\begin{equation}
\label{eq:feoheq}
Fe(OH)_3 + 3H^+ + e^- \leftrightharpoons Fe^{2+} + 3H_2O \qquad E^\circ = 1.057 \text{V}
\end{equation}
Using ({\ref{eq:ehdeffin}}) (Exercise 30) a function
for the concentration of~\(Fe^{2+}\)given~\(\left[H^+\right]\)
and~\(E_h\) can derived as
\begin{equation}
\label{eq:feconc}
[Fe^{2+}] = [H^+]^3 \cdot 10^{\frac{E^\circ_{Fe(OH)_3} - E_h}{0.059}}
\end{equation}
Where~\(E_{Fe\left(OH\right)_3}^{\circ}\) is the standard electrode potential
for~the~\(Fe\left(OH\right)_3 / Fe^{2+}\) redox couple.
({\ref{eq:feconc}}) also assumes T=298K. Using the
definition of ANC as~\(\sum \text{anions to weak acids}-\ \sum \text{cations to weak bases}\) yields
\begin{equation}
\label{eq:ancdef}
0 = [\text{ANC}] + [HCO_3^-] - [R^-] + [H^+] + 3[Al^{3+}] + 2[Fe^{2+}]
\end{equation}
Where~\(\left[R^-\right]\) denotes the concentration of anions to organic
acids. The expression for~\(\left[Al^{3+}\right]\),~\(\left[HCO_3^-\right]\)
and~\(\left[R^-\right]\) have been discussed in previous exercises:
\begin{equation}
\label{eq:aldef}
[Al^{3+}] = K_G \cdot [H^+]^3
\end{equation}
Where $K_R$ is the diassiciation constant for the reaction $Al(OH)_3 + 3H^+ \leftrightharpoons Al^{3+} + 3H_2O$
\begin{equation}
\label{eq:rdef}
[R^-] = 7 \cdot 10^{-6} \cdot [\text{DOC}] \frac{K_R}{K_R + [H^+]}
\end{equation}
In these calculations a simplified value of~\(K_R\) is used
where~\(K_R=5\cdot10^{-5}\) in order to simplify the calculations.
\begin{equation}
\label{eq:hco3def}
[HCO_3^-] = \frac{K_{H_2CO_3} \cdot K_H \cdot P_{CO_2}}{[H^+]}
\end{equation}
Where~\(K_{H_2CO_3}\)is the dissociation constant
for~\(H_2CO_3\) in water,~\(K_H\) is Henry's constant
for~\(CO_2\) in water and~\(P_{CO_2}\)is the partial
pressure of~\(CO_2\). Substituting
({\ref{eq:aldef}})-({\ref{eq:hco3def}})
into the expression for ANC ({\ref{eq:ancdef}}) and
re-arranging the expression to the form of a polynomial yields
\begin{equation}
\label{eq:ancpoly}
\begin{split}
0 = &[H^+]^5(3K_G + 2 \cdot 10^\frac{E^\circ_{Fe(OH)_3 - E_h}}{0.059}) + [H^+]^4K_R(3K_G + 2 \cdot 10^\frac{E^\circ_{Fe(OH)_3 - E_h}}{0.059}) + [H^+]^3 + [H^+]^2([\text{ANC}] + K_R) + \\
& + [H^+](K_R[\text{ANC}] - K_{H_2CO_3}K_HP_{CO_2} - 7\cdot 10^{-6}[\text{DOC}]K_R) - K_RK_{H_2CO_3}K_HP_{CO_2}
\end{split}
\end{equation}\selectlanguage{english}
\begin{figure}[H]
\begin{center}
\includegraphics[width=1.00\columnwidth]{figures/e34/e34}
\caption{{Figure showing the pH vs ANC for a water system including
the~\(Fe\left(OH\right)_3 / Fe^{+2}\) redox couple at different values
of~\(E_h\).
{\label{672239}}%
}}
\end{center}
\end{figure}
Figure~{\ref{672239}} shows the relationship between pH
and ANC for a water system including the~\(Fe\left(OH\right)_3 / Fe^{+2}\) redox couple
at different values of~\(E_h\). It was generated using
({\ref{eq:ancpoly}}) as implemented using matlab and
the script \texttt{e34.m} and the functions \texttt{pheh0.m},
\texttt{pheh010.m}, \texttt{pheh025.m} and \texttt{pheh050.m}. The
calculations use constants for T=298K.
\par\null
It can be seen from Figure~{\ref{672239}} that the ANC
is most affected by iron at lower values of ANC (for ANC \textless{}
1.8~\(E_h\) has a large effect on pH). It can be seen from
({\ref{eq:ancdef}}) that low values of ANC is related
to higher values of~\(\left[Fe^{2+}\right]\) and reversely high values of ANC
is related to iron being in the form of~\(Fe\left(OH\right)_3\). The buffer
capacity of the redox copule is related to its ability to transition
between redox states. At high ANC values the~\(Fe\left(OH\right)_3\) form is
favoured to the extent that it does not transition readily
to~\(Fe^{2+}\)and is thus has little overall contribution to the
overall buffer capacity of the water.
\par\null
Anoxic environments tend to have low~\(E_h\) values and
therefore favour the reduced form of redox couples. An acid and anoxic
solution, such as ketchup, will have a ready supply
of~\(H^+\) ions and will therefore be able to readily
dissolve~\(Fe\left(OH\right)_3\) (the main component in rust). This is
because the reduced (and water soluble) form (\(Fe^{2+}\)) will
be favoured and sufficient~\(H^+\) ions will be present for
the reaction to proceed.
\par\null
The calculations in this exercise make a number of assumptions and
simplifications. The temperature of T=298K, though generally a high
temperature for natural waters in Sweden, was chosen as this temperature
was already baked in to a previously calculated constant (0.059) in
({\ref{eq:feconc}}). It could easily be re-calculated
for other temperatures, however this is outside the scope of this
exercise (and if summer temperatures continue as the summer of 2018
T=298K may not be such an unrealistic temperature for natural waters
even in Sweden).
\par\null
One assumption made is that when calculating~\(\left[R^-\right]\) the
simplified value for~\(K_R=5\cdot10^{-5}\) was used instead of a more
accurate expression that is dependent on the pH. This was done to
simplify the calculations at the expense of accuracy. However as was
examined in Exercise 17 it is a reasonable approximation in this pH
range. Additionally the expression for ANC does not
consider~\(\left[CO_3^{2-}\right]\), however examining Figure 1, Exercise 4
shows that~\(\left[HCO_3^-\right]\) is grater than~\(\left[CO_3^{2-}\right]\) by an
order of magnitude grater than \textasciitilde{}3 for the pH range in
question and therefore this appears to be a reasonable simplification.
The calculations also ignore any other species of aluminium apart
from~\(\left[Al^{3+}\right]\) in order to simplify the calculations.
\par\null
Though several simplifications are made in this exercise in order to
simplify the calculations the results are still useful for a schematic
understanding of the relationship between ANC, pH
and~\(E_h\) for an idealised water system containing
the~\(Fe\left(OH\right)_3 / Fe^{+2}\) redox couple.
\par\null
\textbf{File: e34.m}
\begin{verbatim}
fplot( @pheh0, [-2e-4 1e-3])
hold on
grid on
fplot( @pheh010, [-2e-4 1e-3], '--' )
fplot( @pheh025, [-2e-4 1e-3], '-.')
fplot( @pheh050, [-2e-4 1e-3], ':')
legend('E_h = 0 V', 'E_h = 0.10 V', 'E_h = 0.25 V', 'E_h = 0.50 V')
axis([-2e-4 1e-3 4.3 8.5])
xlabel('ANC / eq/L')
ylabel('pH')
\end{verbatim}
\textbf{File: pheh0.m}
\begin{verbatim}
function phres=pheh0(anc)
% Constants at 25 degC
kg = 10^8.5;
e0 = 1.057; % V
eh = 0; % V
kr = 5e-5;
kh2co3 = 10^-6.366;
kh = 10^-1.462;
pco2 = 400 * 10^(-6); % atm
doc = 5; % mg/L
a(1) = 3*kg + 2*10^((e0-eh)/0.059);
a(2) = kr * (3*kg + 2*10^((e0-eh)/0.059));
a(3) = 1;
a(4) = anc + kr;
a(5) = kr * anc - kh2co3 * kh * pco2 - 7e-6 * doc * kr;
a(6) = - kr * kh2co3 * kh * pco2;
X = roots([a]);
x = max(X(find(imag(X) == 0))); % conc. h+
phres = -log10(x);
end
\end{verbatim}
\textbf{File: pheh010.m}
\begin{verbatim}
function phres=pheh010(anc)
% Constants at 25 degC
kg = 10^8.5;
e0 = 1.057; % V
eh = 0.10; % V
kr = 5e-5;
kh2co3 = 10^-6.366;
kh = 10^-1.462;
pco2 = 400 * 10^(-6); % atm
doc = 5; % mg/L
a(1) = 3*kg + 2*10^((e0-eh)/0.059);
a(2) = kr * (3*kg + 2*10^((e0-eh)/0.059));
a(3) = 1;
a(4) = anc + kr;
a(5) = kr * anc - kh2co3 * kh * pco2 - 7e-6 * doc * kr;
a(6) = - kr * kh2co3 * kh * pco2;
X = roots([a]);
x = max(X(find(imag(X) == 0))); % conc. h+
phres = -log10(x);
end
\end{verbatim}
\textbf{File: pheh025.m}
\begin{verbatim}
function phres=pheh025(anc)
% Constants at 25 degC
kg = 10^8.5;
e0 = 1.057; % V
eh = 0.25; % V
kr = 5e-5;
kh2co3 = 10^-6.366;
kh = 10^-1.462;
pco2 = 400 * 10^(-6); % atm
doc = 5; % mg/L
a(1) = 3*kg + 2*10^((e0-eh)/0.059);
a(2) = kr * (3*kg + 2*10^((e0-eh)/0.059));
a(3) = 1;
a(4) = anc + kr;
a(5) = kr * anc - kh2co3 * kh * pco2 - 7e-6 * doc * kr;
a(6) = - kr * kh2co3 * kh * pco2;
X = roots([a]);
x = max(X(find(imag(X) == 0))); % conc. h+
phres = -log10(x);
end
\end{verbatim}
\textbf{File: pheh050.m}
\begin{verbatim}
function phres=pheh050(anc)
% Constants at 25 degC
kg = 10^8.5;
e0 = 1.057; % V
eh = 0.50; % V
kr = 5e-5;
kh2co3 = 10^-6.366;
kh = 10^-1.462;
pco2 = 400 * 10^(-6); % atm
doc = 5; % mg/L
a(1) = 3*kg + 2*10^((e0-eh)/0.059);
a(2) = kr * (3*kg + 2*10^((e0-eh)/0.059));
a(3) = 1;
a(4) = anc + kr;
a(5) = kr * anc - kh2co3 * kh * pco2 - 7e-6 * doc * kr;
a(6) = - kr * kh2co3 * kh * pco2;
X = roots([a]);
x = max(X(find(imag(X) == 0))); % conc. h+
phres = -log10(x);
end
\end{verbatim}
\selectlanguage{english}
\FloatBarrier
\end{document}