The dissociation of C_{2}H_{5}NH_{3}Cl is as follows:

**C _{2}H_{5}NH_{3}Cl_{(aq)} **

**[C _{2}H_{5}NH_{3}Cl] = [**

The hydrolysis of C_{2}H_{5}NH_{3}^{+}:

**C _{2}H_{5}NH_{3}^{+}_{(aq)} + H_{2}O_{(l)} **

**K _{b} of C_{2}H_{5}NH_{2} = 6.4 × 10^{–4}**

${\mathbf{K}}_{\mathbf{w}}\mathbf{=}{\mathbf{K}}_{\mathbf{a}}{\mathbf{K}}_{\mathbf{b}}\phantom{\rule{0ex}{0ex}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}\mathbf{=}\frac{{\mathbf{K}}_{\mathbf{a}}\overline{){\mathbf{K}}_{\mathbf{b}}}}{\overline{){\mathbf{K}}_{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}\overline{){{\mathbf{K}}}_{{\mathbf{a}}}{\mathbf{=}}\frac{{\mathbf{K}}_{\mathbf{w}}}{{\mathbf{K}}_{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{K}}_{\mathbf{a}}\mathbf{=}\frac{\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{14}}}{\mathbf{6}\mathbf{.}\mathbf{4}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{4}}}$

**K _{a} = 1.56 x 10^{–11} **

Calculate the concentrations of all species present in a 0.25-M solution of ethylammonium chloride (C_{2}H_{5}NH_{3}Cl).

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