\documentclass[12pt,a4paper]{article}
\usepackage{chemfig}

\begin{document}
\title{Buffers}
\author{Philip Thomas K.}
\maketitle

\section{Introduction}
Buffers are solutions that can resist a change in pH when small amounts of acid or base are added to them. They are required in many biological systems such as the circulatory systems.

\newpage

\section{Theory}
Acidic Buffers are made from a weak acid and its conjugate base:\\
\begin{figure}[h]
    \centering
    \schemestart HA\arrow{<<->}H\textsuperscript{$+$} + A\textsuperscript{$-$} \schemestop\par
    \schemestart MA\arrow{->}M\textsuperscript{$+$} + A\textsuperscript{$-$} \schemestop\par
\end{figure}


Upon adding hydroxide ions (OH\textsuperscript{$-$}), the concentration of hydroxonium ions (H\textsuperscript{$+$}) reduces, by L.C.P, the equilibrium will move to the right to resist this change. Because most of the acid has not dissociated, plenty of molecules can still break down to supply hydroxonium ions (H\textsuperscript{$+$}).\\
\medskip


On adding hydroxonium ions (H\textsuperscript{$+$}), by L.C.P, the position of the equilibrium wants to move to the left to use up the extra hydroxonium ions (H\textsuperscript{$+$}). However we need the conjugate base (A\textsuperscript{$-$}) to react with the extra hydroxonium ions (H\textsuperscript{$+$}). The acid does not produce any conjugate base (A\textsuperscript{$-$}) to react with the extra hydroxonium ions (H\textsuperscript{$+$}) added. Therefore we need a secondary source of conjugate base (A\textsuperscript{$-$}), which comes from the salt.
\bigskip


Basic Buffers are made from a weak base and it conjugate acid:
\begin{figure}[h]
    \centering
    \schemestart B + H\textsubscript{2}O\arrow{<<->}BH\textsuperscript{$+$} + OH\textsuperscript{$-$} \schemestop\par
    \schemestart BHX\arrow{->}BH\textsuperscript{$+$} + X\textsuperscript{$-$} \schemestop\par
\end{figure}


Upon adding hydroxonium ions (H\textsuperscript{$+$}), the concentration of hydroxide ions (OH\textsuperscript{$-$}) reduces, by L.C.P, the equilibrium will move to the right to resist this change. Because most of the acid has not dissociated, plenty of molecules can still break down to supply hydroxide ions (OH\textsuperscript{$-$}).\\
\medskip


On adding hydroxide ions (OH\textsuperscript{$-$}), by L.C.P, the position of the equilibrium wants to move to the left to use up the extra hydroxide ions (OH\textsuperscript{$-$}). However we need the conjugate acid (BH\textsuperscript{$+$}) to react with the extra hydroxide ions (OH\textsuperscript{$-$}). The acid does not produce any conjugate acid (BH\textsuperscript{$+$}) to react with the extra hydroxide ions (OH\textsuperscript{$-$}) added. Therefore we need a secondary source of conjugate acid (BH\textsuperscript{$+$}), which comes from the salt.
\newpage
The pH of Buffer Solutions:
\begin{equation}
    K_a = \frac{[H^+][A^-]}{[HA]}
\end{equation}
\medskip

The value of $K_a$ is a representation of how far the weak acid dissociates.With our knowledge of buffers, we can say that:
\begin{equation}
    [HA]\textsubscript{equilibrium} \simeq [HA]\textsubscript{initial}
\end{equation}

because very little of the acid dissociates. We can also say that:
\begin{equation}
    [A^-]\textsubscript{equilibrium} \simeq [A^-]\textsubscript{initial}
\end{equation}

because since there is a weak acid that does not dissociate significantly, only few extra conjugate base (A\textsuperscript{$-$}) at equilibrium are added. We can now say that:
\begin{equation}
    K_a = \frac{[H^+][conc. base]}{[Acid]}
\end{equation}
\medskip
\begin{equation}
    K_a \cdot [Acid] = [H^+][conc. base]
\end{equation}
\medskip
\begin{equation}
    K_a \cdot \frac{[Acid]}{[Base]} = [H^+]
\end{equation}
\medskip
\begin{equation}
    -log[H^+] = -log(K_a \cdot \frac{[Acid]}{[Base]})
\end{equation}
\medskip
\begin{equation}
    pH = -log(K_a) - log(\frac{[Acid]}{[Base]})
\end{equation}
\medskip
\begin{equation}
    pH = pK_a - log\frac{[Acid]}{[Base]}
\end{equation}

\end{document}