# Problem: (b) What is the pH of 0.40 M triethylammonium chloride, (CH3CH2)3NHCl?

###### FREE Expert Solution

The dissociation of (CH3CH2)3NHCl is as follows:

(CH3CH2)3NHCl(aq)  Cl(aq)(CH3CH2)3NH+(aq)

[(CH3CH2)3NHCl] = [(CH3CH2)3NH+] = 0.40 M

The hydrolysis of (CH3CH2)3NH+:

(CH3CH2)3NH+(aq) + H2O(l) H3O+(aq) + (CH3CH2)3N(aq)

Kb of (CH3CH2)3NH = 5.2 × 10–4

${\mathbf{K}}_{\mathbf{w}}\mathbf{=}{\mathbf{K}}_{\mathbf{a}}{\mathbf{K}}_{\mathbf{b}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{a}}\overline{){\mathbf{K}}_{\mathbf{b}}}}{\overline{){\mathbf{K}}_{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{5}\mathbf{.}\mathbf{2}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}$

Ka = 1.92 x 10–11

79% (494 ratings) ###### Problem Details

(b) What is the pH of 0.40 M triethylammonium chloride, (CH3CH2)3NHCl?

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