Chemistry Lab Report

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\title{Lab 6: Acid--Base Titration of Acetic Acid}
\author{First Last \;•\; Section 004 \;•\; TA: Dr. Example}
\date{\today}

\begin{document}
\maketitle

\section*{Objective}
Determine the concentration of commercial vinegar by titration with standardized \ce{NaOH}.

\section*{Introduction}
The neutralization reaction is
\[ \ce{CH3COOH(aq) + NaOH(aq) -> CH3COONa(aq) + H2O(l)}. \]
At the equivalence point, moles \ce{H+} equal moles \ce{OH-}.

\section*{Materials}
\begin{itemize}
  \item Commercial vinegar (\SI{5.0}{\milli\liter} aliquots).
  \item Standardized \SI{0.100}{\Molar} \ce{NaOH}.
  \item Phenolphthalein indicator.
  \item Burette (\SI{\pm 0.05}{\milli\liter}), pipette (\SI{\pm 0.03}{\milli\liter}).
\end{itemize}

\section*{Procedure}
We titrated three aliquots to a faint pink endpoint persisting \SI{30}{\second}.

\section*{Data}
\begin{table}[h]
  \centering
  \caption{Titration data.}
  \begin{tabular}{ccc}
    \toprule
    Trial & $V_{\ce{NaOH}}$ (\si{\milli\liter}) & $[\ce{CH3COOH}]$ (\si{\Molar}) \\
    \midrule
    1 & 41.2 & 0.824 \\
    2 & 41.0 & 0.820 \\
    3 & 41.3 & 0.826 \\
    \bottomrule
  \end{tabular}
\end{table}

\section*{Calculations}
\[ [\ce{CH3COOH}] = \frac{M_{\ce{NaOH}} \cdot V_{\ce{NaOH}}}{V_{\text{vinegar}}}
   = \frac{0.100 \cdot 41.17}{5.00} = \SI{0.823}{\Molar}. \]

\section*{Results}
Mean concentration: $\bar{c} = \SI{0.823(3)}{\Molar}$.
Mass percent: 4.94\% --- consistent with commercial vinegar labeling (5.0\%).

\section*{Discussion}
Sources of uncertainty: (i) titrant volume reading, (ii) endpoint determination (systematic bias low),
(iii) temperature of the reaction mixture.

\section*{Conclusion}
The titration yielded a concentration that agrees with the commercial label within 1\%.

\end{document}