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 \(M \mathrm{NaOH}\) b. \(100.0 \mathrm{mL}\) of \(0.10 \mathrm{M} \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)\) titrated by 0.20 \(M \mathrm{HNO}_{3}\) c. \(100.0 \mathrm{mL}\) of \(0.50 \mathrm{M}\) HCl titrated by \(0.25 \mathrm{M} \mathrm{NaOH}\)

Titration a: Halfway Point Calculations#Region_content#First, let's calculate \(pK_a\): \( pK_a = -\log{K_a} = -\log(6.4 \times 10^{-5}) \) Next, we will plug the values into Henderson-Hasselbalch equation: \( pH = (-\log(6.4 \times 10^{-5})) + \log \frac{[0.10]}{[0.10]} \) \( \)

Titration a: Equivalence Point Calculations#Region_content#Calculate the moles of \(\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}\) and \(\mathrm{NaOH}\): Moles of acid: \((0.1 \textrm{ M})(0.1 \textrm{ L}) = 0.01 \textrm{ mol}\) Amount of \(\mathrm{NaOH}\) required to reach the equivalence point (equal to moles of acid): \(0.01 \textrm{ mol}\) Volume of \(\mathrm{NaOH}\) required to reach the equivalence point: \(\frac{0.01 \textrm{ mol}}{0.1 \textrm{ M}} = 0.1 \textrm{ L}\) Total volume at equivalence point: \(0.1 \textrm{ L} + 0.1 \textrm{ L} = 0.2 \textrm{ L}\) Now we find the pH at the equivalence point using the ion-product of water: \(K_w = [\mathrm{H}^+][\mathrm{OH}^-] \) And the \(K_a\) of the conjugate base (\(\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2}^-\)): \(K_a = [\mathrm{HC}_{7} \mathrm{H}_{5} \mathrm{O}_{2}][\mathrm{OH}^{-}]/[\mathrm{C}_{7} \mathrm{H}_{5} \mathrm{O}_{2}^{-}]\) \( \)

Titration b: Halfway Point Calculations#Region_content#First, let's calculate \(pK_b\): \( pK_b = -\log{K_b} = -\log(5.6 \times 10^{-4}) \) Next, we will plug the values into pOH equation: \( pOH = (-\log(5.6 \times 10^{-4})) + \log \frac{[0.10]}{[0.10]} \) Finally, convert pOH to pH: \( pH = 14 - pOH \) \( \)

Titration b: Equivalence Point Calculations#Region_content#First, calculate the moles of \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) and \(\mathrm{HNO}_{3}\): Moles of base: \((0.1 \textrm{ M})(0.1 \textrm{ L}) = 0.01 \textrm{ mol}\) Amount of \(\mathrm{HNO}_{3}\) required to reach the equivalence point (equal to moles of base): \(0.01 \textrm{ mol}\) Volume of \(\mathrm{HNO}_{3}\) required to reach the equivalence point: \(\frac{0.01 \textrm{ mol}}{0.2 \textrm{ M}} = 0.05 \textrm{ L}\) Total volume at equivalence point: \(0.1 \textrm{ L} + 0.05 \textrm{ L} = 0.15 \textrm{ L}\) Now we find the pH at the equivalence point using the following equation: \(pH = 14 - pOH\) Where pOH can be calculated using the ion-product of water: \(K_w = [\mathrm{H}^+][\mathrm{OH}^-] \) \( \)

Titration c: Halfway Point Calculations#Region_content#Determine the average pKa value of HCl and NaOH: pH = \( \frac{pK_{a, \textrm{HCl}} + pK_{a, \textrm{NaOH}}}{2}\) Since \(\textrm{HCl}\) and \(\textrm{NaOH}\) are strong acids and bases, their pKa values are negligible, and the pH at the halfway point is about 7.0. \( \)

Titration c: Equivalence Point Calculations#Region_content#First, calculate the moles of \(\textrm{HCl}\) and \(\textrm{NaOH}\): Moles of acid: \((0.5 \textrm{ M})(0.1 \textrm{ L}) = 0.05 \textrm{ mol}\) Amount of \(\textrm{NaOH}\) required to reach the equivalence point (equal to moles of acid): \(0.05 \textrm{ mol}\) Volume of \(\textrm{NaOH}\) required to reach the equivalence point: \(\frac{0.05 \textrm{ mol}}{0.25 \textrm{ M}} = 0.2 \textrm{ L}\) Total volume at equivalence point: \(0.1 \textrm{ L} + 0.2 \textrm{ L} = 0.3 \textrm{ L}\) Now we find the concentration of \([\mathrm{H}^+]\) and \([\mathrm{OH}^-]\) ions in the solution: \([\mathrm{H}^+] = \frac{\text{moles of } \mathrm{H}^+}{\textrm{total volume}} = \frac{0.05 \textrm{ mol}}{0.3 \textrm{ L}}\) \([\mathrm{OH}^-] = \frac{\text{moles of } \mathrm{OH}^-}{\textrm{total volume}} = \frac{0.05 \textrm{ mol}}{0.3 \textrm{ L}}\) Finally, calculate pH using the following equation: \(pH = -\log [\mathrm{H}^+]\)