Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. a. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) titrated by \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(100.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{3} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by \(0.20 \mathrm{M} \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{~mL}\) of \(0.50 \mathrm{M} \mathrm{HCl}\) titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

At the halfway point for a weak acid-strong base titration, half of the weak acid has reacted with the strong base. We can calculate the new concentrations at this point and use the acid dissociation constant, \(K_{a}\), to determine the \(\mathrm{pH}\). The stoichiometry of reaction: \(\mathrm{HC}_7\mathrm{H}_5\mathrm{O}_2 + \mathrm{OH}^{-} \rightarrow \mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-} + \mathrm{H}_2\mathrm{O}\) At the halfway point: Half of the weak acid has reacted, we will have: \(\frac{1}{2}\mathrm{M}\cdot 100\mathrm{mL} = 5\mathrm{mmol}\) This means 5 mmol of \(\mathrm{OH}^{-}\) has been added to the solution, and there will remain 5 mmol of the weak acid. The number of moles of the conjugate base, \(\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\): 5 mmol We can calculate the concentration of the weak acid and its conjugate base at the halfway point: \(\left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right] = \frac{5\,\mathrm{mmol}}{100.0\,\mathrm{mL}+50\,\mathrm{mL}} = \frac{5}{150.0}\,\mathrm{M} = 0.03333\,\mathrm{M}\) And since it's the halfway point, \(\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right] = \left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right] = 0.03333\,\mathrm{M}\) Now, using Ka, calculate the \(\left[\mathrm{H}^{+}\right]\): \(K_{a}^{}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right]}{\left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right]}\) Plug in values, \(Ka =6.4 \times 10^{-5}\) and solve for \(\left[\mathrm{H}^{+}\right]\): \(\left[\mathrm{H}^{+}\right] = K_{a}\frac{\left[\mathrm{HC}_{7}\mathrm{H}_{5}\mathrm{O}_{2}\right]}{\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right]}=6.4 \times 10^{-5}\), thus \(\mathrm{pH}\) at halfway point: \(\mathrm{pH} = -\log{(6.4 \times 10^{-5})}= 4.19\)

At equivalence point, the amount of \(\mathrm{OH}^{-}\) added to the solution equals the initial moles of the weak acid which is: \(100.0\mathrm{mL} \times 0.10\mathrm{M} = 10\,\mathrm{mmol}\). So: Total volume: \(100.0\,\mathrm{mL} + 100.0\,\mathrm{mL}(0.1\,\mathrm{M})=200.0\,\mathrm{mL}\) At this point, the weak acid has been completely titrated, and the only remaining species of interest is the conjugate base. The concentration of the conjugate base: \(\left[\mathrm{C}_{7}\mathrm{H}_{5}\mathrm{O}_{2}^{-}\right]=\frac{10\,\mathrm{mmol}}{200\,\mathrm{mL}}=0.05\,\mathrm{M}\) Next, we need to determine \(\left[\mathrm{OH}^{-}\right]\) using Kb: First we need to calculate Kb from Ka: \[K_b = \frac{K_w}{K_a}\] Plug in the values: \(K_w = 1 \times 10^{-14}\) and \(K_a = 6.4 \times 10^{-5}\), and solve for \(K_b\): \(K_{b}=\frac{1 \times 10^{-14}}{6.4 \times 10^{-5}} = 1.56 \times 10^{-10}\) Now, \(K_{b}=\frac{\left[\mathrm{OH}^{-}\right]^2}{\left[\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-}\right]}\). Using the obtained \(K_{b}\) value and the conjugate base concentration \([\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-}]=0.05\,\mathrm{M}\), we can find \(\left[\mathrm{OH}^{-}\right]\): \(\left[\mathrm{OH}^{-}\right] = \sqrt{K_{b}\left[\mathrm{C}_7\mathrm{H}_5\mathrm{O}_2^{-}\right]} = \sqrt{(1.56 \times 10^{-10})\times 0.05} = 7.87 \times 10^{-6}\,\mathrm{M}\) Finally, calculate pH at equivalence point: \[pOH = -\log{[OH^{-}]} = -\log{(7.87 \times 10^{-6})}= 5.10\] Therefore, \(\mathrm{pH} = 14 - \mathrm{pOH}= 14 - 5.10 = 8.90\) Summarizing for part a:\(\) At the halfway point: \(\mathrm{pH} = 4.19\) At the equivalence point: \(\mathrm{pH} = 8.90\) The procedure for parts b and c will follow the same steps as outlined above, but with the appropriate changes depending on whether it's a strong acid-strong base or weak acid-weak base titration.