# Problem: (b) What is the pH of 0.88 M pyridinium chloride, C 5H5NHCl?

🤓 Based on our data, we think this question is relevant for Professor Davis' class at UCF.

###### FREE Expert Solution

C5H5NHCl → C5H5NH+ + Cl-

C5H5NH+ → positively charged amine → weak acid

C5H5NH+ + H2O → C5H5NH + H3O+

Kb for C5H5NH =  1.7x10-9

$\overline{){{\mathbf{K}}}_{{\mathbf{w}}}{\mathbf{=}}{{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{·}}{{\mathbf{K}}}_{{\mathbf{b}}}}$

Assuming the reaction is at 25°C:             Kw = 1.0x10-14

Calculate Kb:

$\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{a}}\mathbf{·}{\mathbf{K}}_{\mathbf{b}}}{\overline{){\mathbf{K}}_{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{a}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{b}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{1}\mathbf{.}\mathbf{7}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{9}}}$

Kb = 5.88x10-6

Kb is also an equilibrium expression and can be calculated as:

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\mathbf{products}}{\mathbf{reactants}}}$

Solids and liquids are ignored in equilibrium expressions

$\overline{){{\mathbf{K}}}_{{\mathbf{b}}}{\mathbf{=}}\frac{\left[{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}\mathbf{N}\right]\left[{\mathbf{H}}_{\mathbf{3}}{\mathbf{O}}^{\mathbf{+}}\right]}{\left[{\mathbf{C}}_{\mathbf{5}}{\mathbf{H}}_{\mathbf{5}}{\mathbf{NH}}^{\mathbf{+}}\right]}}$

$\mathbf{5}\mathbf{.}\mathbf{88}\mathbf{x}{\mathbf{10}}^{\mathbf{-}\mathbf{6}}\mathbf{=}\frac{\left[\mathbf{x}\right]\left[\mathbf{x}\right]}{\left[\mathbf{0}\mathbf{.}\mathbf{88}\mathbf{-}\mathbf{x}\right]}$

###### Problem Details

(b) What is the pH of 0.88 M pyridinium chloride, C 5H5NHCl?