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\begin{document}

\title{Chemistry, O Revision Notes}



\author[1]{Noah Phipps}%
\affil[1]{Affiliation not available}%


\vspace{-1em}



  
  \date{May 08, 2024}


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\tableofcontents
\newpage
\section{O1}
\subsection{Ions in Solution}
\subsubsection{Ionic Solids}
\begin{itemize}
\item Ionic solids are held together by opposite charges
\begin{itemize}
\item Anions and Cations form giant ionic lattice
\item Eg. $NaCl$
\item Each $Na^+$ ion surrounded by six $Cl^-$ ions and vice versa
\end{itemize}
\item Overall attraction within lattice greater than repulsion
\item Strong melting and boiling points
\end{itemize}

\subsubsection{Ionic Substances in Solution}
\begin{itemize}
\item Many ionic substances dissolve readily in water
\item Ions becomes surrounded by water and spread out
\item Ions lose regular arrangement - becomes random
\item Ions behave independently of each other
\end{itemize}

\subsection{Energy Changes in Solutions}
Not all ionic substances dissolve readily in solution, as energy changes are an important factor.

\subsubsection{Lattice Enthalpy}
Before an ionic solid can dissolve, the ions must be separated from their lattice. Energy must be supplied to overcome the electrostatic attractions, making it an endothermic process.\\\\
$\Delta_{LE}H$ is the enthalpy change when one mole of a solid is \textit{\textbf{formed}} from its separated ions in their gaseous states. \\\\
\begin{itemize}
\item Energy required to form a lattice is \textit{negative}
\item Energy required to break a lattice is therefore $-\Delta_{LE}H$
\end{itemize}
Lattice enthalpy is affected by both the charge and radii of the ions.
\begin{itemize}
\item A greater charge will increase the magnitude of the lattice enthalpy
\item A greater atomic radii will decrease the magnitude of the lattice enthalpy
\begin{itemize}
\item Larger separation between charges leads to smaller attraction
\end{itemize}
\item The magnitude of the lattice enthalpy will increase with greater charge density
\item Thus, substances with large enthalpies are normally insoluble
\end{itemize}

\subsubsection{Hydration and Solvation}
Many ionic substances dissolve despite their value of $\Delta_{LE}H$.
\begin{itemize}
\item Many ionics form solutions with water as solvent
\begin{itemize}
\item Covalent bonds in water are polar due to electronegativity differences in oxygen and hydrogen
\item Bent shape means whole molecule is polar and acts as a dipole
\end{itemize}
\item Charges on water are attracted to charges on ions
\begin{itemize}
\item Ion-Dipole interactions
\end{itemize}
\item Molecules interact with surface of ionic solid, separating ions from lattice
\item Ions are now surrounded by water molecules and in solution
\item Ion is now \textbf{hydrated}, as they are bound to water molecules
\end{itemize}
\bigbreak
\begin{itemize}
\item Strength of interaction determines hydration
\item Extensively hydrated ions will bind strongly to water molecules
\item Strongest hydration will lead to many attached water molecules
\item Higher charge density ions will attract more water molecules
\begin{itemize}
\item A small ion can become a large hydrated ion
\end{itemize}
\end{itemize}
\bigbreak
\begin{itemize}
\item Energy is released when ions are hydrated
\item May be sufficient to hydrate another ion from the lattice
\item However, hydrogen bonds between water must be broken
\begin{itemize}
\item Allows water molecules to rearrange around ions
\item Requires energy
\end{itemize}
\end{itemize}
$\Delta_{hyd}H$ is the enthalpy change when one mole of gaseous ions dissolve is sufficient water to form a solution of infinite dilution, and are always negative as it is an exothermic process.
\begin{itemize}
\item $|\Delta_{hyd}H|$ is greatest with a large charge density
\begin{itemize}
\item Smallest radii and largest charge
\end{itemize}
\end{itemize}
When a solvent other than water is used, the enthalpy of solvation, $\Delta_{solv}H$ is used. For example, Ethanol has polar molecules that can bond to ions.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/Ionic-solution-water-bonding1/Ionic-solution-water-bonding1}
\caption{{Polar water molecules interacting with ions in solution%
}}
\end{center}
\end{figure}

\newpage
\subsection{Enthalpy Change of Solution}
The enthalpy change of solution is the enthalpy change when one mole of a solute dissolves to form an infinitely dilute solution, and can be measured experimentally.
\begin{equation*}
\Delta_{solution}H=\Delta_{hyd}H(\text{cation})+\Delta_{hyd}H(\text{anion})-\Delta_{LE}H
\end{equation*}
This can be represented by an enthalpy level diagram or cycle.\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/solution-cycle/solution-cycle}
\caption{{An enthalpy cycle for an ionic solid dissolving%
}}
\end{center}
\end{figure}\selectlanguage{english}
\begin{figure}[h!]
\begin{center}
\includegraphics[width=0.70\columnwidth]{figures/enthalpy-level/enthalpy-level}
\caption{{An enthalpy level diagram for an ionic solid dissolving%
}}
\end{center}
\end{figure}

The above diagram shows a solute for which $\Delta_{solution}H$ is negative. This is because the hydration of the ions provides more energy than is taken up in separating the lattice. As the process is energetically favourable, the substance is likely to dissolve.
\\\\
If $\Delta_{solution}H$ has a large positive value, the solute will not dissolve. However, if $\Delta_{solution}H$ is only slightly positive, the entropy of the system may still allow it to dissolve.

\subsubsection{Non-Polar Solvents}
Molecules in a non-polar solvent do not have charges, so there would be no interaction between the solute and solvent ions - $\Delta_{hyd}H$ would be extremely small, and the solute would not dissolve.

\newpage
\section{O2}
\subsection{Radiation}
\begin{itemize}
\item Objects emit EMR, proportional to their temperature
\item The Earth is heated by the Sun
\item Earth radiates mainly in the IR region
\item Earth is in a balance of absorbing, reflecting and emitting radiation
\item Greenhouse gases can upset this balance
\item Carbon dioxide and methane in the troposphere absorb IR and prevent re-emission
\begin{itemize}
\item Warms earth
\end{itemize}
\end{itemize}
\subsubsection{Absorption of IR}
\begin{itemize}
\item Increases vibrational energy of molecule
\item Vibrational energy transferred to other molecules, increasing kinetic energy
\item Raises temperature of air
\item Small amounts re-emitted to Earth or space
\end{itemize}
\subsubsection{IR Window}
\begin{itemize}
\item Water vapour is most abundant gas in the atmosphere
\item Absorbs only in certain wavelengths
\item Regions that water does not absorb in are collectively known as the IR window
\item Nothing in IR window absorbed by water
\item Regulates temperature on Earth
\item Carbon dioxide absorbs in the IR window
\item Increase in carbon dioxide warms the Earth
\item Increased temperature increases water vapour levels
\item Further increases temperature, creating a positive feedback loop
\end{itemize}

\subsection{Strong and Weak Acids and pH}
\subsubsection{The Bronsted-Lowry Theory}
An acid is donor of $H^+$ ions, whereas a base is an $H^+$ acceptor. This would be known as an acid-base reaction.
\begin{itemize}
\item Acids are proton donors
\item Bases are proton acceptors
\end{itemize}

\subsubsection{Acid-Base Pairs}
Every acid has a conjugate base, and vice-versa. This is known as a conjugate acid-base pair;\selectlanguage{english}
\begin{figure*}[!htb]
\begin{equation*}
HA \Rightarrow H^+ + A^-\\
\end{equation*}
\caption{{An acid, $HA$, and its conjugate base, $A^-$}}
\end{figure*}

\subsection{The pH Scale}
\begin{equation*}
pH= -\log_{10}[H^+]
\end{equation*}
A logarithmic scale gives more reasonable values of pH, making them easier to compare. If an alkaline or acidic solution has a concentration of its respective ions greater than [1], then the pH will be greater/less than 14 or 0.

\subsubsection{Strong Acids}
Strong acids will dissociate fully in an aqueous solution, and have a strong tendency to donate $H^+$ ions. So,
\begin{equation*}
HA \Rightarrow H^+ +A^-
\end{equation*}

\subsubsection{Weak Acids}
A weak acid has a weaker tendency to donate $H^+$ ions, and will not fully dissociate. Thus,
\begin{equation*}
HA \rightleftharpoons H^+ + A^-
\end{equation*}

\subsection{Calculating pH Values}
\begin{equation*}
K_a=\frac{[H^+][A^-]}{[HA]}
\end{equation*}
\subsubsection{pH of a Strong Acid}
\begin{gather*}
[HA] = [H^+] = [A^-]\\
\therefore pH = -\log_{10}[H^+]
\end{gather*}

\subsubsection{pH of a Weak Acid}
Requires $K_a$ since the reaction is an equilibrium. It is assumed that:
\begin{itemize}
\item $[H^+]=[A^-]$
\begin{itemize}
\item Water provides a negligible number of ions
\end{itemize}
\item $[HA]_{\text{initial}} = [HA]_{\text{equilibrium}}$
\begin{itemize}
\item A negligible amount of the acid loses ions
\end{itemize}
\end{itemize}

\subsubsection{pK\textsubscript{a} and Weak Acids}
Since values for $K_a$ are so small, a logarithmic scale can be used:
\begin{equation*}
pK_a=-\log_{10}K_a
\end{equation*}
The weakest acid would have the highest $pK_a$.

\subsubsection{pH of a Strong Alkali}
\begin{gather*}
K_w=\frac{[H^+][OH^-]}{[H_2O]}\\
\text{Where }[H_2O]\text{ is constant}\\
\therefore K_w=[H^+][OH^-]\\
K_w = 1 \times 10^{-14} \text{ at }298K
\end{gather*}
Given further values, this can be used to work out pH as values are substituted in to the above expression.

\subsubsection{Indicators and pH}
Indicators are weak acids, with a different colour to their conjugate base:
\begin{align*}
HIn\rightleftharpoons & H^+ + In^-\\
\text{Colour 1   } & \text{\hspace{0.8cm}Colour 2}
\end{align*}
\begin{itemize}
\item Low pH
\begin{itemize}
\item Equilibrium pushed to the left, giving colour 1
\end{itemize}
\item High pH
\begin{itemize}
\item Equilibrium pushed to the right, giving colour 2
\end{itemize}
\end{itemize}
At the end point of a titration, it is critical that $[HIn]=[In^-]$, giving:
\begin{equation*}
pH=-\log_{10}K_a \tag{pH for indicators}
\end{equation*}

\newpage
\section{O3}
\subsection{Buffer Solutions}
A buffer solution is one that will resist a change in pH when a small amount of acid or alkali is added.\\\\
Buffer solutions are normally made from:
\begin{itemize}
\item A weak acid and its conjugate base
\item A weak base and its conjugate acid
\end{itemize}
\\
The nature and quantities of the substituents will determine its pH:
\begin{equation*}
HA \rightleftharpoons H^+ + A^-
\end{equation*}
Several assumptions are made:
\begin{itemize}
\item All $A^-$ ions are from the salt used, as the $HA$ provides a negligible amount
\item All of the $HA$ molecules in the buffer are unchanged
\end{itemize}
As such, we have the following cases:
\subsubsection{Addition of $H^+$ to a Buffer}
\begin{itemize}
\item $A^-$ ions from the salt will react with the addition $H^+$ ions, forming $HA$ and water, and maintaining the pH
\end{itemize}
\subsubsection{Addition of $OH^-$ to a Buffer}
\begin{itemize}
\item $H^+$ ions are removed, but regenerated from the $HA$, reestablishing the pH of the solution
\end{itemize}\\
For a buffer to function, both the weak acid and salt must be present.

\subsubsection{Buffer Calculations}
When using $K_a$ of the relevant weak acid:
\begin{equation*}
K_a = \frac{[H^+][A^-]}{[HA]}=[H^+]\times \frac{[\text{salt}]}{[\text{acid}]}
\end{equation*}
The pH of the buffer is dependent on the ration $[\text{salt}]:[\text{acid}]$.\\
Dilution of a buffer solution with pure water will have no affect on the pH because it will not affect the above ratio.

\newpage
\section{O4}
\subsection{Solubility Equilibria}
All substances are partially soluble; the dissolving of a substance is an equilibrium. Reactions will have a solubility product, $K_{sp}$.
Equations for $K_{sp}$ are the same as those for $K_c$, but \textit{do not contain any solids.} This is because it is a \textit{heterogeneous equilibrium}, for which solid concentrations are left out of the expression.
\begin{itemize}
\item $K_{sp}$ only applies to a saturated solution
\item If the value calculated is less than the true $K_{sp}$, the substances will remain in solution
\item If the value calculated is greater than the true $K_{sp}$, then substances will be precipitated out
\item Temperature dependent
\end{itemize}
E.g, for Calcium Carbonate:
\begin{gather*}
CaCO_3 _{(s)}\rightleftharpoons Ca^{2+} _{(aq)} + CO_3^{2-} _{(aq)}\\
\text{Omitting state symbols from now,}\\
K_c=\frac{[Ca^{2+}][CO_3^{2-}]}{[CaCO_3]}\\
K_{sp}=\frac{[Ca^{2+}][CO_3^{2-}]}{[\cancel{CaCO_3}]}=[Ca^{2+}][CO_3^{2-}]
\end{gather*}
In general, for an ionic solid $S$, with the ions $A$ and $B$,
\begin{gather*}
S_{(s)}\rightleftharpoons aA_{(aq)}+bB_{(aq)}\\
\text{Omitting state symbols,}\\
\therefore K_{sp}=[A]^a[B]^b\\
\end{gather*}

\newpage
\section{O5}
\subsection{Enthalpy and Entropy}
Entropy is used to explain why endothermic reactions can be spontaneous.
\subsubsection{Entropy}
Entropy is a measure of the disorder in a system - the number of ways of arranging molecules and distributing their energy.
\begin{itemize}
\item Substances are more energetically stable with higher entropy
\item Systems tend to increase entropy
\item Explains spontaneous reactions
\begin{itemize}
\item Endothermic reactions can be spontaneous if they involving spreading out, randomising, or disordering
\end{itemize}
\item Entropy is denoted by $S$
\item Physical state affected entropy
\begin{itemize}
\item Solids least disordered
\end{itemize}
\item More energy a system has, the higher its entropy
\item More particles will mean more entropy
\end{itemize}
\subsubsection{Entropy Calculations}
\begin{gather*}
\Delta_{tot}S=\Delta_{sys}S+\Delta_{surr}S\\
\Delta_{sys}S=S_{prod}-S_{reac}\\
\Delta_{surr}S=\frac{-\Delta H}{T}\\
\end{gather*}
\begin{itemize}
\item Feasible reactions, once started, will continue to completion without any additional energy being supplied to the system
\item For a feasible reaction, $\Delta_{tot}S$ must be greater than zero
\item If $\Delta_{tot}S=0$, the system will be at equilibrium
\end{itemize}
A reaction with a positive entropy change may not be spontaneous if it has a large $E_a$. Hence, the reaction is only termed as feasible.

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