Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M\) pyridine with 0.100\(M\) hydrochloric acid \(\left(K_{\mathrm{b}} \text { for pyridine is } 1.7 \times 10^{-9}\right) .\) Do not calculate the points at 24.9 and 25.1 \(\mathrm{mL}\)

First, we need to calculate the initial concentration of pyridine in the solution before the addition of hydrochloric acid. This is given by the equilibrium expression: \(K_{\mathrm{b}} = \frac{[\mathrm{C_5H_5N}^+][\mathrm{OH^-}]}{[\mathrm{C_5H_5N}]}\) Rearranging to solve for [OH⁻]: \([\mathrm{OH^-}] = \frac{K_{\mathrm{b}}[\mathrm{C_5H_5N}]}{[\mathrm{C_5H_5N}^+]}\) Now we can find the pOH of the solution: \(pOH = -\log{[\mathrm{OH^-}]}\) Finally, the pH of the solution can be found using the relationship: \(pH = 14 - pOH\)

At the equivalence point, the amount of pyridine is equal to the amount of hydrochloric acid added. Given the volumes and concentrations of the two solutions, we can calculate the moles of pyridine and hydrochloric acid: moles of pyridine = (25.0 mL)(0.100 M) = 2.5 mmol moles of hydrochloric acid = (25.0 mL)(0.100 M) = 2.5 mmol As the moles are equal at the equivalence point, we can use the Kb value for the conjugate acid, C5H5NH+, to find the pH. To find the conjugate acid's Ka value, we use the following relationship: \(K_{\mathrm{a}}K_{\mathrm{b}} = K_{\mathrm{w}}\) where Kw is the ion product constant of water, which is equal to 1.0 x 10^-14 at 25°C. First, find the Ka value for the conjugate acid: \(K_{\mathrm{a}} = \frac{K_{\mathrm{w}}}{K_{\mathrm{b}}} = \frac{1.0 \times 10^{-14}}{1.7 \times 10^{-9}}\) Calculate the [H⁺] using the Ka expression: \([\mathrm{H^+}] = \sqrt{K_{\mathrm{a}} [\mathrm{C_5H_5NH^+}]}\) Then, calculate the pH: \(pH = -\log{[\mathrm{H^+}]}\)

After the addition of hydrochloric acid, we can assume that the H⁺ ions from the titrant will combine with the pyridine ions in the solution (C5H5N), forming the conjugate acid (C5H5NH⁺) and increasing the pH. To find the final pH, assume that the reaction is complete and that the amount of remaining pyridine (C5H5N) is equal to the amount of H⁺ ions introduced through the addition of hydrochloric acid: moles of remaining pyridine = moles of added H⁺ - initial moles of pyridine Calculate the final concentration of pyridine, C5H5N: \([\mathrm{C_5H_5N}] = \frac{\text{moles of remaining pyridine}}{\text{total volume}}\) Now we can use the Kb expression to calculate the [OH⁻] concentration: \([\mathrm{OH^-}] = \frac{K_{\mathrm{b}} [\mathrm{C_5H_5N}]}{[\mathrm{C_5H_5N}^+]}\) Find the pOH: \(pOH = -\log{[\mathrm{OH^-}]}\) Finally, calculate the pH: \(pH = 14 - pOH\)