Problem: Find the pH of the equivalence point(s) and the volume (mL) of 0.447 M HNO3 needed to reach the point(s) in titrations of(a) 2.65 L of 0.0750 M pyridine (C5H5N)

We are asked to find the pH of the equivalence point(s) and the volume (mL) of 0.447 M HNO₃ needed to reach the point in titration of 2.65 L of 0.0750 M pyridine (C₅H₅N)

Step 1. Calculate mole pyridine (C₅H₅N)

$\frac{\mathbf{0}\mathbf{.}\mathbf{0750}\mathbf{ }\mathbf{mol}\mathbf{ }{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}}{\mathbf{1}\cancel{\mathbf{ }\mathbf{L}}\mathbf{\hspace{0.17em}}}\mathbf{\times }\mathbf{2}\mathbf{.}\mathbf{65}\cancel{\mathbf{ }\mathbf{L}}\mathbf{\hspace{0.17em}}$

= 0.19875 mol C₅H₅N

Step 2. Calculate the volume of 0.447 M HNO₃

at equivalence point:

mol base = mol acid

mol C₅H₅N = mol HNO₃

$\mathbf{0}\mathbf{.}\mathbf{19875}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{HNO}}_{\mathbf{3}}}\mathbf{ }\mathbf{\times }\mathbf{ }\frac{\mathbf{1}\mathbf{ }\mathbf{L}}{\mathbf{0}\mathbf{.}\mathbf{447}\mathbf{ }\cancel{\mathbf{mol}\mathbf{ }{\mathbf{HNO}}_{\mathbf{3}}}\mathbf{ }}$

= 0.445 L

Step 3. Calculate molarity  C₅H₅NH⁺ 

C₅H₅N + H⁺ → C₅H₅NH⁺ 

This will act as a weak acid in the solution: 

 C₅H₅NH⁺ + H₂O → C₅H₅N + H₃O⁺