Problem: Calculate the pH for each of the following cases in the titration of 25.0 mL of 0.210 M pyridine, C5H5N(aq) with 0.210 M HBr(aq) (Kb = 1.7 x 10-9)after addition of 32.0 mL of HBr

HBr (strong acid) will react with C₅H₅N (weak base).

C₅H₅N_(aq) + HBr_(aq) → C₅H₅NH⁺_(aq) + Br^-_(aq)

Step 1: Calculate the initial amount (in moles) of each species:

$\boxed{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\frac{\mathbf{m}\mathbf{o}\mathbf{l}}{\mathbf{L}}}$

$\mathbf{moles}\mathbf{ }{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}\mathbf{=}\left(\mathbf{0}\mathbf{.}\mathbf{210}\frac{\mathbf{mol}\mathbf{ }{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}}{\cancel{\mathbf{L}}}\right)\mathbf{ }\left(\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{ }\mathbf{x}\mathbf{ }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\right)$

moles C₅H₅N = 0.00525 mol C₅H₅N

$\mathbf{moles}\mathbf{ }\mathbf{HBr}\mathbf{=}\left(\mathbf{0}\mathbf{.}\mathbf{210}\frac{\mathbf{mol}\mathbf{ }\mathbf{HBr}}{\cancel{\mathbf{L}}}\right)\mathbf{ }\left(\mathbf{32}\mathbf{.}\mathbf{00}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{ }\mathbf{x}\mathbf{ }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\cancel{\mathbf{L}}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\right)$

moles HBr = 0.00672 mol HBr

Step 2: Construct an ICF chart to determine the final amount of each species.

After the reaction is complete the solution contains:

• 0.00147 mol HBr → strong acid
• 0.00525 mol C₅H₅NH⁺ → conjugate acid
• pH will depend on the strong acid

Step 3: Calculate the pH of the solution 

$\boxed{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{t}\mathbf{y}\mathbf{ }\mathbf{\left(}\mathbf{M}\mathbf{\right)}\mathbf{=}\frac{\mathbf{m}\mathbf{o}\mathbf{l}}{\mathbf{L}}}$

$$\begin{gathered} \mathbf{final}\mathbf{ }\mathbf{volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{solution}\mathbf{=}\mathbf{volume}\mathbf{ }{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}\mathbf{\hspace{0.17em}}\mathbf{+}\mathbf{volume}\mathbf{\hspace{0.17em}}\mathbf{HBr} \\ \mathbf{final}\mathbf{ }\mathbf{volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{solution}\mathbf{=}\left(\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{ }\mathbf{x}\mathbf{ }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{L}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\right)\mathbf{+}\left(\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{ }\cancel{\mathbf{mL}}\mathbf{ }\mathbf{x}\mathbf{ }\frac{{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{ }\mathbf{L}}{\mathbf{1}\mathbf{ }\cancel{\mathbf{mL}}}\right) \\ \mathbf{final}\mathbf{ }\mathbf{volume}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{solution}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{025}\mathbf{ }\mathbf{L}\mathbf{+}\mathbf{0}\mathbf{.}\mathbf{032}\mathbf{ }\mathbf{L} \end{gathered}$$

final volume of solution = 0.057 L