A classic experiment in equilibrium studies dating from 1862 involved the
reaction in solution of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5}
\mathrm{OH}\right)\) and acetic acid \(\left(\mathrm{CH}_{3}
\mathrm{COOH}\right)\) to produce
ethyl acetate and water.
$$\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}+\mathrm{CH}_{3} \mathrm{COOH}
\rightleftharpoons \mathrm{CH}_{3} \mathrm{COOC}_{2}
\mathrm{H}_{5}+\mathrm{H}_{2} \mathrm{O}$$
The reaction can be followed by analyzing the equilibrium mixture for its
acetic acid content.
$$\begin{array}{r}
2 \mathrm{CH}_{3}
\mathrm{COOH}(\mathrm{aq})+\mathrm{Ba}(\mathrm{OH})_{2}(\mathrm{aq})
\rightleftharpoons \\
\mathrm{Ba}\left(\mathrm{CH}_{3} \mathrm{COO}\right)_{2}(\mathrm{aq})+2
\mathrm{H}_{2} \mathrm{O}(\mathrm{l})
\end{array}$$
In one experiment, a mixture of 1.000 mol acetic acid and 0.5000 mol ethanol
is brought to equilibrium. A sample containing exactly one-hundredth of the
equilibrium mixture requires \(28.85 \mathrm{mL} 0.1000 \mathrm{M}\)
\(\mathrm{Ba}(\mathrm{OH})_{2}\) for its titration. Calculate the equilibrium
constant, \(K_{c}\), for the ethanol-acetic acid reaction based on this
experiment.