# Problem: Calculate the concentrations of all species present in a 0.25-M solution of ethylammonium chloride (C2H5NH3Cl).

###### FREE Expert Solution

The dissociation of C2H5NH3Cl is as follows:

C2H5NH3Cl(aq)  Cl(aq) + C2H5NH3+(aq)

[C2H5NH3Cl] = [C2H5NH3+] = 0.25 M

The hydrolysis of C2H5NH3+:

C2H5NH3+(aq) + H2O(l) H3O+(aq) + C2H5NH2(aq)

Kb of C2H5NH2 = 6.4 × 10–4

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

Ka = 1.56 x 10–11

###### Problem Details

Calculate the concentrations of all species present in a 0.25-M solution of ethylammonium chloride (C2H5NH3Cl).