Calculate the \(\mathrm{pH}\) of each of the following buffered solutions. a. \(0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} / 0.25 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) b. \(0.25 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2} / 0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) c. \(0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} / 0.50 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\)

For each case, identify the weak acid and weak base present in the solution. In all three cases here, ethylamine (\( \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} \)) is the weak base and the ethylamine hydrochloride (\( \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl} \)) is the weak acid.

The \(\mathrm{p}K_\mathrm{a}\) of the weak acid can be found using the \(\mathrm{K}_{\mathrm{b}}\) of its conjugate base, which is ethylamine. Given that the \(\mathrm{K}_{\mathrm{b}}\) for ethylamine is \(5.6\times10^{-4}\), we can use the relationship between \(\mathrm{K}_{\mathrm{a}}\) and \(\mathrm{K}_{\mathrm{b}}\): \[\mathrm{K}_{\mathrm{a}}\times\mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{w}}\] Where \(\mathrm{K}_{\mathrm{w}}=1.0\times10^{-14}\) at 25°C. Calculate the \(\mathrm{p}K_\mathrm{a}\) value of the weak acid: \[\mathrm{p}K_\mathrm{a}=-\log(\mathrm{K}_\mathrm{a})\]

For each buffered solution, use the concentrations of the weak acid and weak base, along with the calculated \(\mathrm{p}K_\mathrm{a}\) value, in the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\): \[\mathrm{pH}=\mathrm{p}K_\mathrm{a}+\log\frac{[\mathrm{base}]}{[\mathrm{acid}]}\]

a. For the buffered solution with \(0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} / 0.25\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\), plug in the concentrations of the weak base and weak acid, and the calculated \(\mathrm{p}K_\mathrm{a}\) value, into the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\). b. For the buffered solution with \(0.25\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{3}\mathrm{NH}_{2} / 0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\), plug in the concentrations of the weak base and weak acid, and the calculated \(\mathrm{p}K_\mathrm{a}\) value, into the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\). c. For the buffered solution with \(0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{2} / 0.50\ \mathrm{M}\ \mathrm{C}_{2}\mathrm{H}_{5}\mathrm{NH}_{3}\mathrm{Cl}\), plug in the concentrations of the weak base and weak acid, and the calculated \(\mathrm{p}K_\mathrm{a}\) value, into the Henderson-Hasselbalch equation to find the \(\mathrm{pH}\). After completing these steps, the \(\mathrm{pH}\) values of the given buffered solutions will be calculated.