(Solved): 1) A 25.0 mL sample of the weak base trimethylamine, \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} ...

1) A 25.0 mL sample of the weak base trimethylamine, \( \left(\mathrm{CH}_{3}\right)_{3} \mathrm{~N} \), requires 50.0 mL of 0.200 M HCl to reach equivalence. The \( \mathrm{pK}_{\mathrm{b}} \) of trimethylamine is 4.20 , and the titration was carried out at \( 25^{\circ} \mathrm{C} \). a) Calculate the pH of the solution at the start of the titration. [10.75] b) Calculate the pH of the solution when 15.0 mL of the acid solution have been added. [10.17] c) Calculate the pH at the equivalence point of this titration. [5.34] d) Calculate the pH when 10.0 mL beyond the equivalence point have been added. [1.63] e) If you were to select an indicator for the above titration, approximately what should be its \( \mathrm{pK}_{\mathrm{a}} \) ? EXPLAIN YOUR ANSWER. [choose an indicator with a \( \mathbf{p K} \mathbf{K}_{\mathrm{a}}=\mathbf{p H} \) at equivalence]